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Question Number 104265 by mohammad17 last updated on 20/Jul/20

$$\frac{dy}{dx}-y=sinx$$

Answered by bemath last updated on 20/Jul/20

$$Homogenoussolution$$$$\lambda -1=0\Rightarrow \lambda =1\Rightarrow y_h=C_1{e}^x$$$$particularsolution$$$$y_p=A\cos x+B\sin x$$$$y_p^{\prime }=-A\sin x+B\cos x$$$$comparingcoefficient$$$$(-A\sin x+B\cos x)-(A\cos x$$$$+B\sin x)=\sin x$$$$\Rightarrow (B-A)\cos x+(-B-A)\sin x$$$$=\sin x$$$$B=A\wedge -B-A=1\Rightarrow A=-\frac{1}{2}$$$$y_p=-\frac{1}{2}\cos x-\frac{1}{2}\sin x$$$$generalsolution$$$$\therefore y={Ce}^x-\frac{1}{2}(\cos x+\sin x)$$

Commented by mohammad17 last updated on 20/Jul/20

$$$$$$$$$$sorysirhowthisithinkthisy=\frac{\int e^{-x}sinxdx}{e^{-x}}$$

Commented by mohammad17 last updated on 20/Jul/20

Commented by mohammad17 last updated on 20/Jul/20

$$isthesolutionitsright?$$

Commented by bemath last updated on 20/Jul/20

$$yessir$$

Commented by mohammad17 last updated on 20/Jul/20

$$thankyou$$

Answered by Dwaipayan Shikari last updated on 20/Jul/20

$$\mathrm{IF}={\mathrm{e}}^{\int -1\mathrm{dx}}={\mathrm{e}}^{-\mathrm{x}}$$$$\mathrm{y}.{\mathrm{e}}^{-\mathrm{x}}=\int {\mathrm{e}}^{-\mathrm{x}}\mathrm{sinxdx}=\mathrm{I}$$$${\mathrm{ye}}^{-\mathrm{x}}=-{\mathrm{sinxe}}^{-\mathrm{x}}+\int {\mathrm{e}}^{-\mathrm{x}}\mathrm{cosxdx}$$$${\mathrm{ye}}^{-\mathrm{x}}=-\mathrm{sinx}{\mathrm{e}}^{-\mathrm{x}}-\mathrm{cosx}{\mathrm{e}}^{-\mathrm{x}}-\int \mathrm{sinx}{\mathrm{e}}^{-\mathrm{x}}=\mathrm{I}$$$${\mathrm{ye}}^{-\mathrm{x}}=-\frac{1}{2}{\mathrm{e}}^{-\mathrm{x}}(\mathrm{sinx}+\mathrm{cosx})+\mathrm{C}$$$$\mathrm{y}=\frac{-1}{2}(\mathrm{sinx}+\mathrm{cosx})+{\mathrm{Ce}}^{\mathrm{x}}$$$$$$

Answered by Ar Brandon last updated on 20/Jul/20

$$\{\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}=\mathrm{sinx}\}\cdot {\mathrm{e}}^{-\mathrm{x}}$$$$\Rightarrow {\mathrm{e}}^{-\mathrm{x}}\frac{\mathrm{dy}}{\mathrm{dx}}-{\mathrm{ye}}^{-\mathrm{x}}={\mathrm{e}}^{-\mathrm{x}}\mathrm{sinx}$$$$\Rightarrow \frac{\mathrm{d}\left({\mathrm{ye}}^{-\mathrm{x}}\right)}{\mathrm{dx}}={\mathrm{e}}^{-\mathrm{x}}\mathrm{sinx}$$$$\Rightarrow {\mathrm{ye}}^{-\mathrm{x}}=\int {\mathrm{e}}^{-\mathrm{x}}\mathrm{sinxdx}$$$$=\mathrm{sinx}\int {\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}-\int \{\frac{\mathrm{dsinx}}{\mathrm{dx}}\int {\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}\}\mathrm{dx}$$$$=-{\mathrm{e}}^{-\mathrm{x}}\mathrm{sinx}+\int {\mathrm{e}}^{-\mathrm{x}}\mathrm{cosxdx}$$$$=-{\mathrm{e}}^{-\mathrm{x}}\mathrm{sinx}+\{\mathrm{cosx}\int {\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}-\int \{\frac{\mathrm{dcosx}}{\mathrm{dx}}\int {\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}\}\mathrm{dx}\}$$$$=-{\mathrm{e}}^{-\mathrm{x}}\mathrm{sinx}-{\mathrm{e}}^{-\mathrm{x}}\mathrm{cosx}-\int {\mathrm{e}}^{-\mathrm{x}}\mathrm{sinxdx}$$$$=\frac{-(\mathrm{sinx}+\mathrm{cosx}){\mathrm{e}}^{-\mathrm{x}}}{2}+\mathcal{C}$$$$\Rightarrow \mathrm{y}=-\frac{(\mathrm{sinx}+\mathrm{cosx})}{2}+\mathcal{C}{\mathrm{e}}^{\mathrm{x}}$$

