JEE Main & Advanced JEE Main Paper (Held On 19 May 2012)

  • question_answer
    The general solution of the differential equation \[\frac{dy}{dx}+\frac{2}{x}y={{x}^{2}}\]is     JEE Main  Online Paper (Held On 19  May  2012)

    A) \[y=c{{x}^{-3}}-\frac{{{x}^{2}}}{4}\]        

    B)                        \[y=c{{x}^{3}}-\frac{{{x}^{2}}}{4}\]

    C)                        \[y=c{{x}^{2}}+\frac{{{x}^{3}}}{5}\]                         

    D)                        \[y=c{{x}^{-2}}+\frac{{{x}^{3}}}{5}\]

    Correct Answer: D

    Solution :

                    Given differential equation is\[\frac{dy}{dx}+\frac{2}{x}.y={{x}^{2}}\] This is of the linear form. \[\therefore \]\[P=\frac{2}{x},Q={{x}^{2}}\] \[I.F={{e}^{\int_{{}}^{{}}{\frac{2}{x}dx}}}={{e}^{\log {{x}^{2}}={{x}^{2}}}}\] Solution is\[y.{{x}^{2}}=\int_{{}}^{{}}{{{x}^{2}}.{{x}^{2}}dx+c=\frac{{{x}^{5}}}{5}+c}\] \[y=\frac{{{x}^{3}}}{5}+c{{x}^{-2}}\]


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