Find the general solution of the given differential equation.
(x + 1) dy/dx + (x + 2)y = 2xe^−x
(x+1)y′+(x+2)y=2xe−xHomogeneous equation:(x+1)y′+(x+2)y=0dyy=−x+2x+1dxln∣y∣=−x−ln∣x+1∣+C′y=Ce−xx+1y=C(x)e−xx+1y′=C′(x)e−xx+1+C(x)(e−xx+1)′(x+1)C′(x)e−xx+1=2xe−xC′(x)=2xC(x)=x2+Cy=(x2+C)e−xx+1\left( x+1 \right) y'+\left( x+2 \right) y=2xe^{-x}\\Homogeneous\,\,equation:\\\left( x+1 \right) y'+\left( x+2 \right) y=0\\\frac{dy}{y}=-\frac{x+2}{x+1}dx\\\ln \left| y \right|=-x-\ln \left| x+1 \right|+C'\\y=C\frac{e^{-x}}{x+1}\\y=C\left( x \right) \frac{e^{-x}}{x+1}\\y'=C'\left( x \right) \frac{e^{-x}}{x+1}+C\left( x \right) \left( \frac{e^{-x}}{x+1} \right) '\\\left( x+1 \right) C'\left( x \right) \frac{e^{-x}}{x+1}=2xe^{-x}\\C'\left( x \right) =2x\\C\left( x \right) =x^2+C\\y=\left( x^2+C \right) \frac{e^{-x}}{x+1}(x+1)y′+(x+2)y=2xe−xHomogeneousequation:(x+1)y′+(x+2)y=0ydy=−x+1x+2dxln∣y∣=−x−ln∣x+1∣+C′y=Cx+1e−xy=C(x)x+1e−xy′=C′(x)x+1e−x+C(x)(x+1e−x)′(x+1)C′(x)x+1e−x=2xe−xC′(x)=2xC(x)=x2+Cy=(x2+C)x+1e−x
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