JEE Main & Advanced JEE Main Paper (Held On 10 April 2016)

  • question_answer
    For \[x\,\in \,R,\,x\,\ne 0\], if y(x) is a differentiable function such that\[x\int\limits_{1}^{x}{y(t)dt=(x+1)\int\limits_{1}^{x}{ty(t)dt}}\], then y(x) equals(where C is a constant)   JEE Main Online Paper (Held On 10 April 2016)

    A) \[C{{x}^{3}}{{e}^{\frac{1}{x}}}\]                   

    B) \[\frac{C}{x}{{e}^{\frac{-1}{x}}}\]

    C) \[\frac{C}{{{x}^{2}}}{{e}^{-\frac{1}{x}}}\]                 

    D) \[\frac{C}{{{x}^{3}}}{{e}^{-\frac{1}{x}}}\]

    Correct Answer: B

    Solution :

                 \[x\int\limits_{1}^{x}{y(t)dt=(x+1)\int\limits_{1}^{x}{ty(t)dt}}\] .......(i) differentiate equation (a) \[xy(1)+\int\limits_{1}^{x}{y(t)dt=(x+1)xy(x)-\int\limits_{1}^{x}{ty(t)dt}}\]              \[\int\limits_{1}^{x}{y(t)dt={{x}^{2}}y(x)-\int\limits_{1}^{x}{ty(t)dt}}\] again differentiate \[y(x)=2xy(x)+{{x}^{2}}y(x)-xy(x)\] \[y=xy+{{x}^{2}}\frac{dy}{dx}\] \[y(1-x)={{x}^{2}}\frac{dy}{dx}\] \[\frac{(1-x)}{{{x}^{2}}}dx=\frac{dy}{y}\] solve differential equation \[-\frac{1}{x}-\ell nx=\ell ny+\ell nc\] \[-\frac{1}{x}=\ell n\,xy+\ell nc\] \[xyc={{e}^{-1/x}}\] \[y=\frac{c}{x}{{e}^{-1/x}}\]


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