A problem asked by Archit Mittal

Solve $\frac{dy}{dx} + \frac{2y}{3} = \frac{x}{\sqrt{y}} $

Answer:

$\frac{dy}{dx} + \frac{2y}{3} = \frac{x}{\sqrt{y}} $

$= \sqrt{y}\frac{dy}{dx} + \frac {2y\sqrt{y}}{3} = x $

Substitute $u = y\sqrt{y} $, and you get $\frac{du}{dx} = \frac{3}{2}\sqrt{y} \frac{dy}{dx} $

Or, $ \sqrt{y}\frac{dy}{dx} = \frac{2}{3} \frac{du}{dx} $

And, the equation becomes

$= \frac{2}{3} \frac{du}{dx} + \frac {2}{3} u = x $

or, $ \frac{du}{dx} + u = \frac{3x}{2} $

or, $ e^x \frac{du}{dx} + ue^x = \frac{3x}{2} e^x $

or, $ \frac{d}{dx} \left( ue^x \right) = \frac{3x}{2} e^x $

or, $ue^x = \frac{3}{2} \int xe^x \; dx = \frac{3}{2} \left( xe^x - e^x \right) + C$

or, $u = \frac{3}{2} (x - 1) + Ce^{-x} $

or, $y\sqrt{y} = \frac{3}{2} (x - 1) + Ce^{-x} $

or $y = \left( \frac{3}{2} (x - 1) + Ce^{-x} \right)^{\frac{2}{3}} $