Answer to Question #284739 in Quantitative Methods for Vaibhav

Question

Given that y'=x+y with y(0) = 1 find y(0.1).h=0.05 by Euler's modified method.

Answer 1

Solution:

Given $y^{\prime}=x+y, y(0)=1, h=0.05, y(0.1)=?$

Here, $x_{0}=0, y_{0}=1, h=0.05, x_{n}=0.1$

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

Modified Euler method

$\begin{aligned} &y_{m+1}=y_{m}+h f\left(x_{m}+\frac{1}{2} h, y_{m}+\frac{1}{2} h f\left(x_{m}, y_{m}\right)\right) \\ &f\left(x_{0}, y_{0}\right)=f(0,1)=1 \\ &\begin{array}{l} 1 \\ x_{0}+\frac{1}{2} h=0+\frac{0.05}{2}=0.025 \\ 1 \\ y_{0}+\frac{1}{2} h f\left(x_{0}, y_{0}\right)=1+\frac{0.05}{2} \cdot 1=1.025 \\ f\left(x_{0}+\frac{1}{2} h, y_{0}+\frac{1}{2} h f\left(x_{0}, y_{0}\right)=f(0.025,1.025)=1.05\right. \\ \therefore y(0.05)=1.0525 \end{array} \end{aligned}$

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

$\begin{aligned} &f\left(x_{1}, y_{1}\right)=f(0.05,1.0525)=1.1025 \\ &x_{1}+\frac{1}{2} h=0.05+\frac{0.05}{2}=0.075 \\ &y_{1}+\frac{1}{2} h f\left(x_{1}, y_{1}\right)=1.0525+\frac{0.05}{2} \cdot 1.1025=1.0801 \\ &f\left(x_{1}+\frac{1}{2} h, y_{1}+\frac{1}{2} h f\left(x_{1}, y_{1}\right)=f(0.075,1.0801)=1.1551\right. \\ &y_{2}=y_{1}+h f\left(x_{1}+\frac{1}{2} h, y_{1}+\frac{1}{2} h f\left(x_{1}, y_{1}\right)\right)=1.0525+0.05 \cdot 1.1551=1.1103 \\ &\therefore y(0.1)=1.1103 \end{aligned}$