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Question Number 61178 by mathsolverby Abdo last updated on 30/May/19

$$solve(1+x^2)y^{{}^{\prime }}+(1-x^2)y=x{e}^{-3x}$$

Commented by maxmathsup by imad last updated on 31/May/19

$$\left(he\right)\to (1+x^2)y^{{}^{\prime }}+(1-x^2)y=0\Rightarrow (1+x^2)y^{{}^{\prime }}=(x^2-1)y\Rightarrow$$$$\frac{y^{{}^{\prime }}}{y}=\frac{x^2-1}{x^2+1}\Rightarrow ln\mid y\mid =\int \frac{x^2-1}{x^2+1}dx+c$$$$=\int \frac{x^2+1-2}{x^2+1}dx+c=x-2arctan\left(x\right)+c\Rightarrow$$$$y\left(x\right)=K{e}^{x-2arctan\left(x\right)}mvcmethodgive$$$$y^{{}^{\prime }}=K^{{}^{\prime }}{e}^{x-2arctan\left(x\right)}+K(1-\frac{2}{1+x^2})e^{x-2arctan\left(x\right)}$$$$=(K^{{}^{\prime }}+K\frac{x^2-1}{x^2+2})e^{x-2arctan\left(x\right)}$$$$\left(e\right)\Rightarrow (1+x^2)K^{{}^{\prime }}{e}^{x-2arctan\left(x\right)}+K(x^2-1)e^{x-2arctan\left(x\right)}$$$$+(1-x^2)K{e}^{x-2arctan\left(x\right)}=x{e}^{-3x}\Rightarrow$$$$(1+x^2)K^{{}^{\prime }}{e}^{x-2arctan\left(x\right)}=x{e}^{-3x}\Rightarrow K^{{}^{\prime }}=\frac{x}{1+x^2}{e}^{-3x-x+2arctan\left(x\right)}$$$$=\frac{x}{1+x^2}{e}^{-4x+2arctan\left(x\right)}\Rightarrow K\left(x\right)=\int \frac{x{e}^{-4x+2arctan\left(x\right)}}{1+x^2}dx+\lambda \Rightarrow$$$$y\left(x\right)=(\int \frac{x{e}^{-4x+2arctan\left(x\right)}}{1+x^2}dx+\lambda )e^{x-2arctan\left(x\right)}$$$$=\lambda e^{x-2arctan\left(x\right)}+e^{x-2arctan\left(x\right)}{\int }_{.}^x\frac{t{e}^{-4t+2arctan\left(t\right)}}{1+t^2}dt(\lambda \in R).$$

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