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DIFFERENTIAL EQUATIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS1 Mark
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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