Question

The differential equation $\frac{dy}{dx}=0.25y^2$ is to be solved using the backward (implicit) Euler's method with the boundary condition y = 1 at x = 0 and with astep size of 1. What would be the value of y at x = 1?

• A. 1.33
• B. 1.67
• C. 2.00
• D. 2.33

Answer 1

The correct option is C 2.00

Backward (implicit) Euler's method :
$y_{n+1}=y_n+hf(x_{n+1},y_{n+1})$
$\Rightarrow f(x_{n+1},y_{n+1})=\frac{y_{n+1}-y_n}{h}$ ...(i)

Given, $\frac{dy}{dx}=0.25y^2=f(x,y)$ (let)
and h = 1
so, by (i)
$0.25y_{n+1}^2=\frac{y_{n+1}-y_n}{1}$
$\Rightarrow 0.25y_{n+1}^2-y_{n+1}+y_n=0$
Putting, $n=0$ & $y_0=1$
$0.25y_1^2-y_1+y_0=0$
$\frac{y_1^2}{4}-y_1+1=0$
$\Rightarrow$ $(y_1-2{)}^2=0$
$\Rightarrow$ $y_1=2$
$\Rightarrow$ $y\left(1\right)=2$