Q. Solve the y' = x²+ y² , Given that y = 0 for x = 0 . Find the value of y(0.2) up to second approximation by using Picard Method.

Solution:

Given differential equation is,

y' = x²+ y² , 

Here \(x_1 = x_0 + h \) 

We have to do it up to second approximation so h = 0.1 

so, \(x_1 = x_0 +h  = 0 + 0.1 = 0.1\) 

By using Picard Method,

 \(y(x_{i+1}) = y(x_i)+\int_{x_i}^{x_{i+1}}f(x,y(x_i))dx\)

or, \(y(x_1) = y(x_0)+\int_{x_0}^{x_1}f(x,y(x_0))dx\)

or, \(y(x_1) = [(x^2 + 0^2)]_{x_0}^{x_1}\)

here \(x_0 = 0 \) and \(x_1 = 0.1 \) 

so, \(y(x_1) = [\frac{x^3}{3}]_0^{0.1} \)

finally, \(y(x_1) = \frac{0.1^3}{3} = 0.0003\)


Now for the Second Iteration, 

\(x_2 = 0.1+0.1 = 0.2\) ,  \(x_1 = 0.1\)  and \(y(x_1) = 0.0003\)

\(y(x_2) = y(x_1)+\int_{x_1}^{x_2}f(x,y(x_1))dx\)

or, \(y(x_2) = 0.0003+\int_{0.1}^{0.2}f(x,0.0003)dx\)

or, \(y(x_2) = 0.0003+[x^2 + 0.0003^2]_{0.1}^{0.2}\)

or , \(y(x_2) = 0.002633\)

which is the required answer.