JEE Main & Advanced Mathematics Differential Equations Question Bank Self Evaluation Test - Differential Equations

  • question_answer
    What is the solution of the differential equation\[a\left( x\frac{dy}{dx}+2y \right)=xy\frac{dy}{dx}\]?

    A) \[{{x}^{2}}=ky{{e}^{\frac{y}{a}}}\]

    B) \[y{{x}^{2}}=ky{{e}^{\frac{y}{a}}}\]

    C) \[{{y}^{2}}{{x}^{2}}=ky{{e}^{\frac{{{y}^{2}}}{a}}}\]   

    D) None of the above

    Correct Answer: D

    Solution :

    [d] Given differential equation is
    \[a\left( x\frac{dy}{dx}+2y \right)=xy\frac{dy}{dx}\]
    \[\Rightarrow ax\frac{dy}{dx}-xy\frac{dy}{dx}=-2ay\]
    \[\Rightarrow (xy-ax)\frac{dy}{dx}=2ay\Rightarrow x(y-a)\frac{dy}{dx}=2ay\]
    \[\Rightarrow x(y-a)dy=2aydx\]
    \[\Rightarrow \frac{(y-a)}{y}dy=\frac{2a}{x}dx\Rightarrow \left( 1-\frac{a}{y} \right)dy=\frac{2a}{x}dx.\]
    \[dy-\frac{a}{y}dy=\frac{2a}{x}dx\]
    Integrate on both side
    \[\int{dy-a\int{\frac{1}{y}dy=2a\int{\frac{1}{x}dx}}}\]
    \[y-a\log \,y=2alogx+logc\]
    \[\Rightarrow y=a\log {{x}^{2}}yc\Rightarrow {{x}^{2}}y=k{{e}^{y/a}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner