Differential equations and Euler's method – IB Maths exam questions with mark scheme

1StandardNon-calculatorHL only4 marks

Solve the differential equation $\dfrac{dy}{dx}=2xy$, given that $y=3$ when $x=0$ and $y>0$. Give $y$ in terms of $x$.

Separates variables: $\int\dfrac1y\,dy=\int2x\,dx$M1
$\ln y=x^2+c$A1
Uses $x=0,\ y=3$: $c=\ln3$M1
$y=3e^{x^2}$A1
Answer: $y=3e^{x^2}$
Examiner tip: Exponentiate carefully: $e^{x^2+c}=Ae^{x^2}$, not $e^{x^2}+e^c$.

2StandardCalculatorHL only6 marks

Consider the differential equation $\dfrac{dy}{dx}=x+y$ with $y=1$ when $x=0$.

(a) Use Euler's method with step length $h=0.1$ to estimate $y(0.3)$. Show each step. [4]

(b) The exact solution is $y=2e^x-x-1$. Use it to find the value of $y(0.3)$ correct to 4 decimal places, and state, with a reason, whether your Euler estimate is an under-estimate or an over-estimate. [2]

Uses $y_{n+1}=y_n+h\,f(x_n,y_n)$ with $f=x+y$, $h=0.1$M1
$y(0.1)\approx1+0.1(0+1)=1.1$A1
$y(0.2)\approx1.1+0.1(0.1+1.1)=1.22$A1
$y(0.3)\approx1.22+0.1(0.2+1.22)=1.362$A1
Exact $y(0.3)=2e^{0.3}-1.3=1.3997$ (4 d.p.) (accept $1.39972$)A1
Under-estimate, with reason: $y''=1+y'=1+x+y>0$ on the interval (curve concave up), so each tangent-line step lies below the curve (accept: Euler value $1.362<1.3997$)R1
Answer: (a) $1.362$; (b) $1.3997$, so Euler under-estimates
Examiner tip: Increase $x$ by $h$ at every step and use the current $x_n$ and $y_n$ together; forgetting to update $x$ (e.g. keeping $x=0$) is the common slip.

3HardNon-calculatorHL only5 marks

Given that $\dfrac{dy}{dx}=\dfrac{x+y}{x}$ for $x>0$ and $y=2$ when $x=1$, use the substitution $y=vx$ to find $y$ in terms of $x$.

$y=vx\Rightarrow\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$M1
$v+x\dfrac{dv}{dx}=1+v$, so $x\dfrac{dv}{dx}=1$A1
Separates and integrates: $\int dv=\int\dfrac1x\,dx$M1
$v=\ln x+c$A1
Uses $x=1,\ y=2$ (so $v=2$): $y=x\ln x+2x$A1
Answer: $y=x\ln x+2x$
Examiner tip: Apply the initial condition to $v=y/x$ or return to $y$ first, but do not mix the two.

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4HardNon-calculatorHL only6 marks

Solve $\dfrac{dy}{dx}+\dfrac{2}{x}y=3x$ for $x>0$, given that $y=2$ when $x=1$. Give $y$ in terms of $x$.

Integrating factor $e^{\int 2/x\,dx}$M1
$I=x^2$A1
Multiplies through: $\dfrac{d}{dx}(x^2y)=3x^3$M1
$x^2y=\dfrac{3x^4}{4}+c$A1
Uses $x=1,\ y=2$: $c=\dfrac54$M1
$y=\dfrac{3x^2}{4}+\dfrac{5}{4x^2}$A1
Answer: $y=\dfrac{3x^2}{4}+\dfrac{5}{4x^2}$
Examiner tip: Divide the whole equation, including $c$, by $x^2$ at the end.

5CoreCalculatorHL only6 marks

The function $y=f(x)$ satisfies the differential equation $\dfrac{dy}{dx}=\sqrt{x+y}$, with $f(1)=3$.

