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Question Number 126683 by bramlexs22 last updated on 23/Dec/20

$$solve(D^2+a)y=\tan ax$$

Answered by liberty last updated on 23/Dec/20

$$\left(1\right)Homogenoussolution$$$$y_h=A\cos ax+B\sin ax$$$$\left(2\right)Particularsolution$$$$y_1=\cos ax\to y_1^{\prime }=-a\sin ax$$$$y_2=\sin ax\to y_2^{\prime }=a\cos ax$$$$WronskianW(y_1,y_2)=\left|\begin{array}{c}\cos ax\sin ax\\ -a\sin axa\cos ax\end{array}\right|=a$$$$putu=-\int \frac{y_{2}g\left(x\right)}{W}dx=-\int \frac{\sin ax\tan ax}{a}dx$$$$=\frac{1}{a^2}(\sin ax-\ln \mid \sec ax+\tan ax\mid )$$$$putv=\int \frac{y_{1}g\left(x\right)}{W}dx=\int \frac{\cos ax\tan ax}{a}dx$$$$=-\frac{\cos ax}{a^2}$$$$y_p={uy}_1+{vy}_2=-\frac{1}{a^2}\cos ax\ln \mid \sec ax+\tan ax\mid$$$$\left(3\right)y_g=A\cos ax+B\sin ax-\frac{\cos ax\ln \mid \sec ax+\tan ax\mid }{a^2}$$$$$$

Answered by mathmax by abdo last updated on 24/Dec/20

$${\mathrm{y}}^{{}^{″}}+\mathrm{ay}=\mathrm{tanx}$$$$\mathrm{h}\to {\mathrm{r}}^2+\mathrm{a}=0$$$$\mathrm{case}1\mathrm{a}>0\Rightarrow {\mathrm{r}}^2=-\mathrm{a}\Rightarrow \mathrm{r}=\stackrel{-}{+}\sqrt{-\mathrm{a}}=\stackrel{-}{+}\mathrm{i}\sqrt{\mathrm{a}}\Rightarrow$$$${\mathrm{y}}_{\mathrm{h}}={\mathrm{ae}}^{\mathrm{i}\sqrt{\mathrm{a}}\mathrm{x}}+{\mathrm{be}}^{-\mathrm{i}\sqrt{\mathrm{a}}\mathrm{x}}=\alpha \cos \left(\sqrt{\mathrm{a}}\mathrm{x}\right)+\beta \sin \left(\sqrt{\mathrm{a}}\mathrm{x}\right)=\alpha {\mathrm{u}}_1+\beta {\mathrm{u}}_2$$$$\mathrm{w}({\mathrm{u}}_1,{\mathrm{u}}_2)=\left|\begin{array}{c}\cos \left(\sqrt{\mathrm{a}}\mathrm{x}\right)\sin \left(\sqrt{\mathrm{a}}\mathrm{x}\right)\\ -\sqrt{\mathrm{a}}\sin \left(\sqrt{\mathrm{a}}\mathrm{x}\right)\sqrt{\mathrm{a}}\cos \left(\sqrt{\mathrm{a}}\mathrm{x}\right)\end{array}\right|=\sqrt{\mathrm{a}}$$$${\mathrm{w}}_1=\left|\begin{array}{c}\mathrm{o}\sin \left(\sqrt{\mathrm{a}}\mathrm{x}\right)\\ \tan \left(\mathrm{ax}\right)\sqrt{\mathrm{a}}\cos \left(\sqrt{\mathrm{a}}\mathrm{x}\right)\end{array}\right|=-\sin \left(\sqrt{\mathrm{a}}\mathrm{x}\right)\tan \left(\mathrm{ax}\right))$$$${\mathrm{w}}_2=\left|\begin{array}{c}\cos \left(\sqrt{\mathrm{a}}\mathrm{x}\right)0\\ -\sqrt{\mathrm{a}}\sin \left(\sqrt{\mathrm{a}}\mathrm{x}\right)\tan \left(\mathrm{ax}\right)\end{array}\right|=\cos \left(\sqrt{\mathrm{a}}\mathrm{x}\right)\tan \left(\mathrm{ax}\right)$$$${\mathrm{v}}_1=\int \frac{{\mathrm{w}}_1}{\mathrm{w}}\mathrm{dx}=-\int \frac{\sin \left(\sqrt{\mathrm{a}}\mathrm{x}\right)\tan \left(\mathrm{ax}\right)}{\sqrt{\mathrm{a}}}\mathrm{dx}=-\frac{1}{\sqrt{\mathrm{a}}}\int \sin \left(\sqrt{\mathrm{a}}\mathrm{x}\right)\tan \left(\mathrm{ax}\right)\mathrm{dx}$$$${\mathrm{v}}_2=\int \frac{{\mathrm{w}}_2}{\mathrm{w}}\mathrm{dx}=\int \frac{\cos \left(\sqrt{\mathrm{a}}\mathrm{x}\right)\tan \left(\mathrm{ax}\right)}{\sqrt{\mathrm{a}}}\mathrm{dx}$$$$\Rightarrow {\mathrm{y}}_{\mathrm{p}}={\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2=-\frac{1}{\sqrt{\mathrm{a}}}\cos \left(\sqrt{\mathrm{a}}\mathrm{x}\right)\int \sin \left(\sqrt{\mathrm{a}}\mathrm{x}\right)\tan \left(\mathrm{ax}\right)\mathrm{dx}$$$$+\frac{1}{\sqrt{\mathrm{a}}}\sin \left(\sqrt{\mathrm{a}}\mathrm{x}\right)\int \cos \left(\sqrt{\mathrm{a}}\mathrm{x}\right)\tan \left(\mathrm{ax}\right)\mathrm{dx}\Rightarrow$$$$\mathrm{general}\mathrm{solution}\mathrm{is}\mathrm{y}={\mathrm{y}}_{\mathrm{p}}+{\mathrm{h}}_{\mathrm{h}}$$$$\mathrm{case}2<0{\mathrm{r}}^2+\mathrm{a}=0\Rightarrow {\mathrm{r}}^2=-\mathrm{a}\Rightarrow \mathrm{r}=\stackrel{-}{+}\sqrt{\mathrm{a}}\Rightarrow$$$${\mathrm{y}}_{\mathrm{h}}=\mathrm{a}{\mathrm{e}}^{\sqrt{\mathrm{a}}\mathrm{x}}+{\mathrm{be}}^{-\sqrt{\mathrm{a}}\mathrm{x}}={\mathrm{au}}_1+{\mathrm{bu}}_2\mathrm{and}\mathrm{we}\mathrm{follow}\mathrm{the}\mathrm{same}\mathrm{way}...$$

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