Exercise 1: Recognize the type, then solve: five first-order equations

None of these equations comes in its textbook form. Before any computation, rewrite it until you recognize its TYPE: separable $\frac{dy}{dx}=f(x)g(y)$ , linear $y'+P(x)y=Q(x)$ , homogeneous $y'=F\left(\frac{y}{x}\right)$ , or linear if $x$ is viewed as a function of $y$ . Then solve through the given point: the point $(x_{0}, y_{0})$ means $y(x_{0})=y_{0}$ .

a) $\dfrac{dy}{dx}=e^{2x+y}+e^{y}$ , through the point $(0, 0)$ . Compute $y\left(-\frac{1}{2}\right)$ and the right endpoint of the interval on which the solution exists.
b) $\dfrac{dy}{dx}=\dfrac{y+x^{2}\sin x}{x}$ for $x>0$ , through the point $(\pi, 2\pi)$ . What type of equation is it? Compute $y\left(\frac{\pi}{2}\right)$ .
c) $x^{2}\,\dfrac{dy}{dx}=y^{2}+xy+x^{2}$ for $x>0$ , through the point $(1, 0)$ . Compute $y\left(e^{\pi/4}\right)$ and the right endpoint of the interval on which the solution exists.
d) $\dfrac{dy}{dx}=\dfrac{1}{x+y^{2}}$ , through the point $(0, 0)$ . Show that the equation is neither separable nor linear in $y$ , but that it becomes linear when $x$ is viewed as a function of $y$ . Compute the coordinate $x$ of the point on the curve where $y=1$ .
e) $x\,\dfrac{dy}{dx}+y=x^{2}y^{2}$ for $x>0$ , through the point $(1, \frac{1}{2})$ . Solve by recognizing the derivative of a product, then compute $y\left(\frac{3}{2}\right)$ and the right endpoint of the interval on which the solution exists.

Type your answers, the page tells you right or wrong 0/10

a)

y\left(-\frac{1}{2}\right)

Right endpoint

b)

Type of equation

y\left(\frac{\pi}{2}\right)

c)

y\left(e^{\pi/4}\right)

Right endpoint

d)

Type of equation

x

for

y=1

e)

y\left(\frac{3}{2}\right)

Right endpoint

Show the solution

Answers

a) Separable: $y=-\ln\left(\frac{3}{2}-x-\frac{e^{2x}}{2}\right)$ , $y\left(-\frac{1}{2}\right)\approx -0.597$ , right endpoint $\approx 0.396$
b) Linear: $y=x-x\cos x$ , $y\left(\frac{\pi}{2}\right)=\frac{\pi}{2}\approx 1.571$
c) Homogeneous: $y=x\tan(\ln x)$ , $y\left(e^{\pi/4}\right)=e^{\pi/4}\approx 2.193$ , endpoint $e^{\pi/2}\approx 4.810$
d) Linear in $x$ : $x=2e^{y}-y^{2}-2y-2$ ; for $y=1$ , $x=2e-5\approx 0.437$
e) $(xy)'=(xy)^{2}$ : $y=\frac{1}{x(3-x)}$ on $(0, 3)$ , $y\left(\frac{3}{2}\right)=\frac{4}{9}$

a)

The trap is in the exponent: $e^{2x+y}=e^{2x}\,e^{y}$ , not $e^{2x}+e^{y}$ . Factor out $e^{y}$ : $\dfrac{dy}{dx}=e^{y}\left(e^{2x}+1\right)$ , a SEPARABLE equation. $e^{-y}\,dy=\left(e^{2x}+1\right)dx$ , so $-e^{-y}=\dfrac{e^{2x}}{2}+x+C$ . At the point $(0, 0)$ : $-1=\frac{1}{2}+C$ , $C=-\frac{3}{2}$ . Hence $e^{-y}=\dfrac{3}{2}-x-\dfrac{e^{2x}}{2}$ and $y=-\ln\left(\dfrac{3}{2}-x-\dfrac{e^{2x}}{2}\right)$ . At $x=-\frac{1}{2}$ : $\frac{3}{2}+\frac{1}{2}-\frac{e^{-1}}{2}=2-\frac{1}{2e}\approx 1.816$ , so $y\approx -0.597$ . The solution exists only as long as the argument of the logarithm is positive: $3-2x-e^{2x}>0$ . This function is strictly decreasing, equals $1$ at $0$ and vanishes between $0.39$ and $0.40$ ; by bisection, the right endpoint is about $0.396$ . As you approach it, $e^{-y}\to 0$ and $y\to+\infty$ : the solution blows up in finite time, although the right-hand side looked harmless.

b)

Split the fraction: $\dfrac{dy}{dx}=\dfrac{y}{x}+x\sin x$ . The right-hand side is a SUM of a term in $y$ and a term in $x$ : the equation is not separable, it is LINEAR, $y'-\dfrac{1}{x}\,y=x\sin x$ . Here $P(x)=-\frac{1}{x}$ , so $\mu=e^{-\ln x}=\dfrac{1}{x}$ . Sign pitfall: taking $\mu=x$ by forgetting the minus sign of $P$ . Multiply: $\left(\dfrac{y}{x}\right)'=\sin x$ , so $\dfrac{y}{x}=-\cos x+C$ and $y=-x\cos x+Cx$ . At the point $(\pi, 2\pi)$ : $2\pi=\pi+C\pi$ , so $C=1$ and $y=x-x\cos x$ . Check: $y'=1-\cos x+x\sin x$ , and $\frac{y}{x}+x\sin x=1-\cos x+x\sin x$ . Finally $y\left(\frac{\pi}{2}\right)=\frac{\pi}{2}-0=\frac{\pi}{2}\approx 1.571$ .

c)

Divide by $x^{2}$ : $\dfrac{dy}{dx}=\left(\dfrac{y}{x}\right)^{2}+\dfrac{y}{x}+1$ . The right-hand side depends only on $\frac{y}{x}$ : a HOMOGENEOUS equation. Set $v=\dfrac{y}{x}$ , so $y=xv$ and $y'=v+x\,v'$ by the product rule. Forgetting the term $v$ is the classic error: it makes the equation wrong from the very first line. You get $v+x\,v'=v^{2}+v+1$ , that is $x\,v'=v^{2}+1$ , which is separable: $\dfrac{dv}{1+v^{2}}=\dfrac{dx}{x}$ , $\arctan v=\ln x+C$ . At the point $(1, 0)$ : $v=0$ and $\ln 1=0$ , so $C=0$ . Thus $v=\tan(\ln x)$ and $y=x\tan(\ln x)$ . At $x=e^{\pi/4}$ : $\ln x=\frac{\pi}{4}$ , $\tan\frac{\pi}{4}=1$ , so $y=e^{\pi/4}\approx 2.193$ . The solution lives as long as $\ln x<\frac{\pi}{2}$ , that is $x<e^{\pi/2}\approx 4.810$ , where the tangent blows up.

d)

The right-hand side does not factor as $f(x)g(y)$ : not separable. And $y$ appears in the denominator, squared: not linear in $y$ . But a curve can be read both ways: wherever $\frac{dy}{dx}\neq 0$ , $\dfrac{dx}{dy}=\dfrac{1}{dy/dx}=x+y^{2}$ . Viewing $x$ as a function of $y$ , it is a LINEAR equation: $\dfrac{dx}{dy}-x=y^{2}$ . Integrating factor $\mu=e^{-y}$ : $\left(x\,e^{-y}\right)'=y^{2}e^{-y}$ . Two integrations by parts give $\int y^{2}e^{-y}\,dy=-(y^{2}+2y+2)\,e^{-y}$ , which you check by differentiating: $-(2y+2)e^{-y}+(y^{2}+2y+2)e^{-y}=y^{2}e^{-y}$ . So $x\,e^{-y}=-(y^{2}+2y+2)e^{-y}+C$ , that is $x=Ce^{y}-y^{2}-2y-2$ . At the point $(0, 0)$ , meaning $x=0$ when $y=0$ : $0=C-2$ , $C=2$ . The solution is $x=2e^{y}-y^{2}-2y-2$ , a function of $y$ and not of $x$ . For $y=1$ : $x=2e-5\approx 0.437$ . Pitfall: reading the condition “through the point $(0, 0)$ ” backwards, by putting the second coordinate in place of $x$ ; here the point still says $x=0$ AND $y=0$ .

e)

The left-hand side is exactly the derivative of a product: $x\,y'+y=(xy)'$ . The right-hand side, $x^{2}y^{2}$ , equals $(xy)^{2}$ . Setting $w=xy$ , the equation becomes $w'=w^{2}$ , which is separable. For $w\neq 0$ : $\dfrac{dw}{w^{2}}=dx$ , so $-\dfrac{1}{w}=x+C$ . At the point $(1, \frac{1}{2})$ : $w=\frac{1}{2}$ , so $-2=1+C$ and $C=-3$ . Thus $w=\dfrac{1}{3-x}$ and $y=\dfrac{w}{x}=\dfrac{1}{x(3-x)}$ . Another route, longer: it is a Bernoulli equation, $y'+\frac{1}{x}y=x\,y^{2}$ , linearized by $u=\frac{1}{y}$ ; it leads to the same result. Then $y\left(\frac{3}{2}\right)=\dfrac{1}{\frac{3}{2}\cdot\frac{3}{2}}=\dfrac{4}{9}\approx 0.444$ . The solution exists on $(0, 3)$ and blows up at $x=3$ . Pitfall: dividing by $x$ too early and looking for an integrating factor where it was enough to READ the left-hand side.

