Q.Determine the order and degree, if defined, of the differential equation: $\left(\frac{ds}{dt}\right)^4 + 3s \frac{d^2s}{dt^2} = 0$

Concept: Order and Degree of a Differential Equation

The order is the highest derivative present. Here, $\frac{d^2s}{dt^2}$ is the highest, so the order is 2.

The degree is the power of the highest-order derivative, provided the equation is polynomial in derivatives. The term $\frac{d^2s}{dt^2}$ appears to the first power, and no radicals or fractions make it non-polynomial. Hence the degree is 1.

1. What do "order" and "degree" mean here?

Order is the highest derivative present in the equation. Degree is the power of that highest derivative after the equation has been made free of radicals and fractions in the derivatives — essentially, after it is written as a polynomial in the derivatives.

The key point: degree is defined only when the equation is a polynomial in the derivatives. If the highest derivative appears inside a sine, a square root, or a logarithm, degree is not defined.

2. Identify the highest derivative

Look at the terms:

• $\left(\frac{ds}{dt}\right)^4$ — this is the first derivative $\frac{ds}{dt}$ raised to the fourth power.
• $3s \frac{d^2s}{dt^2}$ — this is the second derivative $\frac{d^2s}{dt^2}$ multiplied by $3s$.

The highest derivative present is $\frac{d^2s}{dt^2}$. So the order is $2$.

3. Check if degree is defined

The equation is:

There are no radicals, no fractional powers, no trigonometric or logarithmic functions applied to any derivative. Every derivative appears as a simple power (the first derivative to the power $4$, the second derivative to the power $1$). So the equation is already a polynomial in the derivatives.

Thus, degree is defined.

4. Find the degree

The highest derivative is $\frac{d^2s}{dt^2}$. In the equation, it appears as:

That is, it is raised to the first power (exponent $1$). The exponent of the highest derivative is $1$.

Therefore, the degree is $1$.

5. Final answer

Method: Order and degree when the lower derivative carries a high power

Use this on equations such as $\left(\frac{ds}{dt}\right)^4+3s\frac{d^2s}{dt^2}=0$.

Steps

Step 1: Pick the highest derivative for the order.

$\frac{d^2s}{dt^2}$ is present, so the order is $2$ — the exponent $4$ on the first derivative is irrelevant to the order.

Step 2: Confirm polynomial form.

Every derivative appears as a plain power, no radicals or transcendental wrappers, so the degree is defined.

Step 3: Degree = exponent of the highest derivative only.

$\frac{d^2s}{dt^2}$ sits at power $1$, so the degree is $1$. The fourth power belongs to the lower-order $\frac{ds}{dt}$ and never sets the degree.

Common Mistakes

Mistake 1: Letting the exponent $4$ decide the order.

Why it's wrong: in $\left(\frac{ds}{dt}\right)^4+3s\frac{d^2s}{dt^2}=0$ the power $4$ sits on the first derivative; the order is fixed by the highest derivative $\frac{d^2s}{dt^2}$, so order $=2$. Correct approach: the exponent affects degree, never order.

Mistake 2: Taking the degree as $4$.

Why it's wrong: degree is the power of the highest-order derivative $\frac{d^2s}{dt^2}$, which is $1$ — not the largest exponent in the equation. Correct approach: degree $=1$.