Question

Y''-2y'+y=2(e^x)-3(e^-x)

Answer 1

First find the characteristic solution. The characteristic equation is


`r^2-2r+1=(r-1)^2=0`

which as one root at
`r=1` of multiplicity 2. This means the characteristic solution for this ODE is


`y_c=C_1e^x+C_2xe^x`

For the nonhomogeneous part, you can try a particular solution of the form


`y_p=(a_2x^2+a_1x+a_0)e^x+be^(-x)`

which has derivatives


`{y_p}'=(a_2x^2+(2a_2+a_1)x+a_1+a_0)e^x-be^(-x)`

`{y_p}''=(a_2x^2+(4a_2+a_1)x+2a_2+2a_1+a_0)e^x+be^(-x)`

Substituting into the ODE, the left hand side reduces significantly to


`2a_2e^x+4be^(-x)=2e^x-3e^(-x)`

and it follows that


`\begin{cases}2a_2=2\\4b=-3\end{cases}\implies a_2=1,b=-\frac34`

Therefore the particular solution is


`y_p=x^2e^x-\frac34e^(-x)`

and so the general solution is the sum of the characteristic and particular solutions,


`y=y_c+y_p`

`y=C_1e^x+C_2xe^x+x^2e^x-\frac34e^(-x)`