(a) Use Euler's method with a step length of $0.2$ to find an approximate value for $f(1.6)$. Give your answer correct to 3 significant figures. [4]

(b) By considering $\dfrac{d^2y}{dx^2}$, explain why your answer to (a) is an under-estimate of $f(1.6)$. [2]

Uses $y_{n+1}=y_n+0.2\sqrt{x_n+y_n}$ with $x_{n+1}=x_n+0.2$ (seen at least once with correct values)M1
$y_1=3.4$ (at $x=1.2$)A1
$y_2=3.8289\ldots$ (at $x=1.4$; accept $3.83$)A1
$f(1.6)\approx 4.29$ (accept $4.2863\ldots$; do not accept an answer from carrying 2 sf values that gives a different 3 sf answer)A1
Differentiates implicitly: $\dfrac{d^2y}{dx^2}=\dfrac{1+\frac{dy}{dx}}{2\sqrt{x+y}}$ (or equivalent)M1
Both $\frac{dy}{dx}>0$ and $\sqrt{x+y}>0$, so $\frac{d^2y}{dx^2}>0$: the curve is concave up, so each tangent step lies below the curve and Euler under-estimates. Accept equivalent reasoning about the gradient increasing along each stepR1
Answer: (a) $4.29$ (3 sf) (b) $y''=\dfrac{1+y'}{2\sqrt{x+y}}>0$, curve concave up, tangent steps lie below the curve
Examiner tip: Keep full calculator values between Euler steps: rounding $y_2$ to 3.8 changes the final answer.

6StandardCalculatorHL only7 marks

A cup of tea is left to cool in a room whose temperature is a constant $20^{\circ}\text{C}$. The temperature of the tea, $T\ ^{\circ}\text{C}$, $t$ minutes after it is poured satisfies

$$\frac{dT}{dt}=-k\,(T-20),\qquad k>0.$$

When $t=0$, $T=90$.

(a) By solving the differential equation, show that $T=20+70e^{-kt}$. [4]

(b) When $t=5$, $T=60$. Find the value of $k$. [1]

Separates variables: $\displaystyle\int\frac{1}{T-20}\,dT=\int -k\,dt$M1
$\ln|T-20|=-kt+c$A1
Uses $t=0,\ T=90$ to find the constant: $c=\ln 70$ (or $A=70$ in $T-20=Ae^{-kt}$)M1
AG: Rearranges correctly to $T=20+70e^{-kt}$ (answer given: every step must be seen; award only if no errors in previous working)A1
$k=\frac15\ln\frac74=0.112$ (3 sf; accept $0.111923\ldots$)A1
Sets $20+70e^{-kt}=30$ and solves, e.g. $t=\dfrac{\ln 7}{k}$M1
$t=17.4$ minutes (accept $17.386\ldots$; follow through their $k$)A1
Answer: (b) $k=\frac15\ln\frac74\approx0.112$ (c) $17.4$ minutes
Examiner tip: In a 'show that', write the constant of integration and find it explicitly; jumping from the separated form straight to the given answer loses the method marks.

7StandardNon-calculatorHL only6 marks

Consider the differential equation

$$\frac{dy}{dx}+y\tan x=\cos^2 x,\qquad 0\le x<\frac{\pi}{2}.$$

(a) Show that an integrating factor for this differential equation is $\sec x$. [2]

(b) Hence solve the differential equation, given that $y=1$ when $x=0$. Give your answer in the form $y=f(x)$. [4]

Integrating factor $=e^{\int\tan x\,dx}$M1
AG: $=e^{-\ln(\cos x)}=e^{\ln(\sec x)}=\sec x$ (must show the log step; $\cos x>0$ on the domain so no modulus needed)A1
Multiplies through by $\sec x$: $\dfrac{d}{dx}\left(y\sec x\right)=\cos^2x\sec x=\cos x$M1
$y\sec x=\sin x+c$A1
Substitutes $x=0,\ y=1$ to find $c$: $c=1$M1
$y=\cos x\,(\sin x+1)$ (accept $y=\sin x\cos x+\cos x$, $y=\frac12\sin 2x+\cos x$)A1
Answer: $y=\sin x\cos x+\cos x=\cos x(1+\sin x)$
Examiner tip: The integrating factor must multiply the right-hand side too: $\cos^2x\sec x=\cos x$, not $\cos^2x$.

8HardNon-calculatorHL only7 marks

Consider the differential equation

$$\frac{dy}{dx}=\frac{x^2+y^2}{xy},\qquad x>0,\ y>0,$$

with $y=2$ when $x=1$.

