If x = x(y) is the solution of the differential equation
$y{{dx} \over {dy}} = 2x + {y^3}(y + 1){e^y},\,x(1) = 0$; then x(e) is equal to :
Solution
$\frac{d x}{d y}-\frac{2 x}{y}=y^{2}(y+1) e^{y}$
<br/><br/>
$\text { If }=e^{\int-\frac{2}{y} d y}=e^{-2 \ln y}=\frac{1}{y^{2}}$
<br/><br/>
Solution is given by
<br/><br/>
$$
\begin{aligned}
&x \cdot \frac{1}{y^{2}}=\int y^{2}(y+1) e^{y} \cdot \frac{1}{y^{2}} d y \\\\
\Rightarrow & \frac{x}{y^{2}}=\int(y+1) e^{y} d y \\\\
\Rightarrow & \frac{x}{y^{2}}=y e^{y}+c
\end{aligned}
$$<br/><br/>
$\Rightarrow x=y^{2}\left(y e^{y}+c\right)$ at, $y=1, x=0$
<br/><br/>
$\Rightarrow 0=1\left(1 \cdot e^{1}+c\right) \Rightarrow c=-e$ at $y=e$,
<br/><br/>
$x=e^{2}\left(e . e^{e}-e\right)$
About this question
Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree
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