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  • Solve the differential equation Differential Equations Miscellaneous Exercise 10.

Q10. Solve the differential equation ye^\frac{x}{y}dx = \left(e^\frac{x}{y} + y^2 \right )dy\ (y \neq 0)

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best_answer

Given,

$y e^{\frac{x}{y}} d x=\left(e^{\frac{x}{y}}+y^2\right) d y$
$y e^{\frac{x}{y}} \frac{d x}{d y}=e^{\frac{x}{y}}+y^2$
$\Longrightarrow e^{\frac{x}{y}}\left[y \frac{d x}{d y}-x\right]=y^2$
$\Longrightarrow \frac{e^{\frac{x}{y}}\left[y \frac{d x}{d y}-x\right]}{y^2}=1$
Let $e^{\frac{x}{y}}=t$

Differentiating it w.r.t. $y$, we get,

$\frac{d}{d y} e^{\frac{x}{y}}=\frac{d t}{d y}$
$\Longrightarrow e^{\frac{x}{y}} \cdot \frac{d}{d y}\left(\frac{x}{y}\right)=\frac{d t}{d y}$
$\Longrightarrow \frac{e^{\frac{x}{y}}\left[y \frac{d x}{d y}-x\right]}{y^2}=\frac{d t}{d y}$
Thus from these two equations, we get,
$\frac{d t}{d y}=1$

$\Longrightarrow \int{dt} =\int{dy}$
$\Longrightarrow t=y+C$
$\Longrightarrow \mathrm{e}^{\frac{\mathrm{x}}{\mathrm{y}}}=\mathrm{y}+\mathrm{C}$

Posted by

HARSH KANKARIA

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