Q108 / 200 · TS EAMCET 2001 · Mathematics

The solution of $x^2+y^2 \frac{d y}{d x}=4$ is

  • A
    $x^2-1=C\left(1+y^2\right)$
  • B
    $x^2+1=C\left(1-y^2\right)$
  • C
    $x^2-1=C\left(1-y^2\right)$
  • Correct answer
    D
    $x^2+1=C\left(1-y^2\right)$

Solution

We have, $\begin{aligned} x^2+y^2 \frac{d y}{d x} & =4 \\ y^2 d y & =\left(4-x^2\right) d x\end{aligned}$ On integrating both sides, we get $\begin{array}{rlrl}\frac{y^3}{3} & =4 x-\frac{x^3}{3}+C^{\prime} \\ \Rightarrow \quad & y^3 & =12 x-x^3+C \\ \Rightarrow \quad x^3+y^3 & =12 x+C\end{array}$

Hence, the correct answer is option: 'D'.

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