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The solution of $x^2+y^2 \frac{d y}{d x}=4$ is
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Verified Answer
The correct answer is:
$x^2+1=C\left(1-y^2\right)$
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
$\begin{array}{rlrl}\frac{y^3}{3} & =4 x-\frac{x^3}{3}+C^{\prime} \\ \Rightarrow \quad & y^3 & =12 x-x^3+C \\ \Rightarrow \quad x^3+y^3 & =12 x+C\end{array}$
$\begin{aligned} x^2+y^2 \frac{d y}{d x} & =4 \\ y^2 d y & =\left(4-x^2\right) d x\end{aligned}$
On integrating both sides, we get
$\begin{array}{rlrl}\frac{y^3}{3} & =4 x-\frac{x^3}{3}+C^{\prime} \\ \Rightarrow \quad & y^3 & =12 x-x^3+C \\ \Rightarrow \quad x^3+y^3 & =12 x+C\end{array}$
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