Question

Find the particular sol'n of y"(x) + y(x) = 2^x

Answer 1

I know I need to get the right side in the form\[=Ct^me^{rt}\]But I don't know how to do that

Answer 2

\[y_p=t^se^{rt}(A_mt^m+...+A_0t^0)\]

Answer 3

Anything coming to mind @hartnn

Answer 4

\[r^2+1=0\]\[r=\pm i\]

Answer 5

s=0

Answer 6

m=0

Answer 7

\[e^{rt}=2^x\]\[rt \ln e=\ln 2^x\]\[rt=\ln 2^x\]\[y''(x) + y(x) = e^{\ln 2^x}\]\[y_p=Ae^{\ln 2^x}\]\[y'_p=A (\ln 2) e^{\ln 2^x}\]\[y''_p=A(\ln 2)^2e^{\ln 2^x}\]\[A(\ln 2)^2e^{\ln 2^x} + Ae^{\ln 2^x}\ = e^{\ln 2^x}\]\[equate \space e^{\ln 2^x}:\]\[A(\ln 2)^2+A=1\]\[A=((\ln 2)^2+1)^{-1}\]\[y_p=((\ln 2)^2+1)^{-1}2^x\]

Answer 8

@hartnn in case you were curious I figured it out

Answer 9

nice!

Answer 10

Thanks. It was easy once I figured out how to convert the right side