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

  • question_answer
    The particular solution of the differential equation \[{{\sin }^{-1}}\left( \frac{{{d}^{2}}y}{d{{x}^{2}}}-1 \right)=x\], where\[y=\frac{dy}{dx}=0\] when\[x=0\], is

    A) \[y={{x}^{2}}+x-\sin x\]

    B) \[y=\frac{{{x}^{2}}}{2}+x-\sin x\]

    C) \[y=\frac{{{x}^{2}}}{2}+\frac{x}{2}-\sin x\]

    D) \[2y={{x}^{2}}+x-\sin x\]

    Correct Answer: B

    Solution :

    [b] The differential equation is \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=1+\sin x\]                          ?. (i) Integrating we get \[\frac{dy}{dx}=x-\cos x+c\]     ? (ii) When \[x=0,\frac{dy}{dx}=0\Rightarrow c=1\] \[\therefore \] Equation (ii) is \[\frac{dy}{dx}=x-\cos x+1\] Integrating again we get \[y=\frac{{{x}^{2}}}{2}-\sin x+x+D\]                         ? (iii) When \[x=0,y=0\Rightarrow D=0\] \[\therefore \] The particular solution is \[y=\frac{{{x}^{2}}}{2}+x-\operatorname{sinx}\]


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