JEE Main & Advanced Sample Paper JEE Main Sample Paper-5

  • 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)  \[Cy={{e}^{\sqrt{2x/y}}}\]

    Correct Answer: B

    Solution :

     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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