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

H Harsh Kankaria

Given,

$ye^\frac{x}{y}dx = (e^\frac{x}{y} + y^2)dy$

$\\ ye^\frac{x}{y}\frac{dx}{dy} = e^\frac{x}{y} + y^2 \\ \implies e^\frac{x}{y}[y\frac{dx}{dy} -x] = y^2 \\ \implies \frac{e^\frac{x}{y}[y\frac{dx}{dy} -x]}{y^2} = 1$

Let $\large e^\frac{x}{y} = t$

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

$\\ \frac{d}{dy}e^\frac{x}{y} = \frac{dt}{dy} \\ \implies e^\frac{x}{y}.\frac{d}{dy}(\frac{x}{y}) = \frac{dt}{dy} \\ \implies \frac{e^\frac{x}{y}[y\frac{dx}{dy} -x]}{y^2} =\frac{dt}{dy}$

Thus from these two equations,we get,

$\\ \frac{dt}{dy} = 1 \\ \implies \int dt = \int dy \\ \implies t = y + C$

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