Answer to Question #301670 in Quantitative Methods for Sayan Hazra

Question

If dy/dx = y − x, y(0) = 1 2 . Use Modified Euler’s method with h = 0.1 to obtain an approximation to y(0.2).

Answer 1

Solution:

Given $y^{\prime}=y-x, y(0)=0.5, h=0.1, y(0.2)=?$

Here, $x_{0}=0, y_{0}=0.5, h=0.1, x_{n}=0.2$

$\begin{aligned} &y^{\prime}=y-x \\ &\therefore f(x, y)=y-x \end{aligned}$

Modified Euler method

$y_{m+1}=y_{m}+h f\left(x_{m}+\frac{1}{2} h, y_{m}+\frac{1}{2} h f\left(x_{m}, y_{m}\right)\right) \\x_{0}+\frac{1}{2} h=0+\frac{0.1}{2}=0.05 \\y_{0}+\frac{1}{2} h f\left(x_{0}, y_{0}\right)=0.5+\frac{0.1}{2} \cdot 0.5=0.525 \\f\left(x_{0}+\frac{1}{2} h, y_{0}+\frac{1}{2} h f\left(x_{0}, y_{0}\right)=f(0.05,0.525)=0.475\right.$

$\begin{aligned} &y_{1}=y_{0}+h f\left(x_{0}+\frac{1}{2} h, y_{0}+\frac{1}{2} h f\left(x_{0}, y_{0}\right)\right)=0.5+0.1 \cdot 0.475=0.5475 \\ &\therefore y(0.1)=0.5475 \end{aligned}$

Again taking $\left(x_{1}, y_{1}\right)$  in place of $\left(x_{0}, y_{0}\right)$  and repeat the process

$\begin{aligned} &f\left(x_{1}, y_{1}\right)=f(0.1,0.5475)=0.4475 \\ &x_{1}+\frac{1}{2} h=0.1+\frac{0.1}{2}=0.15 \\ &y_{1}+\frac{1}{2} h f\left(x_{1}, y_{1}\right)=0.5475+\frac{0.1}{2} \cdot 0.4475=0.5699 \\ &f\left(x_{1}+\frac{1}{2} h, y_{1}+\frac{1}{2} h f\left(x_{1}, y_{1}\right)=f(0.15,0.5699)=0.4199\right. \\ &y_{2}=y_{1}+h f\left(x_{1}+\frac{1}{2} h, y_{1}+\frac{1}{2} h f\left(x_{1}, y_{1}\right)\right)=0.5475+0.1 \cdot 0.4199=0.5895 \\ &\therefore y(0.2)=0.5895 \end{aligned}$