How to solve $y'' -3y'+2y = \sin x$ with partial fraction decomposition?

Question

I have $$y''-3y'+2y = \sin x$$

I do not want to use Lagrange's method here (only partial fraction decomposition)

so $\sin x$ can be represented as $\sin x = \frac{e^{ix}-e^{-ix}}{2i}$

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solving second-ordered equation:

$$y'' -3y'+2y = 0$$

$$3\lambda^2-3\lambda+2 = 0$$

$$\lambda_1 =1; \lambda_2 = 2$$

$$y = C_1e^x+C_2e^{2x}$$

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What should I do next?

Answer 1

The only partial fraction decomposition I can imagine here is if you write your DE in the differential operator $D$ $$ (D^2-3D+2)y=\sin x\quad\Leftrightarrow\quad y=\frac{1}{D^2-3D+2}\sin x=\frac{1}{D-1}\cdot \underbrace{\frac{1}{D-2}\sin x}_{z} $$ and would like to solve two first order DE:

1. $z=\frac{1}{D-2}\sin x$ $\Leftrightarrow$ $z'-2z=\sin x$, and
2. $y=\frac{1}{D-1}z$ $\Leftrightarrow$ $y'-y=z$.

When you do the partial fraction decomposition $$ \frac{1}{(D-2)(D-1)}=\frac{1}{D-2}-\frac{1}{D-1} $$ then it is possible to solve instead $$ y=\underbrace{\frac{1}{D-2}\sin x}_{y_1}-\underbrace{\frac{1}{D-1}\sin x}_{y_2} $$ where $y_1'-2y_1=\sin x$ and $y_2'-y_2=\sin x$.