It is easy to check that for any value of c, the function y=x2+cx2 is solution of equation xy′+2y=4x2, (x>0). Find the value of c for which the solution satisfies the initial condition y(10)=2.

Therefore is a solution of .

Step-by-step explanation:

The given differential equation is

Here  and Q= 4x

The integrating factor is

Multiplying the integrating factor both sides of the DE

Integrating both sides

         [ c is an arbitrary constant]

            [ dividing by x²]

Therefore is a solution of .

The initial condition is y(10)=2

Putting x=10 and y=2

  [ multiply by 100]

Putting the value of c in the solution we get,