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Exercise 2: From second order to first: five equations to integrate twice or to reduce

An equation of order $2$ always comes down to two equations of order $1$ . If the right-hand side depends only on the variable, integrate twice, with two constants. If it contains $y'$ , set $v=y'$ and first solve a first-order equation in $v$ , separable or linear. And if the left-hand side is already a derivative, integrate it in one step. In every case, two conditions fix the two constants.

a) $\dfrac{d^{2}s}{dt^{2}}=\dfrac{(t+1)^{2}}{t^{4}}$ for $t>0$ , given that for $t=1$ you have $s=0$ and $\dfrac{ds}{dt}=0$ . Find $s(t)$ and compute $s(2)$ .
b) $y''=\sin^{2}x$ with $y(0)=0$ and $y'(0)=1$ . Compute $y(\pi)$ .
c) $x^{2}\,y''=1$ for $x>0$ , with $y(1)=0$ and $y(e)=0$ . Compute $y'(1)$ , then the coordinate $x$ of the point where the tangent line is horizontal.
d) $y''+\dfrac{y'}{x}=4$ for $x>0$ , with $y(1)=1$ and $y'(1)=3$ . Compute $y(e)$ .
e) $y''=2y\,y'$ with $y(0)=1$ and $y'(0)=2$ . Compute $y\left(\frac{\pi}{12}\right)$ and the positive value of $x$ at which the solution ceases to exist.

Type your answers, the page tells you right or wrong 0/8

a)

s(2)

b)

y(\pi)

c)

y'(1)

Value of

x

d)

Integrating factor of the equation in

v=y'

y(e)

e)

y\left(\frac{\pi}{12}\right)

Blow-up point

x

Show the solution

Answers

a) $s(t)=-\ln t+\frac{1}{t}+\frac{1}{6t^{2}}+\frac{7t}{3}-\frac{7}{2}$ , $s(2)=\frac{41}{24}-\ln 2\approx 1.015$
b) $y=\frac{x^{2}}{4}+\frac{\cos 2x-1}{8}+x$ , $y(\pi)=\frac{\pi^{2}}{4}+\pi\approx 5.609$
c) $y=-\ln x+\frac{x-1}{e-1}$ , $y'(1)=\frac{2-e}{e-1}\approx -0.418$ , horizontal tangent at $x=e-1\approx 1.718$
d) $\mu=x$ ; $y=x^{2}+\ln x$ , $y(e)=e^{2}+1\approx 8.389$
e) $y=\tan\left(x+\frac{\pi}{4}\right)$ , $y\left(\frac{\pi}{12}\right)=\sqrt{3}\approx 1.732$ , blow-up at $x=\frac{\pi}{4}$

a)

EXPAND the square, then split: $\dfrac{(t+1)^{2}}{t^{4}}=\dfrac{t^{2}+2t+1}{t^{4}}=t^{-2}+2t^{-3}+t^{-4}$ . First integration: $\dfrac{ds}{dt}=-t^{-1}-t^{-2}-\dfrac{t^{-3}}{3}+C_{1}$ . Condition $\frac{ds}{dt}=0$ at $t=1$ : $-1-1-\frac{1}{3}+C_{1}=0$ , so $C_{1}=\frac{7}{3}$ . Pitfall: a zero velocity does not give $C_{1}=0$ , since the antiderivative equals $-\frac{7}{3}$ at $t=1$ . Second integration, with $\int t^{-1}\,dt=\ln t$ (and not $\frac{t^{0}}{0}$ ): $s=-\ln t+\dfrac{1}{t}+\dfrac{1}{6t^{2}}+\dfrac{7t}{3}+C_{2}$ . At $t=1$ : $0+1+\frac{1}{6}+\frac{7}{3}+C_{2}=0$ , so $C_{2}=-\frac{7}{2}$ . Thus $s(t)=-\ln t+\dfrac{1}{t}+\dfrac{1}{6t^{2}}+\dfrac{7t}{3}-\dfrac{7}{2}$ , and $s(2)=-\ln 2+\frac{1}{2}+\frac{1}{24}+\frac{14}{3}-\frac{7}{2}=\frac{41}{24}-\ln 2\approx 1.015$ .

b)

$\sin^{2}x$ has no immediate antiderivative: LINEARIZE it, $\sin^{2}x=\dfrac{1-\cos 2x}{2}$ . First integration: $y'=\dfrac{x}{2}-\dfrac{\sin 2x}{4}+C_{1}$ . At $x=0$ : $C_{1}=1$ , since $\sin 0=0$ . Second integration: $y=\dfrac{x^{2}}{4}+\dfrac{\cos 2x}{8}+x+C_{2}$ ; watch the sign, the antiderivative of $-\sin 2x$ is $+\frac{\cos 2x}{2}$ . At $x=0$ : $\frac{1}{8}+C_{2}=0$ , so $C_{2}=-\frac{1}{8}$ , and NOT $0$ , since $\cos 0=1$ . Thus $y=\dfrac{x^{2}}{4}+\dfrac{\cos 2x-1}{8}+x$ . Check: $y''=\frac{1}{2}-\frac{\cos 2x}{2}=\sin^{2}x$ . At $x=\pi$ : $\cos 2\pi=1$ , so $y(\pi)=\dfrac{\pi^{2}}{4}+\pi\approx 5.609$ . Common pitfall: writing $\int\sin^{2}x\,dx=\frac{\sin^{3}x}{3}$ , whose derivative is $\sin^{2}x\cos x$ .

c)

Divide by $x^{2}$ : $y''=\dfrac{1}{x^{2}}$ . First integration: $y'=-\dfrac{1}{x}+C_{1}$ ; second: $y=-\ln x+C_{1}x+C_{2}$ . Both conditions are on $y$ : no constant can be fixed before the end, you have to solve a system. $y(1)=0$ gives $C_{1}+C_{2}=0$ . $y(e)=0$ gives $-1+C_{1}e+C_{2}=0$ . Subtracting: $C_{1}(e-1)=1$ , so $C_{1}=\dfrac{1}{e-1}\approx 0.582$ and $C_{2}=-C_{1}$ . Thus $y=-\ln x+\dfrac{x-1}{e-1}$ . Then $y'(1)=-1+\dfrac{1}{e-1}=\dfrac{2-e}{e-1}\approx -0.418$ : the initial slope was not given, it follows from the two points. The tangent line is horizontal when $y'=0$ , that is $\frac{1}{x}=\frac{1}{e-1}$ and $x=e-1\approx 1.718$ . Since $y''=\frac{1}{x^{2}}>0$ , it is a minimum, equal to $y(e-1)=1-\ln(e-1)-\frac{1}{e-1}\approx -0.123$ : the curve dips below the axis between its two zeros, consistent with a graph that is concave up.

d)

The right-hand side contains $y'$ : you cannot integrate twice directly. Set $v=y'$ , so $y''=v'$ : $v'+\dfrac{1}{x}\,v=4$ , a first-order LINEAR equation in $v$ . Integrating factor $\mu=e^{\int\frac{dx}{x}}=e^{\ln x}=x$ , and not $e^{x}$ . You get $(x\,v)'=4x$ , so $x\,v=2x^{2}+C_{1}$ and $v=2x+\dfrac{C_{1}}{x}$ . Condition $y'(1)=3$ : $2+C_{1}=3$ , $C_{1}=1$ . Back to $y$ : $y=x^{2}+\ln x+C_{2}$ , and $y(1)=1+C_{2}=1$ gives $C_{2}=0$ . Thus $y=x^{2}+\ln x$ . Check: $y'=2x+\frac{1}{x}$ , $y''=2-\frac{1}{x^{2}}$ , and $y''+\frac{y'}{x}=2-\frac{1}{x^{2}}+2+\frac{1}{x^{2}}=4$ . Finally $y(e)=e^{2}+1\approx 8.389$ . You could also multiply the equation by $x$ : $x\,y''+y'=(x\,y')'$ , which integrates in one step to $x\,y'=2x^{2}+C_{1}$ ; it is the same integrating factor, read off directly.

e)

The right-hand side $2y\,y'$ is the derivative of $y^{2}$ by the chain rule. The equation therefore reads $(y')'=(y^{2})'$ , and a first integration gives $y'=y^{2}+C_{1}$ . Condition at $x=0$ : $2=1+C_{1}$ , so $C_{1}=1$ . Pitfall: setting $C_{1}=0$ by reflex, which would give $y'=y^{2}$ and contradict $y'(0)=2$ . You get $y'=1+y^{2}$ , which is separable: $\dfrac{dy}{1+y^{2}}=dx$ , so $\arctan y=x+C_{2}$ . At $x=0$ : $\arctan 1=\frac{\pi}{4}=C_{2}$ . Thus $y=\tan\left(x+\dfrac{\pi}{4}\right)$ . Check: $y'=1+y^{2}$ , then $y''=2y\,y'$ . At $x=\frac{\pi}{12}$ : $\frac{\pi}{12}+\frac{\pi}{4}=\frac{\pi}{3}$ , so $y=\tan\frac{\pi}{3}=\sqrt{3}\approx 1.732$ . The solution ceases to exist when $x+\frac{\pi}{4}=\frac{\pi}{2}$ , that is $x=\frac{\pi}{4}\approx 0.785$ : an equation with no visible singularity at all produces a solution that blows up in finite time.

Exercise 3: Verify, separate, solve

The figure shows the family of solution curves of the equation in question d): concentric circles, one for each value of the constant.

a) Verify that $y=Ce^{-3x}+2$ is a solution of $y'+3y=6$ for every constant $C$ , then find the particular solution satisfying $y(0)=5$ . What does $y=2$ represent?
b) Solve the separable equation $\frac{dy}{dx}=xy$ with $y(0)=3$ .
c) Solve $\frac{dy}{dx}=-y^{2}$ with $y(0)=1$ , and comment on the interval of validity of the solution.
d) Solve $\frac{dy}{dx}=-\frac{x}{y}$ and show that the solution curves are the circles of the figure. Why does the solution remain implicit?

