← 2023 Paper 2

UPSC 2023 Maths Optional Paper 2 Q5b — Step-by-Step Solution

10 marks · Section B

Euler's method (and modified Euler) · Numerical Analysis · asked 2× in 13 yrs · Read the full method →

Question

Given $\dfrac{dy}{dx}=\dfrac{y^2-x}{y^2+x}$ with initial condition $y=1$ at $x=0$. Find the value of $y$ for $x=0.4$ by Euler’s method, correct to 4 decimal places, taking step length $h=0.1$.

Technique

Euler’s method (forward, first-order). Four steps of $h=0.1$ from $x=0$ to $x=0.4$.

Solution

Euler’s recurrence. With $f(x,y)=\dfrac{y^{2}-x}{y^{2}+x}$ and $h=0.1$, the iteration is

Starting from $(x_0,y_0)=(0,1)$, perform four steps to reach $x=0.4$.

Step 1 — From $(0,1)$ to $(0.1,y_1)$

$f(0,1)=\dfrac{1-0}{1+0}=1.0000.$ $y_1=1+0.1(1.0000)=1.1000.$

Step 2 — From $(0.1,1.1000)$ to $(0.2,y_2)$

$y_1^{2}=1.2100$. $f(0.1,1.1)=\dfrac{1.2100-0.1}{1.2100+0.1}=\dfrac{1.1100}{1.3100}=0.84733.$ $y_2=1.1000+0.1(0.84733)=1.1847.$

Step 3 — From $(0.2,1.1847)$ to $(0.3,y_3)$

$y_2^{2}=1.4035$ (since $1.1847^{2}=1.40352\ldots$). $f(0.2,1.1847)=\dfrac{1.4035-0.2}{1.4035+0.2}=\dfrac{1.2035}{1.6035}=0.75055.$ $y_3=1.1847+0.1(0.75055)=1.2598.$

Step 4 — From $(0.3,1.2598)$ to $(0.4,y_4)$

$y_3^{2}=1.5871$ (since $1.2598^{2}=1.58708\ldots$). $f(0.3,1.2598)=\dfrac{1.5871-0.3}{1.5871+0.3}=\dfrac{1.2871}{1.8871}=0.68205.$ $y_4=1.2598+0.1(0.68205)=1.3280.$

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