Question

The solution of x² + y² $\frac{dy}{dx}$ = 4, is
(a) x² + y² = 12x + C
(b) x² + y² = 3x + C
(c) x³ + y³ = 3x + C
(d) x³ + y³ = 12x + C

Answer 1

(d) x³ + y³ = 12x + C

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ {x}^2+y^2\frac{dy}{dx}=4 \\ \Rightarrow y^2\frac{dy}{dx}=4-x^2 \\ \Rightarrow y^{2}dy=\left(4-x^2\right)dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int y^{2}dy=\int \left(4-x^2\right)dx \\ \Rightarrow \frac{y^3}{3}=4x-\frac{x^3}{3}+D \\ \Rightarrow y^3=12x-x^3+3D \\ \Rightarrow x^3+y^3=12x+C,\mathrm{where}C=3D \end{gathered}$$