Question

3y''-6y'+6y=e*x sexcx

Answer 1

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is


`3r^2-6r+6=0\iff r^2-2r+2=0`

which has roots at
`r=1\pm i`. This admits the two fundamental solutions


`y_1=e^x\cos x`

`y_2=e^x\sin x`

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form


`y_p=u_1y_1+u_2y_2`

where


`u_1=-\displaystyle\frac13\int(y_2e^x\sec x)/(W(y_1,y_2))\,\mathrm dx`

`u_2=\displaystyle\frac13\int(y_1e^x\sec x)/(W(y_1,y_2))\,\mathrm dx`

and
`W(y_1,y_2)` is the Wronskian of the fundamental solutions. We have


`W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^(2x)`

and so


`u_1=-\displaystyle\frac13\int(e^(2x)\sin x\sec x)/(e^(2x))\,\mathrm dx=-\int\tan x\,\mathrm dx`

`u_1=\frac13\ln|\cos x|`


`u_2=\displaystyle\frac13\int(e^(2x)\cos x\sec x)/(e^(2x))\,\mathrm dx=\int\mathrm dx`

`u_2=\frac13x`

Therefore the particular solution is


`y_p=\frac13e^x\cos x\ln|\cos x|+\frac13xe^x\sin x`

so that the general solution to the ODE is


`y=C_1e^x\cos x+C_2e^x\sin x+\frac13e^x\cos x\ln|\cos x|+\frac13xe^x\sin x`