Question

Differential Equations Undetermined Coefficients
y''(x)+y(x)=2^x

Answer 1

would my trial solution be Yp=A^x?

Answer 2

Y'p=xA^x-1

Answer 3

I don't think what im doing is right....

Answer 4

or A^x+B?

Answer 5

Unfortunately I don't have much time to check into this problem, although it seems very interesting.

What you're doing seems reasonable to me, I would also start with your trial solution to be:
\[\Large y_p=A^x \]
if that wont work try adding factors and constants to it, such that:
\[\Large y_p=A^{cx}+(b) \]
although I believe the constant wont do much, however make sure that you differentiate your trial right:

\[\Large y=2^x=\left(e^{\ln 2}\right)^x=e^{ \ln 2 x} \]

Answer 6

yeah I thought that adding constants wouldn't do much since when we derive they go away...

Answer 7

So if you differentiate your trial you should get something like:
\[\Large y=A^x=e^{\ln Ax} \]
differentiation will lead to:
\[\Large y'= \ln Ae^{\ln Ax}=\ln A \cdot A^x \]

I believe this is where you went wrong in your differentiation.

Answer 8

yeah that's what maple gave me ill try working it again

Answer 9

sorry guys got physics class....thanks for help though!