Q.Find the order and degree, if defined, of each of the following differential equations:

Concept: Order is the highest derivative present; degree is the power of the highest derivative after removing radicals/fractions, defined only when the equation is polynomial in derivatives.

(i) $\frac{dy}{dx} - \cos x = 0$

Highest derivative is $\frac{dy}{dx}$ (first order). The equation is polynomial in $\frac{dy}{dx}$ with exponent 1.

Order = 1, Degree = 1.

(ii) $xy\frac{d^2y}{dx^2} + x\left(\frac{dy}{dx}\right)^2 - y\frac{dy}{dx} = 0$

Highest derivative is $\frac{d^2y}{dx^2}$ (second order). The equation is polynomial in all derivatives; the power of $\frac{d^2y}{dx^2}$ is 1.

Order = 2, Degree = 1.

(iii) $y''' + y^2 + e^{y'} = 0$ …

The Core Idea: Order and Degree

Before we touch a single equation, let's be crystal clear on what "order" and "degree" mean. These are two of the most basic labels we put on a differential equation.

Order is the easy one. It's simply the highest derivative that appears in the equation. If you see $\frac{dy}{dx}$, the order is 1. If you see $\frac{d^2y}{dx^2}$, the order is 2. If you see $y'''$ (which is the same as $\frac{d^3y}{dx^3}$), the order is 3. No tricks here — just find the derivative with the most dashes.

Degree is where students often slip. The degree is defined only when the differential equation is a polynomial in the derivatives. That means every derivative term (like $y'$, $y''$, etc.) must appear with a whole-number exponent, and there can be no functions like $\sin(y')$, $e^{y'}$, or $\log(y'')$ wrapping around them. If the equation is polynomial in derivatives, the degree is the power of the highest-order derivative.

Now let's apply this to each equation.

(i) $\frac{dy}{dx} - \cos x = 0$

1. Identify the highest derivative.

The only derivative here is $\frac{dy}{dx}$, which is a first derivative. So the order is 1.

2. Check if the equation is polynomial in derivatives.

Rewrite it as $\frac{dy}{dx} = \cos x$. The derivative $\frac{dy}{dx}$ appears with an implied exponent of 1 (it's just $y'$). There are no functions like $\sin(y')$ or $e^{y'}$. The $\cos x$ on the right is a function of $x$ alone, which is fine — it doesn't involve any derivatives. So the equation is polynomial in derivatives.

3. Find the degree.

The highest-order derivative is $\frac{dy}{dx}$, and its power is 1. Therefore, the degree is 1.

(ii) $xy\frac{d^2y}{dx^2} + x\left(\frac{dy}{dx}\right)^2 - y\frac{dy}{dx} = 0$

1. Identify the highest derivative.

We see $\frac{d^2y}{dx^2}$ (the second derivative) and $\frac{dy}{dx}$ (the first derivative). The highest is the second derivative, so the order is 2.

2. Check if the equation is polynomial in derivatives.

Look at each term:

• $xy\frac{d^2y}{dx^2}$: the second derivative appears with exponent 1.
• $x\left(\frac{dy}{dx}\right)^2$: the first derivative appears with exponent 2.
• $- y\frac{dy}{dx}$: the first derivative appears with exponent 1.

All derivatives have whole-number exponents. There are no trigonometric, exponential, or logarithmic functions applied to any derivative. The equation is a polynomial in $y''$ and $y'$. So the degree is defined.

3. Find the degree. …

Method: Order and degree by inspection

Determine order and degree using the two standard rules, and know exactly when the degree fails to exist.

Steps

Step 1: Order = highest derivative present.

Scan every term and pick the derivative of greatest order.

Step 2: Is it polynomial in the derivatives?

Degree exists only when each derivative has an integer power and no derivative sits inside $\sin,\cos,\log,e^{(\cdot)}$ or a root.

Step 3: Degree = power of the highest-order derivative. …

Common Mistakes

Mistake 1: Confusing the highest power of a lower derivative with the degree.

Why it's wrong: in (ii), $\left(\tfrac{dy}{dx}\right)^2$ appears, but the degree is the power of the highest-order derivative $\tfrac{d^2y}{dx^2}$, which is $1$. Correct approach: degree looks only at the top-order derivative.

Mistake 2: Giving (iii) a degree. …