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UPSC 2015 Maths Optional Paper 2 Q7b — 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

Solve the initial value problem $\dfrac{dy}{dx}=x(y-x)$, $y(2)=3$ in the interval $[2,2.4]$ using the Runge–Kutta fourth-order method with step size $h=0.2$.

Technique

RK4 with two steps, $h=0.2$; careful tracking of intermediate $f$ evaluations.

Solution

Setup. $f(x,y)=x(y-x)$. $x_0=2$, $y_0=3$, $h=0.2$. Two steps to reach $x=2.4$.

RK4 formula: at each step,

Step 1 — Step 1: $x_0=2,\;y_0=3$ to $x_1=2.2$

$f(2,3)=2(3-2)=2$. $k_1=0.2\cdot 2=0.4$.

$f(2.1,\;3+0.4/2)=f(2.1,3.2)=2.1(3.2-2.1)=2.1\cdot 1.1=2.31$. $k_2=0.2\cdot 2.31=0.462$.

$f(2.1,\;3+0.462/2)=f(2.1,3.231)=2.1(3.231-2.1)=2.1\cdot 1.131=2.3751$. $k_3=0.2\cdot 2.3751=0.47502$.

$f(2.2,\;3+0.47502)=f(2.2,3.47502)=2.2(3.47502-2.2)=2.2\cdot 1.27502=2.805044$. $k_4=0.2\cdot 2.805044=0.561009$.

$y_1=3+\tfrac{1}{6}(0.4+2\cdot 0.462+2\cdot 0.47502+0.561009)$ $=3+\tfrac{1}{6}(0.4+0.924+0.95004+0.561009)$ $=3+\tfrac{1}{6}(2.835049)$ $=3+0.472508$ $\approx 3.4725$.

Step 2 — Step 2: $x_1=2.2,\;y_1=3.4725$ to $x_2=2.4$

$f(2.2,3.4725)=2.2(3.4725-2.2)=2.2\cdot 1.2725=2.7995$. $k_1=0.2\cdot 2.7995=0.5599$.

$f(2.3,\;3.4725+0.5599/2)=f(2.3,\;3.4725+0.27995)=f(2.3,3.75245)=2.3(3.75245-2.3)=2.3\cdot 1.45245=3.340635$. $k_2=0.2\cdot 3.340635=0.668127$.

$f(2.3,\;3.4725+0.668127/2)=f(2.3,\;3.4725+0.3340635)=f(2.3,\;3.8065635)=2.3(3.8065635-2.3)=2.3\cdot 1.5065635=3.465096$. $k_3=0.2\cdot 3.465096=0.693019$.

$f(2.4,\;3.4725+0.693019)=f(2.4,4.165519)=2.4(4.165519-2.4)=2.4\cdot 1.765519=4.237246$. $k_4=0.2\cdot 4.237246=0.847449$.

$y_2=3.4725+\tfrac{1}{6}(0.5599+2\cdot 0.668127+2\cdot 0.693019+0.847449)$ $=3.4725+\tfrac{1}{6}(0.5599+1.336254+1.386038+0.847449)$ $=3.4725+\tfrac{1}{6}(4.129641)$ $=3.4725+0.688274$ $\approx 4.1608$.

Summary

Answer