Question

Using Euler’s modified method, determine the value of y when x = 0.1, given

that y’ = x² + y², y(0) = 1, taking h = 0.05.​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given:

Differential equation

$y'=x^2+y^2$

with initial condition $y(0)=1$.

To FInd:

$y(0.1)$ using Euler's modified method.

Solution:

According to Euler's modified method,

        $y_{n+1}^*=y_n+hf(x_n,y_n)$

and  $y_{n+1}=y_n+\frac{h}{2} [f(x_n,y_n)+f(x_{n+1},y_{n+1}^*)]$,

where, $f(x,y)=y'(x,y)$ and $h$ is the step size

Here we have to take the step size $h=0.05$

So in the first step, we will find the value of $y(0.05)$ and in the next step we will find $y(0.1)$

Step-1:

$y(0.05)^*=y(0)+0.05*(0^2+1^2)$

             $=1.05$

$y(0.05)=y(0)+\frac{0.05}{2} [(0^2+1^2)+(0.05^2+1.05^2)]$

           $=1.052625$

So, we obtained $y_{n+1}=y(0.05)=1.052625$ in this step.

Step-2:

$y(0.1)^*=y(0.05)+0.05*(0.05^2+1.052625^2)$

           $=1.10815097$

$y(0.1)=y(0.05)+\frac{0.05}{2} [(0.05^2+1.052625^2)+(0.1^2+1.10815097^2)]$

         $=1.11133795$

So, we obtained $y_{n+1}=y(0.1)=1.11133795$ in this step.

Hence the value of $y$ at $x=0.1$ is found using Euler's modified method which is equal to 1.11133795.