Answer to Question #320453 in Differential Equations for anna

Question #320453

Find the general solution of the given differential equation.


(x + 1) dy/dx + (x + 2)y = 2xe^−x


1
Expert's answer
2022-03-30T11:22:02-0400

(x+1)y+(x+2)y=2xexHomogeneous  equation:(x+1)y+(x+2)y=0dyy=x+2x+1dxlny=xlnx+1+Cy=Cexx+1y=C(x)exx+1y=C(x)exx+1+C(x)(exx+1)(x+1)C(x)exx+1=2xexC(x)=2xC(x)=x2+Cy=(x2+C)exx+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}


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