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: Understanding the Concept
Rewrite as \(\dfrac{dx}{dy}=\dfrac{x^{3}+y^{3}}{x^{2}y}\). Put \(z=x^{3}\), so \(\dfrac{dz}{dy}=3x^{2}\dfrac{dx}{dy}\).

Step 2: Linear Form
\(\dfrac{dz}{dy}=\dfrac{3(x^{3}+y^{3})}{y}=\dfrac{3z}{y}+3y^{2}\), i.e. \(\dfrac{dz}{dy}-\dfrac{3}{y}z=3y^{2}\).

Step 3: Integrating Factor
\(\text{IF}=e^{-3\int\frac{dy}{y}}=y^{-3}\).

Step 4: Solve
\(z\,y^{-3}=\int3y^{2}y^{-3}dy=3\log|y|+C\). So \(\dfrac{x^{3}}{y^{3}}=3\log|y|+C\), which gives \(\log|y|-\dfrac{x^{3}}{3y^{3}}=c\), option (D).

Final Answer:
The solution is log of modulus y minus x cubed over 3 y cubed equals c. \[ \boxed{\text{(D) }\log|y|-\dfrac{x^{3}}{3y^{3}}=c} \]