Mathematics

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

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

##### 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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

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