JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer The solution of the differential equation \[x\frac{dy}{dx}+2y={{x}^{2}}\]\[(x\ne 0)\]with y(1) = 1, is                         [JEE Main 9-4-2019 Morning]

    A) \[y=\frac{{{x}^{3}}}{5}+\frac{1}{5{{x}^{2}}}\]                   

    B) \[y=\frac{4}{5}{{x}^{3}}+\frac{1}{5{{x}^{2}}}\]

    C) \[y=\frac{3}{4}{{x}^{2}}+\frac{1}{4{{x}^{2}}}\]     

    D) \[y=\frac{{{x}^{2}}}{4}+\frac{3}{4{{x}^{2}}}\]

    Correct Answer: D

    Solution :

    \[x\frac{dy}{dx}+2y={{x}^{2}}:y(1)=1\]           \[\frac{dy}{dx}\left( \frac{2}{x} \right)y=x\](LDE in y) \[IF={{e}^{\int_{{}}^{{}}{\frac{2}{x}dx}}}={{e}^{2\ell nx}}={{x}^{2}}\] \[y.\left( {{x}^{2}} \right)=\int_{{}}^{{}}{x.{{x}^{2}}}dx=\frac{{{x}^{4}}}{4}+C\] \[y(1)=1\] \[1=\frac{1}{4}+C\Rightarrow C=1-\frac{1}{4}=\frac{3}{4}\] \[y{{x}^{2}}=\frac{{{x}^{4}}}{4}+\frac{3}{4}\] \[y=\frac{{{x}^{2}}}{4}+\frac{3}{4{{x}^{2}}}\]


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