JCECE Engineering JCECE Engineering Solved Paper-2014

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

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

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

    Correct Answer: A

    Solution :

    Given differential equation is                 \[({{x}^{2}}-y{{x}^{2}})\frac{dy}{dx}+{{y}^{2}}=x{{y}^{2}}=0\] \[\Rightarrow \]               \[\frac{1-y}{{{y}^{2}}}dy+\frac{1+x}{{{x}^{2}}}dx=0\] \[\Rightarrow \]               \[\left( \frac{1}{{{y}^{2}}}-\frac{1}{y} \right)dy+\left( \frac{1}{{{x}^{2}}}+\frac{1}{x} \right)dx=0\] On integrating both sides, we get the required solution                 \[\frac{-2}{y}-\log y-\frac{2}{x}+\log x=C\] \[\Rightarrow \]               \[\log \left( \frac{x}{y} \right)=\frac{1}{x}+\frac{1}{y}+C\]


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