Question #203366

(D+3)^2y=sinh2x


1
Expert's answer
2021-06-14T17:44:19-0400

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


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

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


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

CF.=(C1+C2x)e3xSinhx=exex2Particular IntegralP.I.=Sinh2x(D+3)2=12[e2xe2x(D+3)2]=12e2x(D+3)212e2x(D+3)2=12e2x(2+3)212e2x(2+3)2=e2x50e2x2Hence the complete solutionY=C.F.+P.IY=(C1+C2x)e3x+e2x50e2x2\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}\\


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