JEE Main & Advanced AIEEE Solved Paper-2006

  • question_answer
    The differential equation whose solution is \[A{{x}^{2}}+B{{y}^{2}}=1,\]where A and B are arbitrary constants, is of       AIEEE  Solved  Paper-2006

    A) first order and second degree

    B) first order and first degree

    C) second order and first degree

    D) second order and second degree

    Correct Answer: C

    Solution :

    The given equation is\[A{{x}^{2}}+B{{y}^{2}}=1\] On differentiating w:r.t. x, we get \[2Ax+2By\frac{dy}{dx}=0\]                      ...(i) Again differentiating, we get \[2A+2B\left\{ {{\left( \frac{dy}{dx} \right)}^{2}}+y\frac{{{d}^{2}}y}{d{{x}^{2}}} \right\}=0\]         ??..(ii) On solving Eqs. (i) and (ii), we get \[y\frac{{{d}^{2}}y}{d{{x}^{2}}}+{{\left( \frac{dy}{dx} \right)}^{2}}-\frac{y}{x}.\frac{dy}{dx}=0\] This is the required differential equation whose order is 2 and degree is 1.

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