It is easy to check that for any value of c, function y=x^2+c/x^2 is solution of equation xy'+2y=4x^2, (x0).

Find the value of c for which the solution satisfies the initial condition
y(6)=2.

Question

It is easy to check that for any value of c, function y=x^2+c/x^2 is solution of equation xy'+2y=4x^2, (x>0).

Find the value of c for which the solution satisfies the initial condition
y(6)=2.

anonymous · Answer

this is a calculus problem . i just don't know how to go by solving it while putting the initial conditions in

anonymous · Answer

take the derivative and you solve for y', take the y' to the second

anonymous · Answer

it seems not right

anonymous · Answer

is this for the 2nd equation only or both

anonymous · Answer

what seems not right?! im confused

anonymous · Answer

maybe: 6^2+c/(6^2)=2   c=-1224

anonymous · Answer

yeah that is the right answer