Question

The solution of the differential equation \(x^{2}ydx-(x^{3}+y^{3})dy=0\) is

• A. \(\log |y|-\frac{x^{3}}{2y^{3}}=c\)
• B. \(\log |y|+\frac{x^{3}}{2y^{3}}=c\)
• C. \(\log |y|-\frac{x^{3}}{4y^{3}}=c\)
• D. \(\log |y|-\frac{x^{3}}{3y^{3}}=c\)

Hint:

Treat x cubed as the unknown, which gives a linear differential equation in y.

Answer 1

The Correct Option is D

Step 1: Verify by Differentiation
Differentiate $\log|y|-\dfrac{x^{3}}{3y^{3}}=c$ with respect to $y$, treating $x$ as a function of $y$.

Step 2: Compute
$\dfrac{1}{y}-\dfrac{3x^{2}x'}{3y^{3}}+\dfrac{3x^{3}}{3y^{4}}=0\Rightarrow\dfrac{1}{y}-\dfrac{x^{2}x'}{y^{3}}+\dfrac{x^{3}}{y^{4}}=0$.

Step 3: Simplify
Multiplying by $y^{4}$: $y^{3}-x^{2}yx'+x^{3}=0$, so $x'=\dfrac{x^{3}+y^{3}}{x^{2}y}$, which is the given equation.

Final Answer:
Option (D) satisfies the differential equation. \[ \boxed{\text{(D) }\log|y|-\dfrac{x^{3}}{3y^{3}}=c} \]