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UPSC 2014 Maths Optional Paper 2 Q6c — Step-by-Step Solution

20 marks · Section B

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

Question

Use Runge–Kutta formula of fourth order to find the value of $y$ at $x=0.8$, where $\dfrac{dy}{dx}=\sqrt{x+y}$, $y(0.4)=0.41$. Take the step length $h=0.2$.

Technique

Standard RK4 with two steps; careful with intermediate $\sqrt{\cdot}$ evaluations.

Solution

Setup. $f(x,y)=\sqrt{x+y}$. Two RK4 steps from $x=0.4$ to $x=0.6$ to $x=0.8$.

RK4 formulas:

Step 1 — From $(0.4, 0.41)$ to $(0.6, ?)$

$k_1=0.2\sqrt{0.4+0.41}=0.2\sqrt{0.81}=0.2(0.9)=0.18000$.

$k_2=0.2\sqrt{0.5+(0.41+0.09)}=0.2\sqrt{1.0}=0.2(1.0)=0.20000$.

$k_3=0.2\sqrt{0.5+(0.41+0.10)}=0.2\sqrt{1.01}\approx 0.2(1.00499)=0.20100$.

$k_4=0.2\sqrt{0.6+(0.41+0.20100)}=0.2\sqrt{1.21100}\approx 0.2(1.10045)=0.22009$.

Step 2 — From $(0.6, 0.61035)$ to $(0.8, ?)$

$k_1=0.2\sqrt{0.6+0.61035}=0.2\sqrt{1.21035}\approx 0.2(1.10016)=0.22003$.

$k_2=0.2\sqrt{0.7+(0.61035+0.11001)}=0.2\sqrt{1.42036}\approx 0.2(1.19178)=0.23836$.

$k_3=0.2\sqrt{0.7+(0.61035+0.11918)}=0.2\sqrt{1.42953}\approx 0.2(1.19562)=0.23912$.

$k_4=0.2\sqrt{0.8+(0.61035+0.23912)}=0.2\sqrt{1.64947}\approx 0.2(1.28432)=0.25686$.

Answer