Question

The degree of the differential equation \[ \sqrt{1+\left(\frac{dy}{dx}\right)^{1/3}}=\frac{d^2y}{dx^2} \] is:

• A. $6$
• B. $3$
• C. $1$
• D. $2$

Hint:

Degree can be found only after removing radicals and fractional powers of derivatives. If derivatives remain under roots or fractional exponents, the degree is not yet defined.

Answer 1

The Correct Option is A

Concept: The degree of a differential equation is defined only when the differential equation can be expressed as a polynomial in derivatives. To find the degree:
• Remove radicals and fractional powers involving derivatives.
• Rewrite the equation completely in polynomial form.
• Identify the highest order derivative.
• The power of that highest order derivative gives the degree.

Step 1: Write the given equation clearly.
The differential equation is: \[ \sqrt{1+\left(\frac{dy}{dx}\right)^{1/3}} =\frac{d^2y}{dx^2} \] The highest order derivative present is: \[ \frac{d^2y}{dx^2} \] Hence the order is already seen to be $2$. However, we must first convert the equation into polynomial form before determining the degree.

Step 2: Remove the square root.
Squaring both sides: \[ 1+\left(\frac{dy}{dx}\right)^{1/3} = \left(\frac{d^2y}{dx^2}\right)^2 \] Now isolate the fractional power term: \[ \left(\frac{dy}{dx}\right)^{1/3} = \left(\frac{d^2y}{dx^2}\right)^2-1 \]

Step 3: Remove the cube root.
Cube both sides: \[ \left[\left(\frac{dy}{dx}\right)^{1/3}\right]^3 = \left[\left(\frac{d^2y}{dx^2}\right)^2-1\right]^3 \] Therefore, \[ \frac{dy}{dx} = \left[\left(\frac{d^2y}{dx^2}\right)^2-1\right]^3 \] Now the equation is free from radicals and fractional powers.

Step 4: Expand to identify the degree.
Using \[ (a-b)^3=a^3-3a^2b+3ab^2-b^3 \] we get: \[ \frac{dy}{dx} = \left(\frac{d^2y}{dx^2}\right)^6 -3\left(\frac{d^2y}{dx^2}\right)^4 +3\left(\frac{d^2y}{dx^2}\right)^2 -1 \] The highest order derivative is: \[ \frac{d^2y}{dx^2} \] and its highest power is: \[ 6 \] Hence, the degree of the differential equation is: \[ \boxed{6} \]