Answer to Question #203366 in Differential Equations for manu

Question

(D+3)^2y=sinh2x

Answer 1

The auxiliary equation to the homogenous part of the given differential equation is given as:

$$(D+3)^2=0\\$$

Thus, we proceed to get the root of the equation:

$$(D+3)(D+3)=0\\ D=-3 \ (twice)$$

$$\therefore C \cdot F .=\left(C_{1}+C_{2} x\right) e^{-3 x}\\ \text{}\\ \operatorname{Sinh} x=\frac{e^{x}-e^{-x}}{2}\\ \text{}\\ \text{Particular Integral}\\ P.I. =\frac{\operatorname{Sinh} 2 x}{(D+3)^{2}}\\ =\frac{1}{2}\left[\frac{e^{2 x}-e^{-2 x}}{(D+3)^{2}}\right]\\ =\frac{1}{2} \frac{e^{2 x}}{(D+3)^{2}}-\frac{1}{2} \frac{e^{-2 x}}{(D+3)^{2}}\\ =\frac{1}{2} \frac{e^{2 x}}{(2+3)^{2}}-\frac{1}{2} \frac{e^{-2 x}}{(-2+3)^{2}}\\ =\frac{e^{2 x}}{50}-\frac{e^{-2 x}}{2}\\ \text{}\\ \text{Hence the complete solution}\\ Y=C . F .+P . I\\ Y=\left(C_{1}+C_{2} x\right) e^{-3 x}+\frac{e^{2 x}}{50}-\frac{e^{-2 x}}{2}\\$$