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Question Number 117602 by TANMAY PANACEA last updated on 12/Oct/20

$$\frac{d^{2}y}{{dx}^2}+a^{2}y=cosax$$

Answered by TANMAY PANACEA last updated on 12/Oct/20

$$y=e^{mx}$$$$m^2+a^2=0\to m=\pm ia$$$$C.F=c_1{e}^{iax}+c_2{e}^{-iax}$$$$P.I=\frac{cosax}{D^2+a^2}=\frac{e^{iax}}{{(D+ia)}^2+a^2}=\frac{e^{iax}}{1}\times \frac{1}{D^2+2iaD}$$$$=e^{iax}\times \frac{1}{D+i2a}\times \frac{1}{D}$$$$=\frac{e^{iax}}{1}\times \frac{1}{2ia}\times {(1+\frac{D}{2ia})}^{-1}\times x$$$$=\frac{e^{iax}}{2ia}\times (1-\frac{D}{2ia})x$$$$=\frac{cosax+isinax}{2ia}\times (x-\frac{1}{2ia})$$$$=\frac{cosax+isinax(2iax-1)}{-4a^2}$$$$=\frac{i2axcosax-cosax-2axsinax+isinax}{-4a^2}$$$$=\frac{cosax+2axsinax}{4a^2}+\frac{i2axcosax+isinax}{-4a^2}$$$$realpart$$$$=\frac{cosax}{4a^2}+\frac{xsinax}{2a}$$

Answered by mathmax by abdo last updated on 12/Oct/20

$${\mathrm{y}}^{{}^{″}}+{\mathrm{a}}^2\mathrm{y}=\cos \left(\mathrm{ax}\right)$$$$\mathrm{h}\to {\mathrm{r}}^2+{\mathrm{a}}^2=0\Rightarrow \mathrm{r}=\stackrel{-}{+}\mathrm{ia}\Rightarrow {\mathrm{y}}_{\mathrm{h}}={\mathrm{me}}^{\mathrm{iax}}+\mathrm{n}{\mathrm{e}}^{-\mathrm{iax}}$$$$=\alpha \cos \left(\mathrm{ax}\right)+\beta \sin \left(\mathrm{ax}\right)=\alpha {\mathrm{u}}_1\left(\mathrm{x}\right)+\beta {\mathrm{u}}_2\left(\mathrm{x}\right)$$$$\mathrm{W}({\mathrm{u}}_1,{\mathrm{u}}_2)=\left|\begin{array}{c}\cos \left(\mathrm{ax}\right)\sin \left(\mathrm{ax}\right)\\ -\mathrm{asin}\left(\mathrm{ax}\right)\mathrm{acos}\left(\mathrm{ax}\right)\end{array}\right|=\mathrm{a}\mathrm{we}\mathrm{supoose}\mathrm{a}\ne 0$$$${\mathrm{W}}_1=\left|\begin{array}{c}\mathrm{o}\sin \left(\mathrm{ax}\right)\\ \cos \left(\mathrm{ax}\right)\mathrm{acos}\left(\mathrm{ax}\right)\end{array}\right|=-\cos \left(\mathrm{ax}\right)\sin \left(\mathrm{ax}\right)$$$${\mathrm{W}}_2=\left|\begin{array}{c}\cos \left(\mathrm{ax}\right)0\\ -\mathrm{asin}\left(\mathrm{ax}\right)\cos \left(\mathrm{ax}\right)\end{array}\right|={\cos }^2\left(\mathrm{ax}\right)$$$${\mathrm{v}}_1=\int \frac{{\mathrm{W}}_1}{\mathrm{W}}\mathrm{dx}=-\int \frac{\cos \left(\mathrm{ax}\right)\sin \left(\mathrm{ax}\right)}{\mathrm{a}}=-\frac{1}{2\mathrm{a}}\int \sin \left(2\mathrm{ax}\right)\mathrm{dx}$$$$=\frac{1}{{4\mathrm{a}}^2}\cos \left(2\mathrm{ax}\right)$$$${\mathrm{v}}_2=\int \frac{{\mathrm{W}}_2}{\mathrm{W}}\mathrm{dx}=\int \frac{{\cos }^2\left(\mathrm{ax}\right)}{\mathrm{a}}\mathrm{dx}=\frac{1}{2\mathrm{a}}\int (1+\cos \left(2\mathrm{ax}\right))\mathrm{dx}$$$$=\frac{\mathrm{x}}{2\mathrm{a}}+\frac{1}{{4\mathrm{a}}^2}\sin \left(2\mathrm{ax}\right)\Rightarrow$$$${\mathrm{y}}_{\mathrm{p}}={\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2=\cos \left(\mathrm{ax}\right)\left(\frac{1}{{4\mathrm{a}}^2}\cos \left(2\mathrm{ax}\right)\right)+\sin \left(\mathrm{ax}\right)(\frac{\mathrm{x}}{2\mathrm{a}}+\frac{1}{{4\mathrm{a}}^2}\sin \left(2\mathrm{ax}\right)$$$$=\frac{\cos \left(\mathrm{ax}\right)\cos \left(2\mathrm{ax}\right)}{{4\mathrm{a}}^2}+\frac{\mathrm{xsin}\left(\mathrm{ax}\right)}{2\mathrm{a}}+\frac{\sin \left(\mathrm{ax}\right)\sin \left(2\mathrm{ax}\right)}{{4\mathrm{a}}^2}$$$$\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{is}\mathrm{y}={\mathrm{y}}_{\mathrm{h}}+{\mathrm{y}}_{\mathrm{p}}$$

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