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JEE Main & Advanced Mathematics Differential Equations Question Bank Order and degree of differential equations

  • question_answer
    The degree of the differential equation \[\left( \frac{2+\sin x}{1+y} \right)\frac{dy}{dx}=-\cos ,x\ y(0)=1,\] is                [Pb. CET 2003]

    A)                 1             

    B)                 2

    C)                 3             

    D)                 6

    Correct Answer: B

    Solution :

                       \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=-\sqrt{1+{{\left( \frac{dy}{dx} \right)}^{3}}}\]            On squaring, we get \[{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}^{2}}=1+{{\left( \frac{dy}{dx} \right)}^{3}}\]                                 Obviously the degree is 2.


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