Show the solution

Answers

a) $C=3$ ; equilibrium $y=2$
b) $y=3e^{x^{2}/2}$
c) $y=\frac{1}{1+x}$ on $(-1,\infty)$
d) $x^{2}+y^{2}=C$ : circles, implicit solution

a)

$y'=-3Ce^{-3x}$ , so $y'+3y=-3Ce^{-3x}+3Ce^{-3x}+6=6$ : verified for every $C$ , it is the GENERAL solution (one constant, first order). $y(0)=C+2=5$ gives $C=3$ : $y=3e^{-3x}+2$ . The line $y=2$ is the EQUILIBRIUM solution ( $y'=0$ ): all solutions converge to it as $x\to\infty$ , the exponential only damping out the initial gap.

b)

Separation: $\frac{dy}{y}=x\,dx$ , then $\ln|y|=\frac{x^{2}}{2}+C_{1}$ , so $y=Ce^{x^{2}/2}$ . With $y(0)=3$ : $C=3$ , hence $y=3e^{x^{2}/2}$ . Note in passing that dividing by $y$ hid the constant solution $y=0$ : a reflex worth keeping, even though the initial condition does not select it here.

c)

$\frac{dy}{y^{2}}=-dx$ gives $-\frac{1}{y}=-x+C_{1}$ , that is $y=\frac{1}{x+C}$ . With $y(0)=1$ : $C=1$ , so $y=\frac{1}{1+x}$ . The solution is valid only on $(-1,\infty)$ : at $x=-1$ it blows up. A perfectly regular differential equation can produce solutions with a limited lifespan: the interval is part of the answer.

d)

$y\,dy=-x\,dx$ , so $\frac{y^{2}}{2}=-\frac{x^{2}}{2}+C_{1}$ , that is $x^{2}+y^{2}=C$ : the circles centred at the origin in the figure (radii 2, 3, 4 for $C=4, 9, 16$ ). The solution remains IMPLICIT because no single function $y(x)$ describes a whole circle: two branches $y=\pm\sqrt{C-x^{2}}$ are needed. Geometrically, the equation says that the slope $-\frac{x}{y}$ is perpendicular to the radius: the family of circles is the only one with this property at every point.

Exercise 4: The linear equation and the integrating factor

Standard form: $y'+P(x)\,y=Q(x)$ . The integrating factor $\mu(x)=e^{\int P(x)dx}$ turns the left-hand side into an exact derivative $(\mu y)'$ .

a) Prove that the method works: show that multiplying the equation by $\mu=e^{\int P\,dx}$ makes the left-hand side equal to $(\mu y)'$ .
b) Solve $y'+2y=6$ with an integrating factor, and recover the structure “equilibrium + exponential”.
c) Solve $y'+\frac{1}{x}y=x^{2}$ on $(0,\infty)$ with $y(1)=1$ .
d) The RC circuit of physics obeys $R\frac{dq}{dt}+\frac{q}{C}=E$ . Put the equation in standard form, identify the time constant, and solve with $q(0)=0$ for $E=12$ V, $R=1500$ Ω, $C=20$ μF. What do you recover?

Show the solution

Answers

a) $(\mu y)'=\mu(y'+Py)$
b) $y=3+Ce^{-2x}$
c) $y=\frac{x^{3}}{4}+\frac{3}{4x}$
d) $\tau=RC=0.03$ s; $q=CE\left(1-e^{-t/RC}\right)$

a)

By construction $\mu'=P\mu$ (derivative of a composite exponential). Then $(\mu y)'=\mu y'+\mu'y=\mu y'+P\mu y=\mu(y'+Py)$ : multiplying the equation by $\mu$ gives exactly $(\mu y)'=\mu Q$ . That is the whole genius of the method: build a tailor-made factor that glues the left-hand side back into a single derivative, which a simple integration then undoes.

b)

$P=2$ , $\mu=e^{2x}$ : $(e^{2x}y)'=6e^{2x}$ , so $e^{2x}y=3e^{2x}+C$ , that is $y=3+Ce^{-2x}$ . A crystal-clear structure: the equilibrium $y=3$ (constant solution of $y'=0$ ) plus an exponential transient that dies out at the rate $e^{-2x}$ . Every linear equation with constant coefficients tells this same story.

c)

$P=\frac{1}{x}$ , $\mu=e^{\ln x}=x$ : the equation becomes $(xy)'=x^{3}$ , so $xy=\frac{x^{4}}{4}+C$ , that is $y=\frac{x^{3}}{4}+\frac{C}{x}$ . With $y(1)=\frac{1}{4}+C=1$ : $C=\frac{3}{4}$ , hence $y=\frac{x^{3}}{4}+\frac{3}{4x}$ . The integrating factor is not always an exponential: here it is $x$ itself, and the check $(xy)'=x^{3}$ takes one line.

d)

Divide by $R$ : $\frac{dq}{dt}+\frac{1}{RC}q=\frac{E}{R}$ : linear with constant coefficients, with $P=\frac{1}{RC}$ : the time constant is $\tau=RC=(1500)(20\times 10^{-6})=0.030$ s. Solution with $q(0)=0$ : $q(t)=CE\left(1-e^{-t/RC}\right)=2.4\times 10^{-4}\left(1-e^{-t/0.03}\right)$ C. You recover EXACTLY the charging of the capacitor from the physics course 203-SN2-RE (formerly 203-NYB): same equation, same 63% at $t=\tau$ , same equilibrium $q_{\infty}=CE$ . The mathematics of the integrating factor and the physics of the RC circuit are one and the same chapter.

Exercise 5: Exponential growth and decay

The model $\frac{dN}{dt}=kN$ (rate of change proportional to the amount present) governs radioactive decay, populations with unlimited resources and continuously compounded interest. Half-life of carbon 14: 5730 years.

a) Solve the equation by separation and express $k$ in terms of the half-life. A fossil contains only 20% of its initial carbon 14: how old is it?
b) A bacterial culture doubles every 3 hours. How long does it take to grow by a factor of 1000? First give a mental estimate using doublings, then the exact value.
c) A principal of 5000 dollars is invested at 4% interest compounded CONTINUOUSLY ( $A'=0.04A$ ). Value after 10 years? Doubling time? Compare with the financiers' “rule of 72” ( $72/\text{rate in }\%$ ).
d) Name two assumptions of the exponential model that always end up breaking down in a real population, and what the model then becomes.

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Answers

a) $k=-\frac{\ln 2}{5730}$ ; age $\approx 13{,}300$ years
b) $\approx 29.9$ h
c) $7459$ dollars; doubling $17.3$ years
d) Unlimited resources and constant rate: logistic

a)

$\frac{dN}{N}=k\,dt$ gives $N(t)=N_{0}e^{kt}$ . Half-life: $e^{kt_{1/2}}=\frac{1}{2}$ , so $k=-\frac{\ln 2}{t_{1/2}}=-\frac{\ln 2}{5730}$ per year. Age of the fossil: $e^{kt}=0.20$ gives $t=\frac{\ln 5}{\ln 2}\times 5730\approx 13{,}300$ years: the fossil has gone through a little more than two half-lives (two half-lives would leave 25%), and the logarithm pins down the “a little more”.

b)

Estimate: $2^{10}=1024\approx 1000$ , so about 10 doublings, that is $\approx 30$ hours. Exactly: $k=\frac{\ln 2}{3}$ and $t=\frac{\ln 1000}{k}=3\,\frac{\ln 1000}{\ln 2}=3\log_{2}1000\approx 29.9$ h. The mental estimate through $2^{10}\approx 10^{3}$ is an exam tool in its own right: it checks the order of magnitude before any computation.

c)

$A(10)=5000e^{0.4}\approx 7459$ dollars. Doubling: $\frac{\ln 2}{0.04}\approx 17.3$ years. Rule of 72: $\frac{72}{4}=18$ years: the financiers' rule is an approximation (calibrated for annual compounding) of the true $\frac{\ln 2}{r}\approx\frac{69.3}{r\text{ in }\%}$ ; for continuous compounding, the “exact rule” would be the rule of 69.

d)

The model assumes UNLIMITED resources (food, space) and a CONSTANT rate $k$ (no competition, no predation, no saturation). In any real population, density eventually slows growth down: the effective rate decreases with $N$ . The minimal patch is precisely the logistic model of exercise 10: replace $k$ with $k\left(1-\frac{P}{M}\right)$ , and the exponential explosion turns into a plateau.

Exercise 6: Read before solving: slope field, phase line and Euler's method

Many differential equations have no solution that can be written down, and yet their behaviour can be read without solving them. An AUTONOMOUS equation $y'=f(y)$ , where the slope depends only on the value of $y$ , is entirely described by the sign of $f$ : that is the meaning of the “read first” reflex.

a) For the autonomous equation $y'=(y-1)(y-3)$ , find the equilibrium solutions, then draw the PHASE LINE: the sign of $y'$ on each interval. Classify each equilibrium as stable or unstable.
b) Without solving, describe what happens to a solution starting from $y(0)=0$ , from $y(0)=2$ and from $y(0)=4$ . Compare your description with the curves of the figure.
c) Most equations have no exact solution: EULER's method advances step by step through $y_{n+1}=y_{n}+h\,f(x_{n},y_{n})$ . Apply it to $y'=y$ , $y(0)=1$ , with step $h=0.1$ , to estimate $y(0.2)$ ; compare with the exact value $e^{0.2}$ and explain the DIRECTION of the error.
d) Can two distinct solutions of the same autonomous equation cross? Deduce why the phase line is enough to predict everything, then recall, with $y'=y^{2}$ , that a solution can cease to exist in finite time.

