ODE Equation none homogeneous linear differential equation of 2nd order

Exercise:

Solve the following D.E $y''-4y'+3y=1-\sin 3x$

Solution: consider the homogenous D.E $y''-4y'+3y=0$

Put \(y={{e}^{rx}}\Leftrightarrow y'=r{{e}^{rx}}\,\,\And \,\,y''={{r}^{2}}{{e}^{rx}}\)

$\Rightarrow y''-4y'+3y={{r}^{2}}{{e}^{rx}}-4r{{e}^{rx}}+3{{e}^{rx}}={{e}^{rx}}\left( {{r}^{2}}-4r+3 \right)=0$

But ${{e}^{rx}}>0$ so the characteristic equation will be ${{r}^{2}}-4r+3=0$

Hence the roots are $r=1\,\,\And \,\,r=3$

So the homogenous solution will be ${{y}_{H}}={{c}_{1}}{{e}^{x}}+{{c}_{2}}{{e}^{3x}}$

Now the form of the particular solution will be ${{y}_{P}}=A+B\cos 3x+C\sin 3x$

So $y{{'}_{p}}=-3B\sin 3x+3C\cos 3x$ and $y'{{'}_{p}}=-9B\cos 3x-9C\sin 3x$

Hence

$-9B\cos 3x-9C\sin 3x+12B\sin 3x-12C\cos 3x+3A+3B\cos 3x+3C\sin 3x=1-\sin 3x$

$\Rightarrow \cos 3x\left( -9B-12C+3B \right)+\sin 3x\left( -9C+12B+3C \right)+3A=1-\sin 3x$

By compression we get :

$-6B-12C=0\,\,,\,\,-6C+12B=-1\,\,\And \,\,3A=1$

Thus $A=\frac{1}{3}\,\,,\,\,\,B=\frac{-1}{15}\,\,\And \,\,C=\frac{1}{30}$

So $y\left( x \right)={{y}_{H}}+{{y}_{P}}={{c}_{1}}{{e}^{x}}+{{c}_{2}}{{e}^{3x}}+\frac{1}{3}-\frac{1}{15}\cos 3x+\frac{1}{30}\sin 3x$