Question

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`?

Answer 1

The answer is: `y=-t^4+t^2+2t`.

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 `-2t^2`, for the similarity principle the particular solution is a square equation like this:

`beta(t)=at^2+bt+c`.

Now we calculate:

`beta'(t)=2at+b`, and

`beta''(t)=2a`.

Now we have to find `a,b,c`.

`beta(t)` has to satisfy the equation, so:

`t^2(2a)-4t(2at+b)+4at^2+4bt+4c=-2t^2`
`2at^2-8at^2-4bt+4at^2+4bt+4c=-2t^2`
`-2at^2+4c=-2t^2`

Then, for the principle of the identity between polynomials:

`-2a=-2rArra=1`, and
`4c=0rArrc=0`.
`b` is not important, so we can assume that is 0.

So:

`beta(t)=t^2`.

The solution of the non-homogeneous is:

`y=c_1t+c_2t^4+t^2`

and its derivative is:

`y'=c_1+4c_2t^3+2t`

Now we have to find `c_1` and `c_2` using the conditions:

`y(1)=2` and `y'(1)=0`.

`y'=c_1+4c_2t^3+2t`.

So:

`2=c_1+c_2+1rArrc_1+c_2=1` and

`0=c_1+4c_2+2rArrc_1+4c_2=-2`.

We can substitute `c_1=1-c_2` from the first equation to the second:

`1-c_2+4c_2=-2rArr3c_2=-3rArrc_2=-1`

And:

`c_1=1-(-1)rArrc_1=2`.

So, finally:

`y=-t^4+t^2+2t`.