JEE Main & Advanced Mathematics Differential Equations Question Bank Mock Test - Differential Equations

  • question_answer
    The solution of the differential equation\[y(2{{x}^{4}}+y)\frac{dy}{dx}=(1-4x{{y}^{2}}){{x}^{2}}\] is given by

    A) \[3{{({{x}^{2}}y)}^{2}}+{{y}^{3}}-{{x}^{3}}=c\]

    B) \[x{{y}^{2}}+\frac{{{y}^{3}}}{3}-\frac{{{x}^{3}}}{3}+c=0\]

    C) \[\frac{2}{3}y{{x}^{5}}+\frac{{{y}^{3}}}{3}=\frac{{{x}^{3}}}{3}-\frac{4x{{y}^{3}}}{3}+c\]

    D) None of these

    Correct Answer: A

    Solution :

    [a] \[y(2{{x}^{4}}+y)\frac{dy}{dx}=(1-4x{{y}^{2}}){{x}^{2}}\] Or \[2{{x}^{4}}ydy+{{y}^{2}}dy+4{{x}^{3}}{{y}^{2}}dx-{{x}^{2}}dx=0\] Or \[2{{x}^{2}}y({{x}^{2}}dy+2xydx)+{{y}^{2}}dy-{{x}^{2}}dx=0\] Or \[2{{x}^{2}}yd({{x}^{2}}y)+{{y}^{2}}dy-{{x}^{2}}dx=0\] Integrating, we get\[{{({{x}^{2}}y)}^{2}}+\frac{{{y}^{3}}}{3}-\frac{{{x}^{3}}}{3}=c\] Or \[3{{({{x}^{2}}y)}^{2}}+{{y}^{3}}-{{x}^{3}}=c\]


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