Question

Solve: (i) $\frac{dy}{dx}=e^{3x-2y}+x^2{e}^{-2y}$

Answer 1

We have,

$\frac{dy}{dx}=e^{3x-2y}+x^2{e}^{-2y}$

$\Rightarrow \frac{dy}{dx}=e^{-2y}(e^{3x}+x^2)$$\Rightarrow \frac{dy}{e^{-2y}}=(e^{3x}+x^2)dx$

$\Rightarrow e^{2y}dy=(e^{3x}+x^2)dx$

$\Rightarrow e^{2y}dy=e^{3x}dx+x^{2}dx$

On integrating we get,

$\Rightarrow \int e^{2y}dy=\int (e^{3x}dx+x^{2}dx)$

$\Rightarrow \int e^{2y}dy=\int e^{3x}dx+\int x^{2}dx$

$\Rightarrow \frac{e^{2y}}{2}=\frac{e^{3x}}{3}+\frac{x^3}{3}+C$

$\Rightarrow 3e^{2y}=2e^{3x}+2x^3+C$

Hence, this is the answer.