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Answers

a) $y=1$ stable, $y=3$ unstable
b) $0$ and $2$ towards $1$ ; $4$ to infinity
c) $1.21$ against $1.2214$ : underestimate
d) Never any crossing; blow-up possible in finite time

a)

The equilibria make $y'$ zero: $(y-1)(y-3)=0$ gives $y=1$ and $y=3$ , two horizontal lines; the lower two curves of the figure level off along the first one. Phase line, from the sign of the product: for $y<1$ , both factors are negative, their product is POSITIVE, so $y'>0$ and the solution rises; for $1<y<3$ , one factor changes sign, the product is NEGATIVE, $y'<0$ , the solution falls; for $y>3$ , both factors are positive, $y'>0$ , the solution rises. Classification: around $y=1$ , the arrows converge (rising below, falling above), so $y=1$ is STABLE; around $y=3$ , the arrows diverge (falling below, rising above), so $y=3$ is UNSTABLE. This whole analysis holds without writing a single antiderivative.

b)

Starting from $y(0)=0$ : you are below 1, so $y'>0$ , the solution increases, and it is attracted by the stable equilibrium: it rises towards $y=1$ without ever reaching it, flattening out, like the lower curve of the figure. Starting from $y(0)=2$ : you are between the two equilibria, $y'<0$ , the solution DECREASES and it too joins $y=1$ , from above: the middle curve of the figure. Starting from $y(0)=4$ : you are above 3, $y'>0$ , the solution ESCAPES upwards, moving away from the unstable equilibrium, like the upper curve that shoots off to infinity. The split happens exactly at $y=3$ : below it, everything falls back towards 1; above it, everything diverges. The unstable equilibrium is the watershed, and that is the complete qualitative content of the equation.

c)

Euler's method with $f(x,y)=y$ , so $y_{n+1}=y_{n}+0.1\,y_{n}=1.1\,y_{n}$ . First step: $y_{1}=1.1\times 1=1.1$ , an approximation of $y(0.1)$ . Second step: $y_{2}=1.1\times 1.1=1.21$ , an approximation of $y(0.2)$ . Exact value: $e^{0.2}\approx 1.22140$ . The error is about $0.0114$ , and it is an UNDERESTIMATE. The direction of the error has a geometric explanation: Euler advances along the TANGENT LINE, keeping the slope constant over the whole step, whereas the true solution $y=e^{x}$ is concave up and bends upwards. The tangent line therefore stays below the curve, and each step falls slightly behind, a lag that accumulates. Halving the step, $h=0.05$ over four steps gives $1.05^{4}\approx 1.21551$ , noticeably closer: Euler's error decreases in proportion to the step, which makes it a safe but slow method, exactly like the rectangle sums of the numerical integration chapter, of which it is a cousin.

d)

No: two distinct solutions of an autonomous equation (with a sufficiently regular right-hand side) can NEVER cross, by the existence and uniqueness theorem. If two solution curves went through the same point, they would have the same slope $f(y)$ there and would coincide on either side: they would be the same solution. In particular, a solution cannot CROSS an equilibrium line, since that line is itself a solution: it can only approach it asymptotically. That is precisely what gives the phase line its power of complete prediction: the equilibria cut the plane into sealed horizontal strips, and in each strip the sign of $y'$ imposes a single direction of variation, with no surprise possible. One caveat, though: “not escaping towards another equilibrium” does not forbid escaping to infinity, and that can happen in FINITE time. The equation $y'=y^{2}$ with $y(0)=1$ has the solution $y=\frac{1}{1-x}$ , a cousin of the solution $y=\frac{1}{1+x}$ of $y'=-y^{2}$ met in exercise 3: it blows up at $x=1$ and ceases to exist beyond. The phase line correctly says “it rises”, but it does not say how fast, and here the rise is so violent that the solution reaches infinity in one unit of time. Reading the equation tells you the DIRECTION, never by itself the lifespan.

Exercise 7: Substitutions: making linear what was not

Separable and linear do not cover everything. Two important families reduce to them through a well-chosen CHANGE OF FUNCTION: homogeneous equations, through $v=\frac{y}{x}$ , and Bernoulli equations, through $u=y^{1-n}$ . The second one has an unexpected virtue: it proves the logistic solution taken for granted without proof in Part B.

a) An equation is HOMOGENEOUS if $y'$ depends only on the ratio $\frac{y}{x}$ . Solve $x\,y'=x+y$ with the substitution $v=\frac{y}{x}$ , explaining in detail why it makes the equation separable.
b) A BERNOULLI equation has the form $y'+P(x)y=Q(x)y^{n}$ . Solve $y'+y=y^{2}$ with $y(0)=\frac{1}{2}$ using the substitution $u=y^{-1}$ , which linearizes it.
c) Show that the logistic equation $\frac{dP}{dt}=kP\left(1-\frac{P}{M}\right)$ is a Bernoulli equation, and use the substitution $u=\frac{1}{P}$ to recover the solution $P(t)=\frac{M}{1+Ae^{-kt}}$ taken for granted in exercise 10.
d) The ORTHOGONAL TRAJECTORIES of a family of curves cross each of them at right angles. Confirm that the lines $y=Kx$ are orthogonal to the circles $x^{2}+y^{2}=C$ of exercise 3, then find the orthogonal trajectories of the parabolas $y=Cx^{2}$ .

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Answers

a) $y=x(\ln|x|+C)$
b) $y=\frac{1}{1+e^{x}}$
c) Bernoulli: $P=\frac{M}{1+Ae^{-kt}}$
d) Lines $y=Kx$ ; ellipses $x^{2}+2y^{2}=C$

a)

Rewrite in standard form: $y'=\frac{x+y}{x}=1+\frac{y}{x}$ , where the right-hand side indeed depends only on $\frac{y}{x}$ , which signals a homogeneous equation. Set $v=\frac{y}{x}$ , so $y=vx$ and, by the product rule, $y'=v+x\,v'$ . The equation becomes $v+x\,v'=1+v$ , and the term $v$ cancels on both sides: $x\,v'=1$ . That is the whole magic of the substitution: it removes the part that mixed $x$ and $y$ , and what remains is SEPARABLE. Isolate: $dv=\frac{dx}{x}$ , hence $v=\ln\left|x\right|+C$ , then go back to $y$ : $y=vx=x\left(\ln\left|x\right|+C\right)$ . Check: $y'=\ln\left|x\right|+C+x\cdot\frac{1}{x}=\ln\left|x\right|+C+1$ , and $1+\frac{y}{x}=1+\ln\left|x\right|+C$ : identical. The general principle is worth remembering: $v=\frac{y}{x}$ turns every homogeneous equation into a separable equation, without exception.

b)

Here $n=2$ ; the Bernoulli substitution is $u=y^{1-n}=y^{-1}=\frac{1}{y}$ , so $u'=-y^{-2}y'$ . First divide the equation by $y^{2}$ to make these pieces appear: $\frac{y'}{y^{2}}+\frac{1}{y}=1$ , that is $-u'+u=1$ , or $u'-u=-1$ . The equation in $u$ is LINEAR, which was the whole point of the manoeuvre. Integrating factor $\mu=e^{-x}$ : $\left(u e^{-x}\right)'=-e^{-x}$ , so $u e^{-x}=e^{-x}+C$ , that is $u=1+Ce^{x}$ . Going back to $y=\frac{1}{u}=\frac{1}{1+Ce^{x}}$ . Initial condition $y(0)=\frac{1}{2}$ : $\frac{1}{1+C}=\frac{1}{2}$ gives $C=1$ , hence $y=\frac{1}{1+e^{x}}$ . Numerical check: $y(0)=0.5$ and $y'(0.5)=-0.235$ , which matches $y^{2}-y$ evaluated at the same point. Along the way you recognize the shape of a decreasing logistic solution, which is no accident, as the next question shows.

c)

Expand the right-hand side: $\frac{dP}{dt}=kP-\frac{k}{M}P^{2}$ , that is $\frac{dP}{dt}-kP=-\frac{k}{M}P^{2}$ . It is exactly a Bernoulli equation in the unknown $P$ , with $n=2$ : the constant $-k$ plays the role of the coefficient of the unknown in the standard form, and $-\frac{k}{M}$ the role of the coefficient of the power (careful not to confuse the population $P$ with the coefficient named $P(x)$ in the standard form). So apply the same substitution as in b): $u=\frac{1}{P}$ , so $u'=-\frac{P'}{P^{2}}$ . Dividing the equation by $P^{2}$ : $\frac{P'}{P^{2}}-\frac{k}{P}=-\frac{k}{M}$ , that is $-u'-ku=-\frac{k}{M}$ , or $u'+ku=\frac{k}{M}$ . A linear equation, integrating factor $\mu=e^{kt}$ : $\left(u e^{kt}\right)'=\frac{k}{M}e^{kt}$ , so $u e^{kt}=\frac{1}{M}e^{kt}+C$ , that is $u=\frac{1}{M}+Ce^{-kt}$ . Going back to $P=\frac{1}{u}=\frac{1}{\frac{1}{M}+Ce^{-kt}}=\frac{M}{1+CMe^{-kt}}$ . Setting $A=CM$ , you get exactly $P(t)=\frac{M}{1+Ae^{-kt}}$ , the formula taken for granted in exercise 10. The constant $A$ is fixed by $P(0)=P_{0}$ , which gives back $A=\frac{M-P_{0}}{P_{0}}$ ; with $M=6000$ and $P_{0}=500$ , you recover $A=11$ . The logistic solution did not fall from the sky: it is a Bernoulli equation in disguise, and the substitution builds it from start to finish.

d)

