The solution of x^ 2 +y^ 2 dy dx =4 is :

MCQ

The solution of $\text{x}^{2}+\text{y}^{2}\frac{\text{dy}}{\text{dx}}=4$ is :

A
$x^2 + y^2 = 12x + C$
B
$x^2 + y^2 = 3x + C$
C
$x^3 + y^3 = 3x + C$
✓
$x^3 + y^3 = 12x + C$

✓

Answer

Correct option: D.

$x^3 + y^3 = 12x + C$

We have,
$\text{x}^{2}+\text{y}^{2}\frac{\text{dy}}{\text{dx}}=4$
$\Rightarrow\text{y}^{2}\frac{\text{dy}}{\text{dx}}=4-\text{x}^{2}$
$\Rightarrow\text{y}^{2}\frac{\text{dy}}{\text{dx}}=(4-\text{x}^{2})\text{dx}$
Integrating both sides, we get
$\int\text{y}^{2}\frac{\text{dy}}{\text{dx}}=\int(4-\text{x}^{2})\text{dx}$
$\Rightarrow \frac{\text{y}^{3}}{3}=4\text{x}-\frac{\text{x}^{3}}{3}+\text{D} $
$\Rightarrow \text{y}^{3}=12\text{x}-\text{x}^{3}+3\text{D}$
$\Rightarrow \text{x}^{3}+\text{y}^{3}=12\text{x}+\text{C}$

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