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Question Number 104051 by bemath last updated on 19/Jul/20

$$(x^2-1)\frac{dy}{dx}+2y={(x+1)}^2$$

Answered by bramlex last updated on 19/Jul/20

Commented by bramlex last updated on 19/Jul/20

$$typointegratingfactor$$$$u\left(x\right)=e^{\int \frac{2}{(x+1)(x-1)}dx}=\frac{x-1}{x+1}$$$$\frac{d}{dx}(\left(\frac{x-1}{x+1}\right).y)=\int \left(\frac{x-1}{x+1}\right)\left(\frac{x+1}{x-1}\right)dx$$$$\left(\frac{x-1}{x+1}\right)y=\int dx$$$$\left(\frac{x-1}{x+1}\right)y=x+C$$$$\therefore y=\frac{x^2+x}{x-1}+C{(x-1)}^{-1}◼$$

Answered by Dwaipayan Shikari last updated on 19/Jul/20

$$IF=e^{2\int \frac{1}{x^2-1}}=\left(\frac{x-1}{x+1}\right)\{\frac{dy}{dx}+\frac{2y}{x^2-1}=\frac{x+1}{x-1}$$$$y.\left(\frac{x-1}{x+1}\right)=\int dx$$$$y=\frac{x(x+1)}{x-1}+C\left(\frac{x+1}{x-1}\right)$$$$$$

Answered by mathmax by abdo last updated on 19/Jul/20

$$({\mathrm{x}}^2-1){\mathrm{y}}^{{}^{\prime }}+2\mathrm{y}={(\mathrm{x}+1)}^2$$$$\mathrm{h}\to ({\mathrm{x}}^2-1){\mathrm{y}}^{{}^{\prime }}+2\mathrm{y}=0\Rightarrow ({\mathrm{x}}^2-1){\mathrm{y}}^{{}^{\prime }}=-2\mathrm{y}\Rightarrow \frac{{\mathrm{y}}^{{}^{\prime }}}{\mathrm{y}}=\frac{-2}{({\mathrm{x}}^2-1)}\Rightarrow$$$$\ln \mid \mathrm{y}\mid =-2\int \frac{\mathrm{dx}}{{\mathrm{x}}^2-1}=-\int (\frac{1}{\mathrm{x}-1}-\frac{1}{\mathrm{x}+1})\mathrm{dx}=\ln \mid \frac{\mathrm{x}+1}{\mathrm{x}-1}\mid +\mathrm{c}\Rightarrow$$$$\mathrm{y}\left(\mathrm{x}\right)=\mathrm{k}\mid \frac{\mathrm{x}+1}{\mathrm{x}-1}\mid \mathrm{solution}\mathrm{on}\mathrm{w}=\{\mathrm{x}/\frac{\mathrm{x}+1}{\mathrm{x}-1}>0\}\Rightarrow \mathrm{y}=\mathrm{k}\times \frac{\mathrm{x}+1}{\mathrm{x}-1}$$$$\Rightarrow {\mathrm{y}}^{{}^{\prime }}={\mathrm{k}}^{{}^{\prime }}\times \frac{\mathrm{x}+1}{\mathrm{x}-1}+\mathrm{k}\times \frac{\mathrm{x}-1-(\mathrm{x}+1)}{{(\mathrm{x}-1)}^2}={\mathrm{k}}^{{}^{\prime }}\left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)-\frac{2\mathrm{k}}{{(\mathrm{x}-1)}^2}$$$$\mathrm{e}\Rightarrow ({\mathrm{x}}^2-1){\mathrm{k}}^{{}^{\prime }}\left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)+({\mathrm{x}}^2-1)\times \frac{(-2\mathrm{k})}{{(\mathrm{x}-1)}^2}+2\mathrm{k}\left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)={(\mathrm{x}+1)}^2\Rightarrow$$$$({\mathrm{x}}^2-1){\mathrm{k}}^{{}^{\prime }}={(\mathrm{x}+1)}^2\Rightarrow {\mathrm{k}}^{{}^{\prime }}=\frac{{(\mathrm{x}+1)}^2}{(\mathrm{x}+1)(\mathrm{x}-1)}=\frac{\mathrm{x}+1}{\mathrm{x}-1}\Rightarrow \mathrm{k}\left(\mathrm{x}\right)=\int \frac{\mathrm{x}+1}{\mathrm{x}-1}\mathrm{dx}+\mathrm{c}$$$$=\int \frac{\mathrm{x}-1+2}{\mathrm{x}-1}\mathrm{dx}+\mathrm{c}=\mathrm{x}+2\ln \mid \mathrm{x}-1\mid +\mathrm{c}\Rightarrow$$$$\mathrm{y}\left(\mathrm{x}\right)=\frac{\mathrm{x}+1}{\mathrm{x}-1}\{\mathrm{x}+2\ln \mid \mathrm{x}-1\mid +\mathrm{c}\}$$

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