Question

1 point) 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.

Answer 1

c=-0.98

Explanation:

We know that the function

`y=x^2+cx^2`

is solution of the differential equation

`xy'+2y=4x^2`

to find the value of c we use the condition y(10)=2

`y(10)=(10)^2+c(10)^2=2\\y(10)=100+100c=2\\c=(2-100)/(100)=-0.98`

then, c = -0.98

hope this helps!!