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  • Let be the solution of the differential equation <img alt="\mathrm{ 2 y \mathrm{

Let \mathrm{x=x(y)} be the solution of the differential equation \mathrm{ 2 y \mathrm{e}^{x / y^{2}} \mathrm{~d} x+\left(y^{2}-4 x \mathrm{e}^{x / y^{2}}\right) \mathrm{d} y=0} such that \mathrm{ x(1)=0}. Then, \mathrm{x({e})} is equal to :

Option: 1

\mathrm{e \log _{e}(2)}


Option: 2

\mathrm{-e\log _{e}(2)}


Option: 3

\mathrm{\mathrm{e}^{2} \log _{\mathrm{e}}(2)}


Option: 4

-\mathrm{e}^{2} \log _{\mathrm{e}}(2)


Answers (1)

best_answer

\mathrm{2 y e^{x / y^{2}} d x+\left(y^{2}-4 x e^{x / y^{2}}\right) d y=0 }\\

\mathrm{2 y e^{x / y^{2}} d x-4 x e^{x / y^{2}} d y+y^{2} d y=0} \\

\mathrm{2 e^{x / y^{2}}(y d x-2 x d y)+y^{2} d y=0 }

\mathrm{2 e^{x / y^{2}}\left(\frac{y^{2} d x-2 x y d y}{y}\right)+y^{2} d y=0} \\

\mathrm{2 e^{x / y^{2}}\left(\frac{y^{2} d x-2 x y d y}{y^{4}}\right)+\frac{d y}{y}=0} \\

\mathrm{2 e^{x / y^{2}} d\left(\frac{x}{y^{2}}\right)+\frac{d y}{y}=0}

Integrating

\mathrm{2 e^{x / y^{2}}+\ln y+c=0}\\

\mathrm{As \: x(1)=0 \Rightarrow(0,1)\: satisfies\: it}\\

\mathrm{2 \cdot 1+0+c=0 \Rightarrow c=-2}

\mathrm{Putting \: y=e}

\mathrm{2 e^{x / e^{2}}+\ln e-2=0} \\

\mathrm{\Rightarrow 2 e^{x / e^{2}}=1} \\

\mathrm{\Rightarrow e^{x / e^{2}}=\frac{1}{2}} \\

\mathrm{\Rightarrow x=e^{2} \ln \left(\frac{1}{2}\right)=-e^{2} \ln (2) }

Hence the correct answer is option 4.

Posted by

Pankaj Sanodiya

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