The principle of orthogonal trajectories: at a given point, the curve you are looking for must have a slope equal to the NEGATIVE RECIPROCAL of the slope of the original family, since two perpendicular directions have slopes whose product is $-1$ . The case of the circles: the family $x^{2}+y^{2}=C$ satisfies, by implicit differentiation, $2x+2yy'=0$ , so $y'=-\frac{x}{y}$ , the very equation of exercise 3. The orthogonal trajectories then have slope $-\frac{1}{-x/y}=\frac{y}{x}$ , hence the equation $y'=\frac{y}{x}$ , which is separable: $\frac{dy}{y}=\frac{dx}{x}$ gives $\ln\left|y\right|=\ln\left|x\right|+C_{1}$ , that is $y=Kx$ . These are the lines through the origin, which confirms and generalizes the remark of exercise 3: a radius always meets its circle at a right angle. You can check that the product of the slopes at a point such as $(2, 3)$ is $\left(-\frac{2}{3}\right)\left(\frac{3}{2}\right)=-1$ . The case of the parabolas $y=Cx^{2}$ : first eliminate the constant, $C=\frac{y}{x^{2}}$ , then differentiate $y'=2Cx=\frac{2y}{x}$ , the slope of the family. The orthogonal trajectories therefore have slope $-\frac{x}{2y}$ , hence $2y\,dy=-x\,dx$ , which integrates to $y^{2}=-\frac{x^{2}}{2}+C_{1}$ , that is $x^{2}+2y^{2}=C$ : a family of ELLIPSES. Check the orthogonality at a point such as $(2, 1)$ : the parabola has slope $\frac{2\cdot 1}{2}=1$ there, the ellipse $-\frac{2}{2\cdot 1}=-1$ , and the product is indeed $-1$ . These orthogonal networks are not a curiosity: they are the field lines and the equipotentials of electrostatics, or the isotherms and the heat flow lines, always perpendicular, and always obtained by this same flip of the slope.

Exercise 8: Newton's law of cooling: the coffee and the stopwatch

A coffee served at 90 °C sits in a room at 20 °C; five minutes later, it is at 70 °C. Newton's law states that the rate of cooling is proportional to the gap with the room temperature: $\frac{dT}{dt}=-k(T-20)$ . The figure shows the exact curve.

a) Interpret the equation: what does the minus sign say physically, and why is the cooling fast at first and then sluggish? Identify the equilibrium solution.
b) Solve the equation by separation and give the general form $T(t)=20+Ce^{-kt}$ , then fix $C$ with the initial temperature.
c) Find $k$ from the measurement “70 °C at 5 minutes”, then compute when the coffee reaches 40 °C, a comfortable drinking temperature.
d) How long does it take for the coffee to come within one degree of the room temperature? Conclude about the asymptote of the figure, and explain how the same law lets forensic pathologists estimate a time of death.

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Answers

a) Gap that melts away; equilibrium $20$ °C
b) $T=20+70e^{-kt}$
c) $k\approx 0.0673$ ; $40$ °C at $18.6$ min
d) $\approx 63$ min; asymptote never reached

a)

The minus sign says that HOT coffee ( $T>20$ ) cools down ( $T'<0$ ), and that an object colder than the room would warm up: nature erases gaps. The rate is proportional to the gap: at 90 °C, the gap of 70 degrees forces brisk cooling; at 30 °C, the gap of 10 makes it seven times slower: hence the steep start and the dragging end of the curve. Equilibrium: $T=20$ , the room temperature, where $T'=0$ : the only possible outcome.

b)

With $u=T-20$ : $\frac{du}{u}=-k\,dt$ , so $u=Ce^{-kt}$ and $T(t)=20+Ce^{-kt}$ . Initial condition $T(0)=90$ : $C=70$ , the initial GAP: $T(t)=20+70e^{-kt}$ . The solution reads in one sentence: the gap to the room temperature, 70 degrees at the start, melts away exponentially.

c)

$T(5)=70$ gives $20+70e^{-5k}=70$ , that is $e^{-5k}=\frac{50}{70}=\frac{5}{7}$ : $k=\frac{1}{5}\ln\frac{7}{5}\approx 0.0673$ min $^{-1}$ . For 40 °C: $70e^{-kt}=20$ , so $t=\frac{\ln 3.5}{k}\approx 18.6$ minutes: that leaves a quarter of an hour of patience after the five-minute measurement, consistent with the figure (where the curve crosses the 40 °C gridline).

d)

One degree of gap: $70e^{-kt}=1$ , so $t=\frac{\ln 70}{k}\approx 63$ minutes. And the exact equality $T=20$ NEVER happens: the asymptote of the figure is reached only at infinity, even though a real thermometer no longer sees the difference after an hour. Forensic pathologists turn the reasoning around: the body (37 °C at death) cools according to the same law; two temperature readings at known times give $k$ and allow you to trace the curve back to the moment it was at 37: the time of death, estimated with a differential equation.

Exercise 9: The brine tank: the equilibrium that erases the past

A tank contains 100 L of water in which 5 kg of salt is dissolved. Brine at 0.2 kg/L is poured in at a rate of 5 L/min, and the mixture, assumed uniform at every instant, flows out at the same rate of 5 L/min. Let $A(t)$ be the mass of salt (kg) in the tank.

The figure below shows the situation; it is not to scale.

a) Set up the differential equation satisfied by $A(t)$ (balance: inflow minus outflow), and check that the volume stays constant.
b) Identify the equilibrium solution and interpret it physically BEFORE solving.
c) Solve with $A(0)=5$ and compute the mass of salt after 10 minutes.
d) What happens to $A(t)$ as $t\to\infty$ , and in what sense does the limit “forget” the initial condition? After how long does the mass reach 19 kg?

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Answers

a) $A'=1-\frac{A}{20}$
b) Equilibrium $20$ kg, stable
c) $A=20-15e^{-t/20}$ ; $A(10)\approx 10.9$ kg
d) Limit $20$ kg; $19$ kg at $54$ min

a)

Inflow: $5$ L/min $\times\,0.2$ kg/L $=1$ kg/min, constant. Outflow: each litre that leaves carries the CURRENT concentration $\frac{A}{100}$ kg/L, that is $5\times\frac{A}{100}=\frac{A}{20}$ kg/min. Balance: $\frac{dA}{dt}=1-\frac{A}{20}$ . The volume, fed and drained at 5 L/min, stays at 100 L: this is what keeps the equation linear with constant coefficients (different flow rates would give a variable volume, the hard version of the problem).

b)

$\frac{dA}{dt}=0$ for $A=20$ kg. A crystal-clear interpretation: at equilibrium, the tank has everywhere the concentration of the INCOMING brine, $0.2$ kg/L $\times$ 100 L $=20$ kg: what comes in equals what goes out. Below 20, $A'>0$ (the tank gains salt); above, $A'<0$ : the equilibrium is STABLE, everything leads to it.

c)

Separable (or linear) equation: $A(t)=20+(A_{0}-20)e^{-t/20}=20-15e^{-t/20}$ : the gap to equilibrium, $-15$ kg at the start, melts away with the time constant 20 min. $A(10)=20-15e^{-0.5}\approx 10.9$ kg. Check: at $t=10$ min, that is half a time constant, the fraction of the way covered is $1-e^{-0.5}\approx 39$ %, and $5+0.39\times 15\approx 10.9$ kg: the two readings agree (the classic 63%, for their part, would be reached at $t=\tau=20$ min).

d)

$e^{-t/20}\to 0$ : $A\to 20$ kg, the equilibrium limit, WHATEVER the initial mass: starting from 0 kg or from 50 kg, all tanks end up at 20 kg, only the route differs. The exponential term, which carries all the memory of $A_{0}$ , dies out: the steady state erases the past. For 19 kg: $15e^{-t/20}=1$ , so $t=20\ln 15\approx 54$ minutes: getting within one kilogram of equilibrium takes almost an hour, the usual exponential tail.

Exercise 10: The logistic model: the exponential tamed

A lake can support at most $M=6000$ fish. 500 fish are introduced, and the population follows the logistic model $\frac{dP}{dt}=kP\left(1-\frac{P}{M}\right)$ with $k=0.4$ per year. The figure shows the exact solution.

a) Interpret the equation: what does the factor $\left(1-\frac{P}{M}\right)$ do when $P$ is small, then close to $M$ ? Find the equilibrium solutions and their stability.
b) Take for granted the solution $P(t)=\frac{M}{1+Ae^{-kt}}$ with $A=\frac{M-P_{0}}{P_{0}}$ : check the initial condition, compute $A$ , and give the population after 5 years.
c) Show that the growth is MAXIMUM when $P=\frac{M}{2}$ , compute the time at which the lake reaches 3000 fish, and relate it to the figure.
d) Compare the beginning of the curve with a pure exponential and its end with the equilibrium. Why is the logistic curve said to be “the exponential that has learned that limits exist”? Name two fields where this model is genuinely used.