(a) Use the substitution $y=vx$ to show that $v\dfrac{dv}{dx}=\dfrac{1}{x}$. [3]

(b) Hence find $y$ in terms of $x$. [3]

Uses $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$M1
Substitutes $y=vx$ into the right-hand side: $\dfrac{x^2+v^2x^2}{vx^2}=\dfrac{1+v^2}{v}$M1
AG: $v+x\dfrac{dv}{dx}=\dfrac1v+v\Rightarrow x\dfrac{dv}{dx}=\dfrac1v\Rightarrow v\dfrac{dv}{dx}=\dfrac1x$A1
Integrates: $\dfrac{v^2}{2}=\ln x+c$A1
Uses $x=1,\ y=2$ so $v=2$: $c=2$, and replaces $v$ by $\dfrac yx$M1
$y=x\sqrt{2\ln x+4}$ (accept $y^2=2x^2\ln x+4x^2$; positive root must be chosen for a final $y=$ form)A1
$y=\sqrt6\,e$ (accept $e\sqrt6$; do not accept $6.66$ alone as exact value is required)A1
Answer: (b) $y=x\sqrt{2\ln x+4}$ (c) $\sqrt6\,e$
Examiner tip: Find the constant using $v=\frac{y}{x}=2$, not $v=y=2$; this only works here because $x=1$, so always substitute back first.

9HardCalculatorHL only9 marks

A population of fish, $P$, in a lake is modelled by the differential equation

$$\frac{dP}{dt}=0.2P\left(1-\frac{P}{1000}\right),$$

where $t$ is the time in years. When $t=0$, $P=100$.

(a) Write $\dfrac{1000}{P(1000-P)}$ in partial fractions. [1]

(b) Hence show that $P=\dfrac{1000}{1+9e^{-0.2t}}$. [6]

(d) State what happens to the population in the long term, according to this model. [1]

$\dfrac1P+\dfrac1{1000-P}$A1
Separates variables: $\displaystyle\int\frac{1000}{P(1000-P)}\,dP=\int0.2\,dt$ (or equivalent rearrangement)M1
$\ln P-\ln(1000-P)=0.2t+c$ (condone missing modulus since $0A1
Uses $t=0,\ P=100$: $c=\ln\frac{100}{900}=-\ln 9$M1
Combines logs and exponentiates: $\dfrac{P}{1000-P}=\dfrac19e^{0.2t}$M1
Makes $P$ the subject, e.g. $9P=(1000-P)e^{0.2t}\Rightarrow P(9+e^{0.2t})=1000e^{0.2t}$M1
AG: Divides by $e^{0.2t}$ to obtain $P=\dfrac{1000}{1+9e^{-0.2t}}$ (answer given: no errors in working)A1
$t=5\ln 9=11.0$ years (accept $10.986\ldots$, $10\ln3$)A1
As $t\to\infty$, $e^{-0.2t}\to0$ so $P\to1000$ (population approaches but never exceeds 1000)A1
Answer: (a) $\frac1P+\frac1{1000-P}$ (c) $11.0$ years (d) $P\to1000$
Examiner tip: The constant 1000 in the numerator in (a) is there to make the separation tidy: remember the $0.2$ stays on the $t$ side.

10StretchNon-calculatorHL only10 marks

The diagram shows the slope field for the differential equation

$$\frac{dy}{dx}=x-y.$$

(a) Show that the line $y=x-1$ is a solution curve of this differential equation. [2]

(b) Using an integrating factor, solve the differential equation given that $y=2$ when $x=0$. [5]

(d) With reference to the slope field, or otherwise, describe the behaviour of every solution curve as $x\to\infty$. [1]

For $y=x-1$: $\dfrac{dy}{dx}=1$ and $x-y=x-(x-1)$M1
$x-y=1=\dfrac{dy}{dx}$, so the line satisfies the equation (accept: every slope on $y=x-1$ in the field is $1$, along the line itself, supported by algebra)A1
Rearranges to $\dfrac{dy}{dx}+y=x$ and finds integrating factor $e^{x}$M1
$\dfrac{d}{dx}\left(ye^x\right)=xe^x$A1
Integrates $xe^x$ by partsM1
$ye^x=xe^x-e^x+c$A1
Uses $(0,2)$: $c=3$, so $y=x-1+3e^{-x}$A1
Sets $\dfrac{dy}{dx}=0$, i.e. $1-3e^{-x}=0$ (or uses $x-y=0$ on the curve)M1
$(\ln3,\ \ln3)$A1
All solution curves approach the line $y=x-1$ (since $e^{-x}\to0$); accept 'become asymptotic to $y=x-1$'A1
Answer: (b) $y=x-1+3e^{-x}$ (c) $(\ln 3,\ln 3)$ (d) they approach $y=x-1$
Examiner tip: Stationary points of any solution of $\frac{dy}{dx}=x-y$ lie on $y=x$, where the slope field segments are horizontal: use this as a check in (c).

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