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JEE Main & Advanced Mathematics Differential Equations Question Bank Variable Separable type differential equations

  • question_answer
    The solution of the differential equation \[({{x}^{2}}-y{{x}^{2}})\frac{dy}{dx}+{{y}^{2}}+x{{y}^{2}}=0\] is      [Pb. CET 2003]

    A)                 \[\log \left( \frac{x}{y} \right)=\frac{1}{x}+\frac{1}{y}+c\]          

    B)                 \[\log \left( \frac{y}{x} \right)=\frac{1}{x}+\frac{1}{y}+c\]

    C)                 \[\log \left( xy \right)=\frac{1}{x}+\frac{1}{y}+c\]         

    D)                 \[\log \left( xy \right)+\frac{1}{x}+\frac{1}{y}=c\]

    Correct Answer: A

    Solution :

                       The given equation                    \[({{x}^{2}}-y{{x}^{2}})\frac{dy}{dx}+{{y}^{2}}+x{{y}^{2}}=0\]Þ\[\frac{1-y}{{{y}^{2}}}dy+\frac{1+x}{{{x}^{2}}}dx=0\]                    Þ \[\left( \frac{1}{{{y}^{2}}}-\frac{1}{y} \right)dy+\left( \frac{1}{{{x}^{2}}}+\frac{1}{x} \right)dx=0\]                    On integrating, we get the required solution                                 \[\log \left( \frac{x}{y} \right)=\frac{1}{x}+\frac{1}{y}+c\].


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