Question

Find the integrating factor of the different equation: $(e^{{-2}^{\sqrt{x}}}-y)dx=\sqrt{x}dy$

Answer 1

$(e^{{-2}^{\sqrt{x}}}-y)dx=\sqrt{x}dy$

$\Rightarrow \frac{dy}{dx}=\frac{e^{{-2}^{\sqrt{x}}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}$

$\Rightarrow \frac{dy}{dx}+\left(\frac{1}{\sqrt{x}}\right)y=\frac{e^{{-2}^{\sqrt{x}}}}{\sqrt{x}}$

it is in the form of $\frac{dy}{dx}+Py=Q$ where $P$ and $Q$ are the constants or functions of $x$

Thus, the integrating factor of the given differential equation is

$e^{\int Pdx}=e^{\int \frac{dx}{\sqrt{x}}}$

$=e^{\int x^{\frac{-1}{2}}dx}$

$=e^{\frac{\sqrt{x}}{\frac{1}{2}}}$

$=e^{2\sqrt{x}}$