Question

The differential equation of \(3y = \sqrt[3]{x+c}\) is

• A. \(\frac{dy}{dx} = \frac{1}{9y^2}\)
• B. \(\frac{dy}{dx} = \frac{1}{27y^2}\)
• C. \(\frac{dy}{dx} = \frac{1}{81y^2}\)
• D. \(\frac{dy}{dx} = \frac{1}{36y^2}\)

Hint:

Cube both sides to get 27y^3 = x + c, then differentiate once so the constant disappears.

Answer 1

The Correct Option is C

Step 1: Plan
The family has one arbitrary constant \(c\), so we differentiate once and remove \(c\) to get a first order differential equation.

Step 2: Remove the cube root
Cube both sides: \((3y)^3=x+c\), so \(27y^3=x+c\).

Step 3: Differentiate with respect to x
\[ 27\cdot3y^2\frac{dy}{dx}=1 \] \[ 81y^2\frac{dy}{dx}=1 \]

Step 4: Write the result
\[ \frac{dy}{dx}=\frac{1}{81y^2} \]

Step 5: Why the other options are wrong
Options (A), (B) and (D) have \(9\), \(27\) and \(36\) in place of \(81\). They come from forgetting to multiply \(27\) by the power 3, or from differentiating \(3y\) instead of \(27y^3\). Note \(27\cdot3=81\).

Final Answer:
The differential equation is \(\frac{dy}{dx}=\frac{1}{81y^2}\), option (C). \[ \boxed{\dfrac{dy}{dx}=\dfrac{1}{81y^2}} \]