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Answers

a) $0$ unstable, $M$ stable
b) $A=11$ ; $P(5)\approx 2410$
c) Maximum growth at $3000$ , around $6.0$ years
d) Exponential at first, plateau at the end

a)

When $P\ll M$ , the factor is almost 1: the equation becomes $P'=kP$ again, the exponential growth of abundant resources. When $P\to M$ , the factor tends to 0: growth is choked off, the environment is full. Equilibria: $P=0$ and $P=M$ . Between them, $P'>0$ : every population moves away from 0 (UNSTABLE: the slightest introduction takes off) and rises towards $M$ (STABLE: it is approached from below, and an overpopulation $P>M$ would come back down to it, $P'<0$ ).

b)

At $t=0$ : $P(0)=\frac{M}{1+A}=\frac{M}{1+\frac{M-P_{0}}{P_{0}}}=\frac{MP_{0}}{M}=P_{0}$ ✓. Here $A=\frac{6000-500}{500}=11$ . After 5 years: $P(5)=\frac{6000}{1+11e^{-2}}=\frac{6000}{1+1.489}\approx 2410$ fish: the population has almost quintupled but is still below half the capacity of the lake.

c)

$P'=kP\left(1-\frac{P}{M}\right)$ is, for fixed $k$ , a parabola in $P$ that vanishes at 0 and $M$ : its maximum is at the vertex, $P=\frac{M}{2}$ . This is the INFLECTION POINT of the curve: there the population switches from accelerating to slowing down. Time: $\frac{M}{1+Ae^{-kt}}=\frac{M}{2}\iff Ae^{-kt}=1\iff t=\frac{\ln A}{k}=\frac{\ln 11}{0.4}\approx 6.0$ years: exactly the vertical gridline of the figure where the curve crosses the 3000 level. For a fishery, it is the year of the maximum sustainable harvest.

d)

At the start ( $A e^{-kt}\gg 1$ ): $P\approx\frac{M}{A}e^{kt}=P_{0}e^{kt}$ to a good approximation: indistinguishable from the exponential, as on the figure before year 3. At the end: $P\to M$ , flattening out below the asymptote, which the curve never crosses. The logistic IS Malthus's exponential corrected by the carrying capacity: same beginnings, opposite fates. It genuinely models closed-environment populations (bacteria in a Petri dish, the fish of a lake), the spread of epidemics and rumours (the number of people “not yet reached” plays the role of the brake), the cumulative sales of a product that saturates its market, or an autocatalytic reaction.

Exercise 11: Falling with air resistance and terminal velocity

In ideal free fall, the velocity would grow without bound. The air opposes it with a drag force, here proportional to the velocity: Newton's second law gives $m\frac{dv}{dt}=mg-kv$ . A skydiver of mass $m=80$ kg has a drag coefficient $k=16$ kg/s; take $g=9.8$ m/s².

The figure below shows the situation; it is not to scale.

a) Without solving, find the TERMINAL VELOCITY, the one at which the acceleration vanishes. Explain why the fall starts briskly and then levels off, and relate this behaviour to an equilibrium solution.
b) Put the equation in standard form, identify the time constant $\tau=\frac{m}{k}$ , and solve with $v(0)=0$ . Compute the velocity at $t=\tau$ , then at $t=10$ s.
c) After how long does the skydiver reach 90% of the terminal velocity? Then, by integrating $v(t)$ , compute the distance covered during the first 10 seconds.
d) Compare with free fall without air ( $v=gt$ ) at $t=10$ s, and explain why the terminal velocity “forgets” the initial velocity. What changes when the parachute opens, multiplying $k$ by a large factor?

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Answers

a) $v_{\lim}=49$ m/s
b) $\tau=5$ s; $30.97$ then $42.37$ m/s
c) $11.5$ s; $278$ m
d) $98$ m/s without air; the transient dies out

a)

Terminal velocity is reached when the net force vanishes, that is when the drag exactly balances the weight: $mg-kv=0$ , so $v_{\lim}=\frac{mg}{k}=\frac{80\times 9.8}{16}=49$ m/s, about 176 km/h, a realistic order of magnitude for a skydiver spread-eagled before opening the parachute. The behaviour can be read without computation: at the start, at zero velocity, the drag is zero and the acceleration is the full $g$ , the fall is brisk; as $v$ increases, the term $-kv$ eats into the weight and the acceleration decreases; at $49$ m/s, there is no acceleration left at all. This is exactly an EQUILIBRIUM solution of the equation, $v=v_{\lim}$ , towards which every fall converges, just as the coffee settled to room temperature in exercise 8. The differential equation of linear drag and that of Newton's law of cooling are, up to a change of names, the same equation.

b)

Divide by $m$ : $\frac{dv}{dt}=g-\frac{k}{m}v$ , that is $\frac{dv}{dt}+\frac{k}{m}v=g$ , linear with constant coefficients with $P=\frac{k}{m}$ . The time constant is $\tau=\frac{m}{k}=\frac{80}{16}=5$ s. Solve with an integrating factor, or directly with the structure “equilibrium plus transient”: $v(t)=v_{\lim}+Ce^{-t/\tau}$ , and $v(0)=0$ forces $C=-v_{\lim}$ , hence $v(t)=v_{\lim}\left(1-e^{-t/\tau}\right)=49\left(1-e^{-t/5}\right)$ . At $t=\tau=5$ s: $v=49\left(1-e^{-1}\right)\approx 49\times 0.632=30.97$ m/s, the classic 63% of the final value. At $t=10$ s, that is two time constants: $v=49\left(1-e^{-2}\right)\approx 49\times 0.865=42.37$ m/s, already very close to the limit.

c)

You want $v=0.9\,v_{\lim}$ , so $1-e^{-t/\tau}=0.9$ , that is $e^{-t/\tau}=0.1$ , hence $t=\tau\ln 10=5\ln 10\approx 11.5$ s. Distance covered: integrate the velocity, $x(t)=\int_{0}^{t}v_{\lim}\left(1-e^{-s/\tau}\right)ds=v_{\lim}\left[s+\tau e^{-s/\tau}\right]_{0}^{t}=v_{\lim}\left(t-\tau\left(1-e^{-t/\tau}\right)\right)$ . At $t=10$ s: $x=49\left(10-5\left(1-e^{-2}\right)\right)=49\left(10-5\times 0.8647\right)=49\times 5.677\approx 278$ m. You can check consistency by differentiating: $x'(10)\approx 42.37$ m/s, which indeed reproduces $v(10)$ , as it should.

d)

Without air, the velocity would be $v=gt=9.8\times 10=98$ m/s at $t=10$ s, exactly TWICE the actual terminal velocity, and it would keep growing indefinitely. Drag changes the picture radically: instead of a line that rises forever, you get a curve that levels off. As for forgetting the initial velocity, it comes from the transient term $Ce^{-t/\tau}$ : it, and it alone, carries the memory of $v(0)$ , and it dies out exponentially. Two skydivers released at different velocities converge to the same $49$ m/s, only the way of getting there differs, exactly as the brine tank forgot its initial mass of salt in exercise 9. Opening the parachute abruptly multiplies $k$ , say by a factor of 15: the terminal velocity $\frac{mg}{k}$ is divided by as much and drops to a few metres per second, while the time constant $\frac{m}{k}$ drops too, so that the new, slow terminal velocity is reached within a handful of seconds. That is the whole job of the parachute: bring $v_{\lim}$ down to a value that is survivable on landing. An honest caveat to finish: at high speeds, real drag behaves more like $v^{2}$ than $v$ , which makes the equation separable but nonlinear and gives a terminal velocity in $\sqrt{mg/k}$ ; the linear model nonetheless captures the essence of the phenomenon, the existence of a ceiling.

Exercise 12: Fishing and the collapse threshold: the logistic model with harvesting

The lake of exercise 10, with carrying capacity $M=6000$ fish and rate $k=0.4$ per year, is now being fished: $h$ fish are harvested per year. The model becomes $\frac{dP}{dt}=kP\left(1-\frac{P}{M}\right)-h$ . A single constant subtracted, and the whole landscape of equilibria reorganizes itself.

a) The equilibria make the right-hand side zero. Write the equation they satisfy as a quadratic in $P$ , and show that, for a moderate harvest, there are TWO equilibria. State which one is stable.
b) Find the CRITICAL harvest $h_{c}$ beyond which there is no equilibrium at all. Show that it equals exactly $\frac{kM}{4}$ and that it is reached for $P=\frac{M}{2}$ . Relate this result to the maximum growth rate found in exercise 10.
c) Compare three fishing regimes: $h=320$ , $h=560$ and $h=700$ fish per year. Compute the equilibria when they exist, and describe what happens to the population in each case.
d) The critical harvest $\frac{kM}{4}$ is the maximum sustainable yield. Why is a fishery that aims for EXACTLY this yield playing a dangerous game, and what does the analysis of equilibria advise?

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Answers

a) Two equilibria: the high one stable, the low one unstable
b) $h_{c}=\frac{kM}{4}=600$ at $P=3000$
c) $951$ and $5049$ ; $2225$ and $3775$ ; extinction
d) Quota strictly below $h_{c}$

a)

At equilibrium, $kP\left(1-\frac{P}{M}\right)-h=0$ . Expanding: $kP-\frac{k}{M}P^{2}-h=0$ , which you multiply by $-\frac{M}{k}$ to get a monic quadratic: $P^{2}-MP+\frac{hM}{k}=0$ . The equilibria are therefore the roots of this quadratic, $P=\frac{M\pm\sqrt{M^{2}-\frac{4hM}{k}}}{2}$ . For a moderate harvest, the discriminant $M^{2}-\frac{4hM}{k}$ is positive and there are TWO equilibria, a low one $P_{-}$ and a high one $P_{+}$ , both between 0 and $M$ . Their stability is read from the sign of $P'$ : the right-hand side $kP\left(1-\frac{P}{M}\right)-h$ is a parabola in $P$ opening downwards, negative before $P_{-}$ , positive between $P_{-}$ and $P_{+}$ , negative after $P_{+}$ . So below $P_{-}$ the population decreases (towards extinction), between the two it increases, above $P_{+}$ it decreases: the HIGH equilibrium $P_{+}$ is STABLE, the LOW equilibrium $P_{-}$ is UNSTABLE. The latter acts as a critical population threshold: a population that drops below $P_{-}$ is doomed, even without any increase in fishing.

b)

