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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 \[\frac{dy}{dx}=\frac{1+{{y}^{2}}}{1+{{x}^{2}}}\] is [SCRA 1986]

    A)                 \[1+xy+c(y+x)=0\]            

    B)                 \[x+y=c(1-xy)\]

    C)                 \[y-x=c(1+xy)\]

    D)                 \[1+xy=c(x+y)\]

    Correct Answer: C

    Solution :

                       \[\frac{dy}{dx}=\frac{1+{{y}^{2}}}{1+{{x}^{2}}}\Rightarrow \frac{1}{1+{{y}^{2}}}dy=\frac{1}{1+{{x}^{2}}}dx\]        Now on integrating both sides, we get        \[{{\tan }^{-1}}y={{\tan }^{-1}}x+{{\tan }^{-1}}c\]Þ \[{{\tan }^{-1}}y={{\tan }^{-1}}\left( \frac{x+c}{1-cx} \right)\]                 Þ \[y=\frac{x+c}{1-cx}\] Þ \[y-x=c(1+xy)\].


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