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Q)

Solve : $y''+e^{2y}(y')^3=0$

$(a)\;y=e^{2x}+c_1 x+c_2 \\ (b)\;x= e^{2y}+c_1y+c_2 \\ (c)\;y=\frac{1}{4}.e^{2x}+c_1x +c_2 \\ (d)\;x=\frac{1}{4} e^{2y}+c_1y +c_2 $

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A)
$y'=u$
$\large\frac{d^2y}{dx^2}=\frac{du}{dx}$
$\qquad= \large\frac{du}{dy}. \frac{dy}{dx}$
$\qquad= u. \large\frac{du}{dy}$
$u. \large\frac{du}{dy}$$+e^{2y}y^3=0$
$\large\frac{du}{u^2}$$=-e^{2y}dy$
$\large\frac{-1}{u}=\frac{1}{2} $$ e^{2y}+c_1$
$\large\frac{dx}{dy}=\large\frac{1}{2}. e^{2y}+c_1$
$dx=\bigg( \large\frac{1}{2}. e^{2y}+c_1 \bigg)$$dy$
$x= \large\frac{1}{4} .e^{2y}+c_1y+c_2$
Hence d is the correct answer.
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