There is no equilibrium left once the discriminant becomes negative. The critical value makes this discriminant zero: $M^{2}-\frac{4h_{c}M}{k}=0$ , so $h_{c}=\frac{kM}{4}$ . With $k=0.4$ and $M=6000$ : $h_{c}=\frac{0.4\times 6000}{4}=600$ fish per year. At this precise harvest, the discriminant is zero and the two equilibria MERGE into one, the double root $P=\frac{M}{2}=3000$ . The link with exercise 10 is remarkable and no coincidence: there we established that logistic growth WITHOUT harvesting is maximal at the inflection point $P=\frac{M}{2}$ , where it equals $kP\left(1-\frac{P}{M}\right)=k\cdot\frac{M}{2}\cdot\frac{1}{2}=\frac{kM}{4}$ . The critical harvest is therefore exactly the maximum output the population can produce: beyond it, more fish are removed than the lake can generate per unit of time, whatever its size, and collapse is inevitable. The fishing threshold is the top of the growth curve.

c)

Three regimes. For $h=320$ : the quadratic $P^{2}-6000P+\frac{320\times 6000}{0.4}=P^{2}-6000P+4{,}800{,}000=0$ has roots of about $P_{-}\approx 951$ and $P_{+}\approx 5049$ . The population introduced at 500, below the unstable threshold $951$ , is unfortunately doomed to decrease; but any population above 951 would rise towards the stable equilibrium of 5049, an almost full lake that durably supplies 320 fish per year. For $h=560$ , closer to the critical value: $P^{2}-6000P+8{,}400{,}000=0$ gives $P_{-}\approx 2225$ and $P_{+}\approx 3775$ . The two equilibria have moved closer to 3000, and this closeness is the sign of DANGER: the margin between the collapse threshold (2225) and the stable regime (3775) has become thin, so that a bad year, a drought or a counting error can push the population below 2225 and trigger the fall. For $h=700$ , above the critical 600: the discriminant $36\times 10^{6}-42\times 10^{6}$ is NEGATIVE, there is no equilibrium at all, and $P'$ is negative for every value of $P$ , since the maximum growth (600) can no longer make up for the harvest (700). The population decreases relentlessly and dies out in finite time, regardless of its starting size: even a full lake ends up emptied.

d)

Aiming for exactly $h_{c}=\frac{kM}{4}$ amounts to keeping the population on the DOUBLE root $P=\frac{M}{2}=3000$ , which is a semi-stable equilibrium, stable from above but unstable from below. It is a tightrope walk: as long as nothing moves, the yield of 600 is indeed sustained; but at this precise point, the slightest disturbance that pushes $P$ below 3000 puts the population in a zone where $P'<0$ with no equilibrium to catch it, and collapse sets in. There is no safety net any more, because the unstable equilibrium and the stable equilibrium have merged: the warning threshold and the operating point have become the same point. The bifurcation that eliminated the equilibria for $h>h_{c}$ makes the regime $h=h_{c}$ structurally fragile. The analysis of equilibria therefore advises setting the quota STRICTLY below $h_{c}$ , say at 80 or 90 percent, so as to keep two well-separated equilibria and a safety margin between the collapse threshold $P_{-}$ and the operating regime $P_{+}$ . This is exactly the logic of prudent fishing quotas, and the history of the great fisheries that collapsed, the Newfoundland cod in particular, is that of yields held too close to, or beyond, this critical peak. The logistic model with harvesting thus provides, in one equation, the mathematical argument for sustainable resource management.

Exercise 13: Five classic separable equations

For each equation, separate the variables, integrate, then use the initial condition. State the interval of validity when it is not the whole real line.

a) $y'=y\cos x$ with $y(0)=2$ . Compute $y\left(\frac{\pi}{2}\right)$ .
b) $y'=1+y^{2}$ with $y(0)=0$ . On what interval does the solution exist? Compute $y\left(\frac{\pi}{4}\right)$ .
c) $x\,y'=3y$ with $y(1)=2$ , for $x>0$ . Compute $y(2)$ .
d) $y'=2\sqrt{y}$ with $y(0)=0$ . Show that $y=0$ and $y=x^{2}$ (for $x\geq 0$ ) are two solutions. What becomes of uniqueness?
e) $y'=(2x+1)e^{-y}$ with $y(0)=0$ . Compute $y(1)$ .

Show the solution

Answers

a) $y=2e^{\sin x}$ ; $2e$
b) $y=\tan x$ on $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$
c) $y=2x^{3}$ ; $16$
d) Two solutions: uniqueness fails
e) $y=\ln(x^{2}+x+1)$ ; $\ln 3$

a)

$\frac{dy}{y}=\cos x\,dx$ gives $\ln|y|=\sin x+C_{1}$ , that is $y=Ce^{\sin x}$ . With $y(0)=2$ : $C=2$ , so $y=2e^{\sin x}$ , defined on all of $\mathbb{R}$ and periodic. $y\left(\frac{\pi}{2}\right)=2e\approx 5.437$ .

b)

$\frac{dy}{1+y^{2}}=dx$ gives $\arctan y=x+C$ ; with $y(0)=0$ , $C=0$ and $y=\tan x$ . The solution exists only on $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ , that is up to $x\approx 1.571$ : it blows up in finite time although the equation is defined everywhere. $y\left(\frac{\pi}{4}\right)=1$ .

c)

$\frac{dy}{y}=\frac{3\,dx}{x}$ gives $\ln|y|=3\ln x+C_{1}$ , that is $y=Cx^{3}$ . With $y(1)=2$ : $y=2x^{3}$ , and $y(2)=16$ .

d)

$y=0$ gives $y'=0=2\sqrt{0}$ ✓. $y=x^{2}$ gives $y'=2x=2\sqrt{x^{2}}$ for $x\geq 0$ ✓, and $y(0)=0$ in both cases. Two distinct solutions for the same initial condition: UNIQUENESS fails. The cause is the function $\sqrt{y}$ , whose slope is infinite at $0$ : the existence and uniqueness theorem requires a right-hand side that is regular enough in $y$ , which is not the case here. For the nonzero solution, $y(3)=9$ .

e)

$e^{y}\,dy=(2x+1)\,dx$ gives $e^{y}=x^{2}+x+C$ ; with $y(0)=0$ , $C=1$ . So $y=\ln\left(x^{2}+x+1\right)$ , defined everywhere since $x^{2}+x+1>0$ . $y(1)=\ln 3\approx 1.099$ .

Exercise 14: The integrating factor on five linear equations

For each equation, put it in the form $y'+P(x)y=Q(x)$ , compute the integrating factor $\mu=e^{\int P\,dx}$ , solve, then apply the initial condition.

a) $y'-2y=e^{3x}$ , $y(0)=0$ . Compute $y(1)$ .
b) $y'+y=x$ , $y(0)=0$ . Compute $y(2)$ .
c) $y'+y\tan x=\cos x$ on $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ , $y(0)=1$ . Compute $y\left(\frac{\pi}{3}\right)$ .
d) $x\,y'-2y=x^{3}$ for $x>0$ , $y(1)=0$ . Compute $y(2)$ .
e) $y'+2xy=x$ , $y(0)=1$ . What is the limit of $y$ at $+\infty$ ?

Show the solution

Answers

a) $e^{3x}-e^{2x}$
b) $x-1+e^{-x}$
c) $(x+1)\cos x$
d) $x^{3}-x^{2}$
e) $\frac{1}{2}+\frac{1}{2}e^{-x^{2}}\to\frac{1}{2}$

a)

$\mu=e^{-2x}$ : $\left(e^{-2x}y\right)'=e^{x}$ , so $e^{-2x}y=e^{x}+C$ and $y=e^{3x}+Ce^{2x}$ . $y(0)=1+C=0$ : $C=-1$ , $y=e^{3x}-e^{2x}$ . $y(1)=e^{3}-e^{2}\approx 12.696$ .

b)

$\mu=e^{x}$ : $\left(e^{x}y\right)'=xe^{x}$ , and by parts $\int xe^{x}dx=(x-1)e^{x}$ . So $y=x-1+Ce^{-x}$ ; $y(0)=-1+C=0$ gives $C=1$ . $y(2)=1+e^{-2}\approx 1.135$ . The solution joins the line $y=x-1$ : the transient $e^{-x}$ dies out.

c)

$\mu=e^{\int\tan x\,dx}=e^{-\ln(\cos x)}=\frac{1}{\cos x}$ , positive on the interval. The equation becomes $\left(\frac{y}{\cos x}\right)'=1$ , so $y=(x+C)\cos x$ ; $y(0)=C=1$ . $y\left(\frac{\pi}{3}\right)=\left(\frac{\pi}{3}+1\right)\cdot\frac{1}{2}\approx 1.024$ .

d)

Standard form, dividing by $x$ : $y'-\frac{2}{x}y=x^{2}$ . $\mu=e^{-2\ln x}=\frac{1}{x^{2}}$ : $\left(\frac{y}{x^{2}}\right)'=1$ , so $y=x^{3}+Cx^{2}$ ; $y(1)=1+C=0$ , $C=-1$ . $y(2)=8-4=4$ . Forgetting to divide by $x$ before reading off $P$ is the classic error.

e)

$\mu=e^{x^{2}}$ : $\left(e^{x^{2}}y\right)'=xe^{x^{2}}$ , so $e^{x^{2}}y=\frac{1}{2}e^{x^{2}}+C$ and $y=\frac{1}{2}+Ce^{-x^{2}}$ . $y(0)=\frac{1}{2}+C=1$ : $C=\frac{1}{2}$ . As $x\to+\infty$ , $y\to\frac{1}{2}$ , the equilibrium solution $y=\frac{1}{2}$ that can be read directly from the equation.

