JEE Main & Advanced AIEEE Solved Paper-2004

  • question_answer
    The differential equation for the family of curves\[{{x}^{2}}+{{y}^{2}}-2\text{ }ay=0,\]where a is an arbitrary constant, is

    A) \[2({{x}^{2}}-{{y}^{2}})y'=xy\]   

    B) \[2({{x}^{2}}+{{y}^{2}})y'=xy\]

    C)        \[({{x}^{2}}-{{y}^{2}})y'=2xy\]

    D)        \[({{x}^{2}}+{{y}^{2}})y'=2xy\]

    Correct Answer: C

    Solution :

    Given equation of family of curves is \[{{x}^{2}}+{{y}^{2}}-2ay=0\]            ...(i) On differentiating w.r.t.\[x,\]we get \[2x+2yy'-2ay'=0\]           \[\Rightarrow \]               \[2x+2yy'=2ay'\] \[\Rightarrow \]               \[\frac{2x+2yy'}{y'}=2a\]                              ?..(ii) From Eq. (i), we get \[2a=\frac{{{x}^{2}}+{{y}^{2}}}{y}\] On putting this value of 2a in Eq. (ii), we get \[\frac{2x+2yy'}{y'}=\frac{{{x}^{2}}+{{y}^{2}}}{y}\] \[\Rightarrow \]               \[2xy+2{{y}^{2}}y'={{x}^{2}}y+{{y}^{2}}y'\] \[\Rightarrow \]               \[({{x}^{2}}-{{y}^{2}})y'=2xy\]

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