Answer to Question #300553 in Quantitative Methods for Shobika

Question

Apply Runge-Kutta method of 4th order method to find the approximate

value of 𝑦 for 𝑥 = 0.1, if 𝑑𝑦

𝑑𝑥

= 𝑥 + 𝑦

2 given that 𝑦 = 1 where 𝑥 = 0.

Answer 1

Solution:

$\text { Let } f(x, y) : d y / d x=x+y^{2}$

Part I: $h=0.1 x_{0}=0\ and\ y+0=1$  

By Runge kutta method of 4th order. 

$K_{1}=h f\left(x_{0}, y_{0}\right)=0.1 \times f(0,1) \\=0.1 \times\left(0+1^{2}\right)=0.1 \\K_{2}=h f\left(x_{0}+\frac{h}{2}, y_{0}+\frac{k_{1}}{2} \\=0.1 f\left(0+\frac{0.1}{2}, \frac{1+0.1}{2}\right)\right. 0.1 \times\left(\frac{0.1}{2}+\left(\frac{2.1}{2}\right)^{2}\right) \\=0.1 \times\left(0.05+(1.05)^{2}\right) \\=0.1153$

$K_{3}=h f\left(x_{0}+\frac{h}{2}, y_{0}+\frac{k_{2}}{2}\right) \\=0.1 \times f\left(0+\frac{0.1}{2}, 1+\frac{0.1153}{2}\right) \\=0.1 \times\left[\frac{0.1}{2}+\left(\frac{2.1153}{2}\right)^{2}\right] \\=0.1169 K_{4}=h f\left(x_{0}+h, y_{0}+k_{3}\right) \\=0.1 \times f(0+0.1,1+0.1169) \\=0.1\left[0.1+(1.1169)^{2}\right] =0.1347$  

$\therefore K=\frac{1}{6}\left(K_{1}+2 K_{2}+2 K_{3}+K_{4}\right) \\ =\frac{1}{6}(0.1+0.1153 \times 2+0.1169 \times 2+0.1347) \\=0.1165$

$\therefore y=y_{0}+k=1+0.1165=1.1165$

$\therefore$  The correct value of y when x=0.1 is 1.1165

Part 2:

$\begin{aligned} &h=0.1, x=0.1 \& y_{1}=1.1165 \\ &k_{1}=h f\left(x_{1}, y_{1}\right)=0.1 f(0.1,1.1105) \\ &=0.1\left[0.1+(1.1165)^{2}\right] \\ &=0.1347 \\ &K_{2}=h f\left(x_{1}+\frac{h}{2}, y_{1}+\frac{k_{1}}{2}\right) \\ &=0.1 \times f\left(0.1+\frac{0.1}{2}, 1.1165+\frac{0.1347}{2}\right) \\ &=0.1\left[0.15+(1.1838)^{2}\right] \\ &0.1551 \\ &K_{3}=h f\left(x_{1}+\frac{h}{2}, y_{1}+\frac{k_{2}}{2}\right) \\ &=0.1 \times f\left(0.1+\frac{0.1}{2}, 1.1165+\frac{0.1551}{2}\right) \\ &=0.1 \times\left[0.15+(1.1941)^{2}\right] \\ &=0.1576 \\ &K_{4}=h f\left(x_{1}+h_{1}, y_{1}+k_{3}\right) \\ &=0.1 f(0.1+0.1,1.1165+0.1576) \\ &=0.1\left[0.2+(1.2741)^{2}\right] \\ &=0.1823 \end{aligned}$

$\begin{aligned} &K=\frac{1}{6}\left(k_{1}+2 k_{2}+2 k_{3}+k_{4}\right) \\ &=\frac{1}{6}(0.1347+2 \times 0.1551+2 \times 0.1576+0.1823) \\ &=0.1751 \\ &\therefore y=y_{1}+k=1.1165+0.1571 \\ &=1.2736 \end{aligned}$

$K=\frac{1}{6}\left(k_{1}+2 k_{2}+2 k_{3}+k_{4}\right) \\=\frac{1}{6}(0.1347+2 \times 0.1551+2 \times 0.1576+0.1 \\=0.1751$

 $\therefore y=y_{1}+k=1.1165+0.1571 =1.2736$   

$\therefore$  The correct value of y when x=0.2 is 1.2736