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If  \mathrm{x= x\left ( y \right )}  is the solution of the differential equation \mathrm{y \frac{dx}{dy}= 2x+y^{3}\left ( y+1 \right )e^{y},x\left ( 1 \right )= 0;} then \mathrm{x\left ( e \right )} is equal to:

Option: 1

\mathrm{e^{3}\left ( e^{e}-1 \right )}


Option: 2

\mathrm{e^{e}\left ( e^{3}-1 \right )}


Option: 3

\mathrm{e^{2}\left ( e^{e}-1 \right )}


Option: 4

\mathrm{e^{e}\left ( e^{2}-1 \right )}


Answers (1)

\mathrm{y\frac{dx}{dy}= 2x+y^{3}\left ( y+1 \right )e^{y}\quad ,\quad x\left ( 1 \right )= 0}

\mathrm{\frac{dx}{dy}-\left ( \frac{2}{y} \right )x= y^{2}\left ( y+1 \right )e^{y}}

\mathrm{I\cdot F= e^{\int \frac{-2}{y}\, dy}=e^{-2\, ln\, y}=\frac{1}{y^{2}} }
 
\mathrm{\therefore \: \: x\left ( \frac{1}{y^{2}} \right ) = \int \left ( y+1 \right )e^{y}\, dy}

\mathrm{\frac{x}{y^{2}}= \left ( y+1 \right )e^{y}-e^{y}+c}

\mathrm{at\: y= 1, \quad x= 0}
\mathrm{0= 2e-e+c \quad \Rightarrow\quad c= -e }

\mathrm{\therefore \quad x\left ( e \right ) = e^{2}\left ( e+1 \right )e^{e}-e^{e+2}-e^{3}}
          \mathrm{ x\left ( e \right ) = e^{e+3}-e^{3}}
           \mathrm{ x\left ( e \right ) = e^{3}\left ( e^{e}-1 \right )}

The correct answer is option (A)

Posted by

Sumit Saini

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