Question

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

Answer 1

differential equation is
$(\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}})\frac{dx}{dy}=1$
$⟹\frac{e^{-2\sqrt{x}}}{\sqrt{x}}=\frac{dy}{dx}+\frac{y}{\sqrt{x}}$
this first order equation of form
$y^{\prime }+P\left(x\right)y=Q\left(x\right)$
integrating factor would be
$I.F=e^{\int P\left(x\right)dx}$
so integrating factor would be
$I.F=e^{\int \frac{dx}{\sqrt{x}}}=e^{2\sqrt{x}}$
so solution would be
$y\times I.F=\int I.F\times Q\left(x\right)dx$
solving using above method
$y{e}^{2\sqrt{x}}=\int e^{2\sqrt{x}}\frac{e^{-2\sqrt{x}}}{\sqrt{x}}dx$
$⟹y{e}^{2\sqrt{x}}=\int \frac{dx}{\sqrt{x}}$
$⟹y{e}^{2\sqrt{x}}=2\sqrt{x}+C$
hence the solution is
$y=\frac{2\sqrt{x}+C}{e^{2\sqrt{x}}}$