Answered by mathmax by abdo last updated on 20/Jul/20

$${\mathrm{y}}^{{}^{\prime }}-\mathrm{y}=\mathrm{sinx}$$$$\mathrm{h}\to \mathrm{r}-1=0\Rightarrow \mathrm{r}=1\Rightarrow {\mathrm{y}}_{\mathrm{h}}={\mathrm{Ke}}^{\mathrm{x}}$$$$\mathrm{lagrange}\mathrm{method}\to {\mathrm{y}}^{{}^{\prime }}={\mathrm{k}}^{{}^{\prime }}{\mathrm{e}}^{\mathrm{x}}+{\mathrm{ke}}^{\mathrm{x}}$$$$\mathrm{e}\Rightarrow {\mathrm{k}}^{{}^{\prime }}{\mathrm{e}}^{\mathrm{x}}+{\mathrm{ke}}^{\mathrm{x}}-{\mathrm{ke}}^{\mathrm{x}}=\mathrm{sinx}\Rightarrow {\mathrm{k}}^{{}^{\prime }}={\mathrm{e}}^{-\mathrm{x}}\mathrm{sinx}\Rightarrow \mathrm{k}=\int {\mathrm{e}}^{-\mathrm{x}}\mathrm{sinx}\mathrm{dx}$$$$=\mathrm{Im}(\int {\mathrm{e}}^{-\mathrm{x}+\mathrm{ix}}\mathrm{dx})=\mathrm{Im}(\int {\mathrm{e}}^{(-1+\mathrm{i})\mathrm{x}}\mathrm{dx})\mathrm{but}$$$$\int {\mathrm{e}}^{(-1+\mathrm{i})\mathrm{x}}\mathrm{dx}=\frac{1}{-1+\mathrm{i}}{\mathrm{e}}^{(-1+\mathrm{i})\mathrm{x}}+\mathrm{c}=-\frac{1}{1-\mathrm{i}}{\mathrm{e}}^{-\mathrm{x}}(\mathrm{cosx}+\mathrm{sinx})$$$$=-\frac{1+\mathrm{i}}{2}{\mathrm{e}}^{-\mathrm{x}}\{\mathrm{cosx}+\mathrm{isinx}\}=-\frac{{\mathrm{e}}^{-\mathrm{x}}}{2}\{\mathrm{cosx}+\mathrm{isinx}+\mathrm{icosx}-\mathrm{sinx}\}\Rightarrow$$$$\mathrm{Im}(\int ...)=-\frac{{\mathrm{e}}^{-\mathrm{x}}}{2}\{\mathrm{cosx}+\mathrm{sinx}\}\Rightarrow \mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{is}$$$$(-\frac{{\mathrm{e}}^{-\mathrm{x}}}{2}\{\mathrm{cosx}+\mathrm{sinx}\}+\mathrm{c}){\mathrm{e}}^{\mathrm{x}}=-\frac{1}{2}(\mathrm{cosx}+\mathrm{sinx})+{\mathrm{ce}}^{\mathrm{x}}$$

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