Series Solution to Differential Equation

[ChrisVer]

I have to solve the differential equation
[itex] y''+(1-t) y' + y= sin(2t)[/itex]

can someone judge this?
How could I continue it?

[itex] y=\sum_{n=0}^{∞}{a_{n} t^{n}}[/itex]

[itex] y'=\sum_{n=1}^{∞}{a_{n} n t^{n-1}}[/itex]

[itex] y''=\sum_{n=2}^{∞}{a_{n} n(n-1) t^{n-2}}[/itex]

[itex] sin(2t)=\sum_{n=0}^{∞}{\frac{2^{2n}}{2n!} t^{2n}}[/itex][itex] y''+(1-t) y' + y= sin(2t)[/itex]
[itex] \sum_{n=2}^{∞}{a_{n} n(n-1) t^{n-2}}+\sum_{n=1}^{∞}{a_{n} n (1-t)t^{n-1}}+\sum_{n=0}^{∞}{a_{n} t^{n}}=\sum_{n=0}^{∞}{\frac{2^{2n}}{2n!} t^{2n}}[/itex]

[itex] \sum_{n=2}^{∞}{a_{n} n(n-1) t^{n-2}}+\sum_{n=1}^{∞}{a_{n} n t^{n-1}}-\sum_{n=1}^{∞}{a_{n} n t^{n}}+\sum_{n=0}^{∞}{a_{n} t^{n}}=\sum_{n=0}^{∞}{\frac{2^{2n}}{2n!} t^{2n}}[/itex]

[itex] \sum_{k=0}^{∞}{a_{k+2} (k+2)(k+1) t^{k}}+\sum_{k=0}^{∞}{a_{k+1} (k+1) t^{k}}-\sum_{k=0}^{∞}{a_{k} k t^{k}}+\sum_{k=0}^{∞}{a_{k} t^{k}}=\sum_{k=0}^{∞}{\frac{2^{2k}}{2k!} t^{2k}}[/itex]

[itex] \sum_{k=0}^{∞}{(a_{k+2} (k+2)(k+1)+a_{k+1} (k+1) -a_{k} k + a_{k}) t^{k}}=\sum_{k=0}^{∞}{\frac{2^{2k}}{2k!} t^{2k}}[/itex]

[itex] \sum_{k=0}^{∞}{(a_{k+2} (k+2)(k+1)+a_{k+1} (k+1)+ (1-k) a_{k}) t^{k}}=\sum_{k=0}^{∞}{\frac{2^{2k}}{2k!} t^{2k}}[/itex]

[itex] \sum_{m=0}^{∞}{(a_{2m+2} (2m+2)(2m+1)+a_{2m+1} (2m+1)+ (1-2m) a_{2m}) t^{2m}}=\sum_{m=0}^{∞}{\frac{2^{2m}}{2m!} t^{2m}}[/itex][itex]a_{2m+2} (2m+2)(2m+1)+a_{2m+1} (2m+1)+ (1-2m) a_{2m}=\frac{2^{2m}}{2m!}[/itex]

 

[mfb]

Are you sure such a series approach is useful?

This allows to find all coefficients if you know starting values, but finding "nice" solutions based on this can be messy.
Your series for the sin looks wrong - it has even powers of t instead of odd.
Don't forget that you have equations both for even and odd powers of t.

 

[ChrisVer]

hahahaha yes, sorry about the sin, it was my mistake :p
As for the rest, I don't know, I've seen this procedure used in so many applications (from QM, to partial derivative problems eg in cylinder or spherical coords), so I thought of applying it... or maybe try the Fourier transform of [itex]y[/itex]?
yup I know, but because eg of the Fourier transform, I could kill the (corrected) even powers

 

[mfb]

The even powers will have =0 on the right-hand side, but this is still a non-trivial equality. The derivatives mix even and odd coefficients in those equations, you cannot use symmetry as argument to ignore them.

Power series * sin(2t) + Power series * cos(2t) could be an interesting approach. This could work with polynomials.