Question

The differential equation representing the family of curves \(xsinx+y^3 = 4ax\) is

• A. \(\frac{dy}{dx} = \frac{y^3+x^2cosx}{3xy^2}\)
• B. \(\frac{dy}{dx} = \frac{y^3-x^2cosx}{3xy^2}\)
• C. \(\frac{dy}{dx} = \frac{y^3-x^2cosx}{3xy}\)
• D. \(\frac{dy}{dx} = \frac{y^3-x^2cosx}{3x^2y}\)

Hint:

Divide by $x$ to isolate the constant $4a$, then differentiate once.

Answer 1

The Correct Option is B

Step 1: Isolate the constant
\(\sin x+\frac{y^3}{x}=4a\).

Step 2: Differentiate
\(\cos x+\frac{3y^2y'x-y^3}{x^2}=0\).

Step 3: Simplify
Multiply by \(x^2\): \(x^2\cos x+3xy^2y'-y^3=0\).
So \(\frac{dy}{dx}=\frac{y^3-x^2\cos x}{3xy^2}\), option (B).

Final Answer:
\(\frac{dy}{dx}=\frac{y^3-x^2\cos x}{3xy^2}\), option (B). \[ \boxed{\text{(B)}} \]