Question: Apply Runge-kutta Method of fourth order to find an approximate value of y when x = 0.2 given that $\frac{dy}{dx} = x + y$ when y = 1 at x = 0 with step size h = 0.2
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Subject : Applied Mathematics 2

Topic : Differentiation under Integral sign, Numerical Integration and Rectification

Difficulty: High

mumbai university • 22 views
 modified 3 days ago  • written 4 days ago by Ankit Pandey ♦♦ 10
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Solution:

$\begin{array}{ll}{\frac{d y}{d x}=x+y} & {x_{0}=0, y_{0}=1, \mathrm{h}=0.2} \\ {f(x, y)=x+y}\end{array}$

$k_{1}=h . f\left(x_{0}, y_{0}\right)=0.2 \mathrm{f}(0,1)=0.2$ $k_{2}=\mathrm{h.f}\left(x_{0}+\frac{h}{2}, y_{0}+\frac{k_{1}}{2}\right)=0.2 \mathrm{f}(0.1,1.1)=0.24$

$k_{3}=h . f\left(x_{0}+\frac{h}{2}, y_{0}+\frac{k_{2}}{2}\right)=0.2 f(0.1,1.12)=0.244$ $k_{4}=h . f\left(x_{0}+h, y_{0}+k_{3}\right)=0.2 f(0.2,1.244)=0.2888$

$\mathrm{k}=\frac{k_{1}+2 k_{2}+2 k_{3}+k_{4}}{6}=\frac{0.24+0.48+0.488+0.2888}{6}=0.2428$

The value of $y$ at $x=0.2$ is given by, \begin{aligned} y(0.2) &=y_{0}+k=1+0.2428 =1.2428 \end{aligned}