• question_answer The solution of differential equation $ydx+(2\sqrt{xy}-x)$$dy=0$is A)  $Cy={{e}^{\sqrt{x/y}}}$                             B)  $Cy={{e}^{-\sqrt{x/y}}}$ C)  $Cy={{e}^{x/y}}$                          D)  $Cy={{e}^{\sqrt{2x/y}}}$

Idea This is a homogeneous equation, to solve it by putting y = vx, variable separation. Given differential equation is $\frac{dy}{dx}=\frac{y}{x-2\sqrt{xy}}$put$y=vx,$ $\therefore$  $v+\frac{dy}{dx}=\frac{vx}{x-2\sqrt{x(vx)}}$ $\therefore$  $v+x\frac{dv}{dx}=\frac{v}{1-2\sqrt{v}}$ $\therefore$  $x\frac{dv}{dx}=\frac{v}{1-2\sqrt{v}}-v$$x\frac{dv}{dx}=\frac{v-v+2v\sqrt{v}}{1-2\sqrt{v}}$ [by variable separation to separate x and v} $\therefore$  $\frac{1-2\sqrt{v}}{v\sqrt{v}}dv=2\frac{dx}{x}$ $\left( \frac{1}{{{v}^{3/2}}}-\frac{2}{v} \right)dv=2\log x+2\log C$ or$\log (vxC)=-\frac{1}{\sqrt{v}}$          $Cy={{e}^{-\sqrt{x/y}}}$ TEST Edge Linear differential, exact differential equations related questions are asked. To solve such type of question, students are advised to understand the basic concept of differential equation.
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