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UPSC 2022 Maths Optional Paper 2 Q8b — Step-by-Step Solution

15 marks · Section B

Runge-Kutta methods (RK2/RK4) · Numerical Analysis · asked 4× in 13 yrs · Read the full method →

Question

Using RK4, solve $dy/dx=x+y^2$ with $y(0)=1$, at $x=0.2$. Use 4 decimal places, step length $0.1$.

Technique

Standard RK4 with $h=0.1$, two steps.

Solution

Setup. $f(x,y)=x+y^2$. $x_0=0,y_0=1$. $h=0.1$. Two RK4 steps.

Step 1 — RK4 step 1: $x=0\to x=0.1$

$k_1=h f(x_0,y_0)=0.1\cdot(0+1^2)=0.1$.

$k_2=h f(x_0+h/2,y_0+k_1/2)=0.1\cdot f(0.05,1.05)=0.1(0.05+1.05^2)=0.1(0.05+1.1025)=0.1\cdot 1.1525=0.11525$.

$k_3=h f(x_0+h/2,y_0+k_2/2)=0.1\cdot f(0.05,1.057625)=0.1(0.05+1.057625^2)$.

$1.057625^2=1.118571$. So $k_3=0.1(0.05+1.118571)=0.1\cdot 1.168571=0.116857$.

$k_4=h f(x_0+h,y_0+k_3)=0.1\cdot f(0.1,1.116857)=0.1(0.1+1.116857^2)$.

$1.116857^2=1.247370$. So $k_4=0.1(0.1+1.247370)=0.1\cdot 1.347370=0.134737$.

$y_1=y_0+\dfrac{1}{6}(k_1+2k_2+2k_3+k_4)=1+\dfrac{1}{6}(0.1+0.23050+0.23371+0.13474)$ $=1+\dfrac{1}{6}(0.69895)=1+0.11649=1.11649$.

Round to 4 decimals: $y_1=1.1165$.

Step 2 — RK4 step 2: $x=0.1\to x=0.2$

$k_1=0.1\cdot f(0.1,1.1165)=0.1(0.1+1.1165^2)$.

$1.1165^2=1.24657$. $k_1=0.1(0.1+1.24657)=0.1\cdot 1.34657=0.13466$.

$k_2=0.1\cdot f(0.15,1.1165+0.13466/2)=0.1\cdot f(0.15,1.1839)$.

$1.1839^2=1.4016$. $k_2=0.1(0.15+1.4016)=0.1\cdot 1.5516=0.15516$.

$k_3=0.1\cdot f(0.15,1.1165+0.15516/2)=0.1\cdot f(0.15,1.1941)$.

$1.1941^2=1.4259$. $k_3=0.1(0.15+1.4259)=0.1\cdot 1.5759=0.15759$.

$k_4=0.1\cdot f(0.2,1.1165+0.15759)=0.1\cdot f(0.2,1.2741)$.

$1.2741^2=1.6233$. $k_4=0.1(0.2+1.6233)=0.1\cdot 1.8233=0.18233$.

$y_2=1.1165+\dfrac{1}{6}(0.13466+0.31032+0.31518+0.18233)$ $=1.1165+\dfrac{1}{6}(0.94249)$ $=1.1165+0.15708$ $=1.2736$.

Answer