Differential equations, Euler's method, Maclaurin series (HL) – 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 only7 marks

(a) Using the Maclaurin series for $e^x$ and $\sin x$, find the series expansion of $e^x\sin x$ up to and including the term in $x^3$. [5]

(b) Hence estimate $\displaystyle\int_0^{0.5}e^x\sin x\,dx$, giving your answer as a fraction. [2]

$e^x=1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\ldots$ and $\sin x=x-\dfrac{x^3}{6}+\ldots$M1
Multiplies the series, collecting terms up to $x^3$M1
Term in $x$: $x$A1
Term in $x^2$: $x^2$A1
Term in $x^3$: $\dfrac{x^3}{2}-\dfrac{x^3}{6}=\dfrac{x^3}{3}$A1
$\int_0^{0.5}\big(x+x^2+\tfrac{x^3}{3}\big)dx=\Big[\tfrac{x^2}{2}+\tfrac{x^3}{3}+\tfrac{x^4}{12}\Big]_0^{0.5}$M1
$\dfrac{11}{64}$ (accept $0.171875$)A1
Answer: (a) $x+x^2+\dfrac{x^3}{3}$; (b) $\dfrac{11}{64}$
Examiner tip: Only products giving powers up to $x^3$ matter; list them systematically so none is missed.

4HardNon-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.

5HardNon-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.

6StretchNon-calculatorHL only7 marks

(a) Using the standard series for $\cos x$ and $\ln(1+u)$, show that the Maclaurin series for $\ln(\cos x)$ begins $-\dfrac{x^2}{2}-\dfrac{x^4}{12}$. [5]

(b) Hence find $\displaystyle\lim_{x\to0}\dfrac{\ln(\cos x)+\frac{x^2}{2}}{x^4}$. [2]

$\cos x=1-\dfrac{x^2}{2}+\dfrac{x^4}{24}-\ldots$M1
Substitutes into $\ln(1+u)=u-\dfrac{u^2}{2}+\ldots$ with $u=-\dfrac{x^2}{2}+\dfrac{x^4}{24}$M1
$u=-\dfrac{x^2}{2}+\dfrac{x^4}{24}$ retained to $x^4$A1
$-\dfrac{u^2}{2}=-\dfrac{x^4}{8}+\ldots$A1
$\ln\cos x=-\dfrac{x^2}{2}+\dfrac{x^4}{24}-\dfrac{x^4}{8}=-\dfrac{x^2}{2}-\dfrac{x^4}{12}$ (AG)A1
Numerator $=-\dfrac{x^4}{12}+O(x^6)$, divide by $x^4$M1
Limit $=-\dfrac1{12}$A1
Answer: (a) shown: $\ln(\cos x)=-\dfrac{x^2}{2}-\dfrac{x^4}{12}+\ldots$ (b) $-\dfrac{1}{12}$
Examiner tip: Square $u$ and keep only terms to $x^4$; the $x^2$ term of $u$ squared supplies the missing $x^4$ contribution.

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