# Given the general solution to #t^2y'' - 4ty' + 4y = 0# is #y= c_1t + c_2t^4#, how do I solve the problem #t^2y'' - 4ty' + 4y = -2t^2 , y(1) = 2, y'(1) =0#?

##### 1 Answer

The answer is:

If it is known a solution of the homogenous equation, now a particular solution of the non-homogenous equation has to be find.

Since then in the second member there is

Now we calculate:

Now we have to find

Then, for the principle of the identity between polynomials:

So:

The solution of the non-homogeneous is:

and its derivative is:

Now we have to find

So:

We can substitute

And:

So, finally: