Q.(ii) The degree of the differential equation $\sqrt{1+\left(\frac{dy}{dx}\right)^2}=x$ is ______.

The key idea here is that the degree of a differential equation is defined only when the equation is a polynomial in the derivatives. Any radical or non-polynomial form must be removed first.

Step 1: Square both sides to eliminate the square root:

Step 2: Rearrange to get a polynomial in $\frac{dy}{dx}$:

$\left(\frac{dy}{dx}\right)^2 = x^2 - 1$ …

The degree of a differential equation is defined only when the equation is a polynomial in the derivatives. That means we must first remove any square roots, cube roots, or other radicals that involve the derivatives. Once the equation is written as a polynomial in $\frac{dy}{dx}$, $\frac{d^2y}{dx^2}$, etc., the degree is simply the exponent of the highest-order derivative present.

Here, the given equation is:

The left side has a square root that contains the first derivative. To find the degree, we must eliminate this radical.

1. Square both sides to remove the square root:

2. Rearrange to isolate the derivative term:

3. Identify the highest-order derivative — here it is $\frac{dy}{dx}$, which is first order. The equation is now a polynomial in $\frac{dy}{dx}$ (no radicals or fractions involving the derivative).

4. Read the degree: the exponent of $\frac{dy}{dx}$ is 2. So the degree is 2. …

Method: Finding the degree after clearing a radical

Use this when a derivative sits under a square root — you must remove the radical before reading off the degree.

Steps

Step 1: Isolate and remove the radical

Square both sides (or raise to the needed power) so no root remains:

Step 2: Confirm it is now polynomial in the derivatives …

Common Mistakes

Mistake 1: Reading the degree before removing the square root

Why it's wrong: $\sqrt{1+(y')^2}=x$ is not yet polynomial in $y'$; degree is undefined until you square. Correct approach: square first to get $1+(y')^2=x^2$.

Mistake 2: Calling the degree 1 after squaring …