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Question Number 111011 by mohammad17 last updated on 01/Sep/20

$$solve:y^{{}^{″}}+y^{{}^{\prime }}=tanx$$

Answered by mathmax by abdo last updated on 01/Sep/20

$$\mathrm{h})\to {\mathrm{r}}^2+\mathrm{r}=0\Rightarrow \mathrm{r}(\mathrm{r}+1)=0\Rightarrow \mathrm{r}=0\mathrm{or}\mathrm{r}=-1\Rightarrow \mathrm{y}\left(\mathrm{x}\right)={\mathrm{ae}}^{-\mathrm{x}}+\mathrm{b}$$$$={\mathrm{au}}_1+{\mathrm{bu}}_2$$$$\mathrm{W}({\mathrm{u}}_1,{\mathrm{u}}_2)=\left|\begin{array}{c}{\mathrm{e}}^{-\mathrm{x}}1\\ -{\mathrm{e}}^{-\mathrm{x}}0\end{array}\right|={\mathrm{e}}^{-\mathrm{x}}\ne 0$$$${\mathrm{W}}_1=\left|\begin{array}{c}01\\ \mathrm{tanx}0\end{array}\right|=-\mathrm{tanx}$$$${\mathrm{W}}_2=\left|\begin{array}{c}{\mathrm{e}}^{-\mathrm{x}}0\\ -{\mathrm{e}}^{-\mathrm{x}}\mathrm{tanx}\end{array}\right|={\mathrm{e}}^{-\mathrm{x}}\mathrm{tanx}$$$${\mathrm{V}}_1=\int \frac{{\mathrm{w}}_1}{\mathrm{w}}\mathrm{dx}=\int \frac{-\mathrm{tanx}}{{\mathrm{e}}^{-\mathrm{x}}}=-\int {\mathrm{e}}^{-\mathrm{x}}\mathrm{tanx}\mathrm{dx}$$$${\mathrm{V}}_2=\int \frac{{\mathrm{w}}_2}{\mathrm{w}}\mathrm{dx}=\int \frac{{\mathrm{e}}^{-\mathrm{x}}\mathrm{tanx}}{{\mathrm{e}}^{-\mathrm{x}}}\mathrm{dx}=\int \mathrm{tanx}\mathrm{dx}=\int \frac{\mathrm{sinx}}{\mathrm{cosx}}\mathrm{dx}=-\ln \mid \mathrm{cosx}\mid$$$$\Rightarrow {\mathrm{y}}_{\mathrm{h}}={\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2=-{\mathrm{e}}^{-\mathrm{x}}\int {\mathrm{e}}^{-\mathrm{x}}\mathrm{tanx}\mathrm{dx}-\ln \mid \mathrm{cosx}\mid$$$$\mathrm{general}\mathrm{solution}\mathrm{is}\mathrm{y}={\mathrm{y}}_{\mathrm{h}}+{\mathrm{y}}_{\mathrm{p}}={\mathrm{ae}}^{-\mathrm{x}}+\mathrm{b}-{\mathrm{e}}^{-\mathrm{x}}\int {\mathrm{e}}^{-\mathrm{x}}\mathrm{tanxdx}-\ln \mid \mathrm{cosx}\mid$$

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