Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSModel Paper 23 Marks
Question
Solve: $2 y e ^{ x / y } d x+\left(y-2 x e ^{ x / y }\right) dy =0$
✓
Answer
We have, $2 y e^{x / y} d x+\left(y-2 x e^{x / y}\right) d y=0$
$\Rightarrow \frac{d x}{d y}=\frac{2 x e^{x / y}-y}{2 y e^{x / y}} \ldots (i)$
Clearly, the given differential equation is a homogeneous differential equation.
As the right hand side of $(i)$ is expressible as a function of $\frac{x}{y}$.
So, we put $X =$ vy and $\frac{d x}{d y}=v+y \frac{d v}{d y}$ to get
$v+y \frac{d v}{d y}=\frac{2 v e^v-1}{2 e^v}$
$\Rightarrow y \frac{d v}{d y}=\frac{2 v e^v-1}{2 e^v}-v$
$\Rightarrow y \frac{d v}{d y}=-\frac{1}{2 e^v}$
$\Rightarrow 2 ye ^{ v } dv =- dy$
$\Rightarrow 2 e ^{ v } dv =-\frac{1}{y} dy $
Integrating both sides,
$\Rightarrow 2 \int e^v d v=-\int \frac{1}{y} d y$
$\Rightarrow 2 e ^{ v }=-\log | y |+\log C$
$\Rightarrow 2 e ^{ v }=\log \left|\frac{c}{y}\right|$
$\Rightarrow 2 e ^{ x / y }=\log \left|\frac{c}{y}\right|$
Hence, 2 $e ^{ x / y }=\log \left|\frac{c}{y}\right|$ gives the general solution of the given differential equation.
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