CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2003

  • question_answer Solution       of       the       equation\[x{{\left( \frac{dy}{dx} \right)}^{2}}+2\sqrt{xy}\frac{dy}{dx}+y=0\]is:

    A)  \[x+y=a\]          

    B)  \[\sqrt{x}-\sqrt{y}=\sqrt{a}\]

    C)         \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\]   

    D)         \[\sqrt{x}+\sqrt{y}=\sqrt{a}\]

    E)  \[{{x}^{2}}-{{y}^{2}}={{a}^{2}}\]

    Correct Answer: D

    Solution :

    \[x{{\left( \frac{dy}{dx} \right)}^{2}}+2\sqrt{xy}\frac{dy}{dx}+y=0\] \[\Rightarrow \]               \[{{\left( \sqrt{x}\frac{dy}{dx}+\sqrt{y} \right)}^{2}}=0\] On integrating both sides \[\int{\frac{1}{\sqrt{y}}}dy+\int{\frac{1}{\sqrt{x}}}dx=0\] \[\Rightarrow \]               \[2\sqrt{y}+2\sqrt{x}={{c}_{1}}\] \[\Rightarrow \]               \[\sqrt{x}+\sqrt{y}=\frac{{{c}_{1}}}{2}\] which is similar to\[\sqrt{x}+\sqrt{y}=\sqrt{a}.\] \[\therefore \]Solution of given differential equation is \[\sqrt{x}+\sqrt{y}=\sqrt{a}\]

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