NCERT Class 12 Mathematics — Differential Equations, Exercise 9.1. All 12 questions solved.

The whole exercise tests two definitions, and one caveat attached to the second.

Order is the order of the highest derivative appearing in the equation. It is always defined, and it is always a positive integer.

Degree is the power to which that highest derivative is raised — but only after the equation has been written as a polynomial in the derivatives. If it cannot be written that way, the degree is not defined.

Key insight. Look at how the highest derivative appears, not just at what it is. If it sits inside a $\sin$, $\cos$, $\log$ or exponential — anywhere it cannot be expanded into a finite polynomial — the degree is undefined, no matter how simple the rest of the equation looks.

The order is unaffected by any of this. Question 1 has order $4$ and no degree; question 4 has order $2$ and no degree. Both times the only thing spoiling the degree is a derivative trapped inside a trigonometric function.

Watch also for a lower derivative inside a transcendental function. In question 4, $\left(\tfrac{d^2y}{dx^2}\right)^2$ is a clean square, but the $\cos\left(\tfrac{dy}{dx}\right)$ alongside it means the equation is not a polynomial in the derivatives, so the degree is still undefined.

Determine order and degree (if defined) of the differential equations given in Exercises 1 to 10.

Question 1

$\dfrac{d^4y}{dx^4} + \sin\left(y”’\right) = 0$

Solution. The highest derivative is $\dfrac{d^4y}{dx^4}$, so the order is 4.

The equation contains $\sin\left(y”’\right)$, a derivative inside a sine, so it is not a polynomial in the derivatives and the degree is not defined.

Question 2

$y’ + 5y = 0$

Solution. The highest derivative is $y’$, so the order is 1. The equation is a polynomial in $y’$, which appears to the first power, so the degree is 1.

Question 3

$\left(\dfrac{ds}{dt}\right)^4 + 3s\dfrac{d^2s}{dt^2} = 0$

Solution. Here $s$ is the dependent variable and $t$ the independent one. The highest derivative is $\dfrac{d^2s}{dt^2}$, so the order is 2.

The degree is the power of the highest derivative, which is $1$ — the fourth power belongs to $\dfrac{ds}{dt}$, a lower derivative, and does not count. So the degree is 1.

Question 4

$\left(\dfrac{d^2y}{dx^2}\right)^2 + \cos\left(\dfrac{dy}{dx}\right) = 0$

Solution. The highest derivative is $\dfrac{d^2y}{dx^2}$, so the order is 2.

Although that derivative appears as a clean square, the term $\cos\left(\tfrac{dy}{dx}\right)$ contains a derivative inside a cosine. The equation is therefore not a polynomial in its derivatives, and the degree is not defined.

Question 5

$\dfrac{d^2y}{dx^2} = \cos 3x + \sin 3x$

Solution. Order 2. The right-hand side involves only $x$, not any derivative, so it does not affect the degree at all. The highest derivative appears to the first power: degree 1.

Question 6

$\left(y”’\right)^2 + \left(y”\right)^3 + \left(y’\right)^4 + y^5 = 0$

Solution. The highest derivative is $y”’$, so the order is 3. It appears squared, and everything in the equation is a polynomial in the derivatives, so the degree is 2.

The cubes, fourth powers and fifth power belong to lower derivatives and to $y$ itself; none of them is the degree.

Question 7

$y”’ + 2y” + y’ = 0$

Solution. Highest derivative $y”’$, appearing to the first power, in a polynomial equation.

Question 8

$y’ + y = e^x$

Solution. Order 1, degree 1. The exponential is a function of $x$ alone, not of a derivative, so it does not spoil the polynomial form.

Question 9

$y” + \left(y’\right)^2 + 2y = 0$

Solution. Highest derivative $y”$, to the first power: order 2, degree 1. The square on $y’$ is irrelevant to the degree.

Question 10

$y” + 2y’ + \sin y = 0$

Solution. Order 2. The term $\sin y$ is transcendental, but it involves the dependent variable $y$, not a derivative. The definition of degree concerns only whether the equation is a polynomial in the derivatives, so $\sin y$ does not disqualify it. The highest derivative appears to the first power: degree 1.

This is the question that separates students who have memorised “any trig function means no degree” from those who have read the definition.

Question 11

The degree of the differential equation $\left(\dfrac{d^2y}{dx^2}\right)^3 + \left(\dfrac{dy}{dx}\right)^2 + \sin\left(\dfrac{dy}{dx}\right) + 1 = 0$ is

Solution. The cube on $\dfrac{d^2y}{dx^2}$ makes option (A) tempting, but the term $\sin\left(\tfrac{dy}{dx}\right)$ puts a derivative inside a sine. The equation is not a polynomial in its derivatives, so no degree exists.

Question 12

The order of the differential equation $2x^2\dfrac{d^2y}{dx^2} – 3\dfrac{dy}{dx} + y = 0$ is

Solution. The highest derivative is the second, so the order is $2$. The coefficient $2x^2$ is a function of $x$ and has no bearing on the order.

Order is always defined for a differential equation, which rules out (D) without any calculation.

Common mistakes

• Reading the degree off the wrong derivative. In questions 3 and 6 a lower derivative carries a higher power. The degree belongs to the highest-order derivative only.
• Declaring the degree undefined because of $\sin y$. Question 10 has a sine, and its degree is $1$, because the sine acts on $y$ and not on a derivative.
• Declaring the degree undefined because of $e^x$ or $\cos 3x$. Questions 5 and 8 have transcendental functions of $x$. Those are just coefficients.
• Thinking the order can be undefined. Every differential equation has a highest derivative and hence an order. Only the degree can fail to exist.
• Forgetting to clear radicals or fractions first. A degree can only be read once the equation is in polynomial form; if the original has $\sqrt{1 + \left(\tfrac{dy}{dx}\right)^2}$ it must be squared first.

Practise next

• Exercise 9.2 — verifying that a given function solves a given differential equation.
• Miscellaneous Exercise on Chapter 9 — question 1 there is three more order and degree problems, one of them harder than any here.