Question

Solve:

$\frac{dy}{dx}=e^{4x-3y}$

Answer 1

Given,

$\frac{dy}{dx}=e^{4x-3y}$

or, $e^{3y}dy=e^{4x}dx$

Now integrating we have,

$\frac{e^{3y}}{3}=\frac{e^{4x}}{4}+\frac{c}{12}$ [ Where $c$ being integrating constant]

or, $4e^{3y}=3e^{4x}+c$