I need help with a concept. Say I have
y''-2y'-3y=-3te^(-t) and I want to the general solution. I know we're supposed to find the homogeneous roots to get the characteristic equation, but I'm a bit stumped finding the particular solution. I'm not sure how to assume the form of the particular solution. I thought I'd take Yp=Ate^(-t) but it really should be Ae^(-t)+Bte^(-t)... WHY?

Question

I need help with a concept. Say I have 
y''-2y'-3y=-3te^(-t) and I want to the general solution. I know we're supposed to find the homogeneous roots to get the characteristic equation, but I'm a bit stumped finding the particular solution. I'm not sure how to assume the form of the particular solution. I thought I'd take Yp=Ate^(-t) but it really should be Ae^(-t)+Bte^(-t)... WHY?

zepdrix · Answer

The particular solution will be the product of the exponential part $\Large\rm Ae^{-t}$ with the polynomial part $\Large\rm (Bt+C)$.
When you have a polynomial part, you have to include all of the lesser degrees.

zepdrix · Answer

Combine the constants, call them something else,
$$\Large\rm Ae^{-t}(Bt+C)=\mathcal Ate^{-t}+\mathcal Be^{-t}$$