Question

Solve $[\frac{e^{-2\sqrt{x}}}{\sqrt{x}}+\frac{8}{5x}]\frac{dx}{dy}=1$

Answer 1

$\left(\right(\frac{e^{-2\sqrt{x}}}{\sqrt{x}})+(\frac{8}{5x}\left)\right)\left(\frac{dx}{dy}\right)=1or\int \left(\right(\frac{e^{-2\sqrt{x}}}{\sqrt{x}})+(\frac{8}{5x}\left)\right)dx=\int dy\int \left(\frac{e^{-2\sqrt{x}}}{\sqrt{x}}\right)dx+\left(\frac{8}{5}\right)\int \left(\frac{1}{x}\right)dx=y+clet-2\sqrt{x}=tthen-2\times \left(\frac{1}{2\sqrt{x}}\right)dx=dtor\left(\frac{dx}{\sqrt{x}}\right)=-dt\therefore \int e^t(-dt)+\left(\frac{8}{5}\right)log|x|=y+c=-e^t+\left(\frac{8}{5}\right)log|x|=y+cor-e^{-2\sqrt{x}}+\left(\frac{8}{5}\right)log|x|=y+c$