Q:

Find General Solution of $y''-2y'+y=\frac{3e^t}{1+t^2}+7$

Find the general solution of $y''-2y'+y=\frac{3e^t}{1+t^2}+7$
I know I would first have to find the solution to the homogeneous equation, then particular, then add the 2 for the actual solution. I know how to find the homogeneous but have no clue how to approach the particular. We learned about the method of undetermined coefficients but I don't know how to apply it to a rational function like this because there is no general method.

A:

This can be written as 
$$y''e^{-t}-y'e^{-t}-(y'e^{-t}-ye^{-t})=\frac 3{1+t^2} +7e^{-t}$$
$$\frac{d(y'e^{-t})}{dt}-\frac{d(ye^{-t})}{dt}= \frac {3}{1+t^2} + 7e^{-t}$$
$$y'e^{-t}-ye^{-t}=3 \arctan(t) - 7e^{-t} + c_1$$
$$\frac{d(ye^{-t})}{dt}=3 \arctan(t) -7e^{-t}+c_1$$
$$ye^{-t}=3t\cdot\arctan(t)-\frac32 \log (1+x^2)+7e^{-t}+c_1t+c_2$$

