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

  • question_answer
    The gradient of the curve passing through (4, 0) is given by \[\frac{dy}{dx}-\frac{y}{x}+\frac{5x}{(x+2)(x-3)}=0\] if the point (5, a) lies on the curve, then the value of a is

    A) \[\frac{67}{12}\]

    B) \[5\sin \frac{7}{12}\]

    C) \[5\log \frac{7}{12}\]

    D) None of these

    Correct Answer: C

    Solution :

    [c] The differential equation is \[\frac{dy}{dx}-\frac{y}{x}=-\frac{5x}{(x+2)(x-3)}\] I.F = \[{{e}^{\int{\left( \frac{1}{x} \right)dx\,}}}\,={{e}^{-ln\,x}}=\frac{1}{x}\] Solution is \[y\left( \frac{1}{x} \right)=\int{\left( \frac{1}{x} \right)\times \frac{5x}{(x+2)(x-3)}dx=ln\left( \frac{x+2}{x-3} \right)+C}\] It passes through (4, 0), so \[C=-ln\,\,6\] \[\therefore \,\,\,y=xln\left\{ \frac{(x+2)}{6(x-3)} \right\}\] Putting (5, a), we get a = 5 \[\ln \,\left( \frac{7}{12} \right)\]


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