Find the general solution of $4y′′+ 3y′−y=e^{−x}+x$

I do know that my $G(x) = \color{blue}{e^{-x}} + \color{purple}{x}$ but am not too sure how to find $y_p$

Based on the form of $G(x)$ and by linearity you would propose a particular solution of the form: $$y_p=\color{blue}{Ae^{-x}} + \color{purple}{Bx+C}$$ but because $\color{blue}{Ae^{-x}}$ is already contained in the homogeneous solution ($ y_h = Ae^{0.25x} + \color{red}{Be^{-x}}$), you multiply with an extra factor $x$: $$y_p=\color{blue}{A}\color{red}{x}\color{blue}{e^{-x}} + \color{purple}{Bx+C}$$ Now substitute into the differential equation to obtain a linear system in the unknowns ("undetermined coefficients") $A$, $B$ and $C$.

Tags:

Calculus

Ordinary Differential Equations

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