Find the integrating factor of the differential equation  y\frac{\mathrm{d} x}{\mathrm{d} y}-2x=y^3e^{-y}.

 

 

 

 
 
 
 
 

Answers (1)

y\frac{\mathrm{d} x}{\mathrm{d} y}-2x=y^3e^{-y}

\therefore \frac{\mathrm{d} x}{\mathrm{d} y}-\frac{2x}{y}=y^2e^{-y}

\therefore \frac{\mathrm{d} x}{\mathrm{d} y}+\left ( \frac{-2x}{y} \right )=y^2e^{-y}

This is a linear D.E of the form  \frac{\mathrm{d} x}{\mathrm{d} y}+Px=Q

P=\frac{-2}{y}\:\: \: \: and\: \: \: \: Q=y^2e^{-y}

Now,  I\!F=e^{\int Pdy}=e^{\int \frac{-2}{y}dy}

                  =e^{-2\log y}=e^{ \log \left ( \frac{1}{y^2} \right )}

                  =\frac{1}{y^2}

 

 

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