Exercise 15: Euler and improved Euler against the exact solution

Consider $y'=x-y$ with $y(0)=1$ , whose exact solution is known, so as to measure honestly the error of two numerical methods.

a) Solve the equation with an integrating factor and compute the exact value $y(1)$ .
b) Apply Euler's method $y_{n+1}=y_{n}+h\,f(x_{n},y_{n})$ with $h=0.5$ to estimate $y(1)$ .
c) Start again with $h=0.25$ .
d) Compute the two errors and their ratio. What does this ratio say about the order of Euler's method?
e) The improved Euler method (Heun's method) predicts $\tilde{y}=y_{n}+h\,f(x_{n},y_{n})$ then corrects with the average of the slopes: $y_{n+1}=y_{n}+\frac{h}{2}\left[f(x_{n},y_{n})+f(x_{n+1},\tilde{y})\right]$ . Apply it with $h=0.5$ and compare the error.

Show the solution

Answers

a) $y=x-1+2e^{-x}$ ; $y(1)=\frac{2}{e}$
b) $0.5$
c) $0.6328$
d) Ratio $\approx 2$ : order $1$
e) Heun: $0.78125$ , error $0.0455$

a)

$y'+y=x$ : $\mu=e^{x}$ , $\left(e^{x}y\right)'=xe^{x}$ , so $y=x-1+Ce^{-x}$ . $y(0)=-1+C=1$ : $C=2$ and $y=x-1+2e^{-x}$ . $y(1)=\frac{2}{e}\approx 0.7358$ .

b)

$f(x,y)=x-y$ . Step 1: $y_{1}=1+0.5\,(0-1)=0.5$ at $x=0.5$ . Step 2: $y_{2}=0.5+0.5\,(0.5-0.5)=0.5$ at $x=1$ . Estimate: $y(1)\approx 0.5$ .

c)

Four steps: $y_{1}=1+0.25(0-1)=0.75$ ; $y_{2}=0.75+0.25(0.25-0.75)=0.625$ ; $y_{3}=0.625+0.25(0.5-0.625)=0.59375$ ; $y_{4}=0.59375+0.25(0.75-0.59375)\approx 0.6328$ .

d)

Errors: $0.7358-0.5=0.2358$ and $0.7358-0.6328=0.1030$ . Ratio: $\frac{0.2358}{0.1030}\approx 2.3$ , close to $2$ : halving the step roughly halves the error. Euler's error is proportional to the step: the method is of order $1$ . Both estimates are too low: the solution is concave up ( $y''=1-y'=y-x+1=2e^{-x}>0$ ) and the tangent line passes below the curve.

e)

Step 1: slope $f(0, 1)=-1$ , prediction $\tilde{y}=0.5$ , slope $f(0.5, 0.5)=0$ , average $-0.5$ : $y_{1}=1-0.25=0.75$ . Step 2: slope $f(0.5, 0.75)=-0.25$ , prediction $\tilde{y}=0.625$ , slope $f(1, 0.625)=0.375$ , average $0.0625$ : $y_{2}=0.75+0.03125=0.78125$ . Error $\left|0.7358-0.7813\right|\approx 0.0455$ , five times smaller than Euler with the same step, for only two slope evaluations per step. Heun's method is of order $2$ : it is the trapezoidal rule applied to differential equations.

Exercise 16: Draining a tank: Torricelli's law

A cylindrical tank drains through a small hole drilled in its bottom. Torricelli's law gives the rate at which the level $h$ (in m) drops: $\frac{dh}{dt}=-k\sqrt{h}$ , with time $t$ in minutes. At the start, $h(0)=4$ m; after $5$ minutes, the measured level is $h(5)=2.25$ m.

The figure below shows the situation; it is not to scale.

a) Solve the equation by separation and express $\sqrt{h}$ in terms of $t$ , $k$ and $h(0)$ .
b) Find $k$ from the measurement at $5$ minutes.
c) When is the tank empty?
d) When does the level reach half its initial height? Comment on the fact that this time is well under half the draining time.
e) A tank obeying $\frac{dh}{dt}=-kh$ would never empty. Explain the difference.

Show the solution

Answers

a) $\sqrt{h}=2-\frac{kt}{2}$
b) $k=0.2$
c) Empty at $20$ min
d) Half height at $5.86$ min
e) $\sqrt{h}$ dominates $h$ near $0$ : finite time

a)

$\frac{dh}{\sqrt{h}}=-k\,dt$ gives $2\sqrt{h}=-kt+C$ , and $C=2\sqrt{h(0)}=4$ . So $\sqrt{h}=2-\frac{kt}{2}$ , as long as this number is positive.

b)

$\sqrt{2.25}=1.5=2-\frac{5k}{2}$ , so $k=0.2$ ( $\text{m}^{1/2}$ per minute). Hence $\sqrt{h}=2-0.1\,t$ and $h(t)=(2-0.1\,t)^{2}$ .

c)

$h=0$ when $2-0.1\,t=0$ , that is $t=20$ minutes. Beyond that, the physical solution is $h=0$ : the tank stays empty.

d)

$h=2$ gives $\sqrt{2}=2-0.1\,t$ , that is $t=10\left(2-\sqrt{2}\right)\approx 5.86$ minutes. The first half of the height drains in under six minutes, the second in more than fourteen: the outflow, proportional to $\sqrt{h}$ , weakens as the pressure of the water column decreases.

e)

With $h'=-kh$ , the solution $h=4e^{-kt}$ never reaches $0$ : the rate of drop is proportional to $h$ , and fades as fast as $h$ does. With $\sqrt{h}$ , when $h$ is small, $\sqrt{h}$ is MUCH larger than $h$ (for example $\sqrt{0.01}=0.1$ ): the draining slows down too little to keep the level from reaching zero, and it gets there in finite time. It is the same lack of regularity of $\sqrt{h}$ at $0$ that, read backwards, breaks uniqueness: seeing an empty tank, you cannot tell how long it has been empty.

Exercise 17: A chemical reaction and partial fractions

In the reaction $\text{A}+\text{B}\to\text{C}$ , the initial concentrations are $[\text{A}]_{0}=2$ mol/L and $[\text{B}]_{0}=3$ mol/L. The concentration $x(t)$ of product formed (mol/L, $t$ in minutes) satisfies the rate law $\frac{dx}{dt}=k(2-x)(3-x)$ , with $k=0.1$ L/(mol·min) and $x(0)=0$ .

a) Compute the initial rate of the reaction and decompose $\frac{1}{(2-x)(3-x)}$ into partial fractions.
b) Solve the equation by separation and show that $\frac{3-x}{2-x}=\frac{3}{2}e^{kt}$ .
c) After how long does the concentration of the product reach $1$ mol/L?
d) Express $x(t)$ and compute $x(10)$ .
e) What value does $x(t)$ tend to? Interpret chemically.

Show the solution

Answers

a) $0.6$ ; $\frac{1}{2-x}-\frac{1}{3-x}$
b) $\frac{3-x}{2-x}=\frac{3}{2}e^{kt}$
c) $10\ln\frac{4}{3}\approx 2.88$ min
d) $x(10)\approx 1.675$
e) Limit $2$ : limiting reactant

a)

$x'(0)=0.1\times 2\times 3=0.6$ mol/(L·min). Decomposition: $\frac{1}{(2-x)(3-x)}=\frac{a}{2-x}+\frac{b}{3-x}$ , that is $1=a(3-x)+b(2-x)$ ; at $x=2$ , $a=1$ ; at $x=3$ , $b=-1$ . So $\frac{1}{(2-x)(3-x)}=\frac{1}{2-x}-\frac{1}{3-x}$ .

b)

$\int\left(\frac{1}{2-x}-\frac{1}{3-x}\right)dx=-\ln(2-x)+\ln(3-x)=\ln\frac{3-x}{2-x}$ , for $0\leq x<2$ . Hence $\ln\frac{3-x}{2-x}=kt+C$ , and $x(0)=0$ gives $C=\ln\frac{3}{2}$ . Taking the exponential: $\frac{3-x}{2-x}=\frac{3}{2}e^{kt}$ .

c)

$x=1$ gives $\frac{2}{1}=\frac{3}{2}e^{0.1t}$ , that is $t=10\ln\frac{4}{3}\approx 2.88$ minutes.

d)

Let $R=\frac{3}{2}e^{kt}$ . From $3-x=R(2-x)$ you get $x(R-1)=2R-3$ , so $x(t)=\frac{2R-3}{R-1}$ . At $t=10$ : $R=1.5\,e\approx 4.077$ and $x(10)\approx\frac{5.155}{3.077}\approx 1.675$ mol/L.

e)

As $t\to\infty$ , $R\to\infty$ and $x=\frac{2-3/R}{1-1/R}\to 2$ . The reaction stops when A, the LIMITING REACTANT, is used up: $1$ mol/L of B is then left in excess. The equation says so too: $x'=0$ for $x=2$ , a stable equilibrium, whereas $x=3$ is never reached.

Corrected exercises: differential equations (201-SN3-RE)

Course recap

Calculations: solving in quick succession (/20)

Exercise 1: Recognize the type, then solve: five first-order equations

Exercise 2: From second order to first: five equations to integrate twice or to reduce

Part A: techniques, qualitative analysis and substitutions (/50)

Exercise 3: Verify, separate, solve

Exercise 4: The linear equation and the integrating factor

Exercise 5: Exponential growth and decay

Exercise 6: Read before solving: slope field, phase line and Euler's method

Exercise 7: Substitutions: making linear what was not

Part B: modelling problems (/50)

Exercise 8: Newton's law of cooling: the coffee and the stopwatch

Exercise 9: The brine tank: the equilibrium that erases the past

Exercise 10: The logistic model: the exponential tamed

Exercise 11: Falling with air resistance and terminal velocity

Exercise 12: Fishing and the collapse threshold: the logistic model with harvesting

Part C: the classics (/50)

Exercise 13: Five classic separable equations

Exercise 14: The integrating factor on five linear equations

Exercise 15: Euler and improved Euler against the exact solution

Exercise 16: Draining a tank: Torricelli's law

Exercise 17: A chemical reaction and partial fractions

See also

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