Question

Solve the differential equation $[\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}]\frac{dx}{dy}=1(x≠0)$

Answer 1

The correct answer is: $ye^{2\sqrt{x}}=2\sqrt{x}+C$
$[\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}]\frac{dx}{dy}=1$
$\implies \frac{dy}{dx}=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}$
$⇒\frac{dy}{dx}+\frac{y}{\sqrt{x}}=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}$
This equation is a linear differential equation of the form
$\frac{dy}{dx}+py=Q$,where $p=\frac{1}{\sqrt{x}}$ and $Q=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}.$
Now,$I.F.=e^{∫pdx}=e^{∫\frac{1}{\sqrt{x}}dx}=e^{2\sqrt{x}}$
The general solution of the given differential equation is given by,
$y(I.F.)=∫(Q×I.F.)dx+C$
$⇒ye^{2\sqrt{x}}=∫(\frac{e^{-2\sqrt{x}}}{\sqrt{x}}×e^{2\sqrt{x}})dx+C$
$⇒ye^{2\sqrt{x}}=∫\frac{1}{\sqrt{x}}dx+C$
$⇒ye^{2\sqrt{x}}=2\sqrt{x}+C$