Mathematics

The general solution of the equation $$\dfrac{dy}{dx} = \dfrac{y}{x + \sqrt{xy}}$$ is


ANSWER

$$\sqrt{x} = \dfrac {1}{2}  \sqrt{y}(log \,y + c)$$


SOLUTION
$$\dfrac{dy}{dx}=\dfrac{y}{x+\sqrt (xy)}$$
$$(x+\sqrt(xy))dy=ydx$$
$$\dfrac{x}{y}+\sqrt(\frac{x}{y})=\dfrac{dx}{dy} $$   $$..(a)$$
let $$x=ty$$
$$\dfrac{dx}{dy}=t+y\dfrac{dt}{dx}$$
$$t+\sqrt t=t+y\dfrac{dt}{dy}$$
$$\dfrac{dy}{y}=\dfrac{dt}{\sqrt t}$$
$$lny+c=2\sqrt t$$
$$\sqrt x=\frac{1}{2}\sqrt y(lny+c)$$
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Single Correct Hard Published on 17th 09, 2020
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