What is the product of the order and degree of the differential equation $\frac{d^2 y}{dx^2} \sin y + \left(\frac{dy}{dx}\right)^3 \cos y = \sqrt{y}$ ? (a) $3$ (b) $2$ (c) $6$ (d) not defined

What is the product of the order and degree of the differential equation (d² y)/(dx²) sin y + ((dy)/(dx))³ cos

Question

\frac{d^2 y}{dx^2} \sin y + \left(\frac{dy}{dx}\right)^3 \cos y = \sqrt{y}

(a)

3

(b)

2

(c)

6

(d) not defined

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The order of the differential equation is 2 (highest derivative is $\frac{d^2 y}{dx^2}$ ), and its degree is 1 (the power of the highest order derivative). The product of the order and degree is $\boxed{2}$ .

When we talk about the "order" and "degree" of a differential equation, we are essentially classifying it based on the derivatives it contains. This classification helps us understand the nature of the equation and often guides us in choosing appropriate methods for solving it.

The order of a differential equation is straightforward: it's simply the order of the highest derivative present in the equation. For example, if the highest derivative is $\frac{d^3 y}{dx^3}$ , the order is 3.

The degree is a bit more nuanced. It is defined as the power of the highest order derivative, provided the differential equation can be expressed as a polynomial in its derivatives. This means there should be no fractional powers of derivatives, no radicals involving derivatives, and no transcendental functions (like $\sin(\frac{dy}{dx})$ or $e^{\frac{d^2 y}{dx^2}}$ ) of the derivatives. If the equation cannot be written as a polynomial in its derivatives, its degree is said to be "not defined". It's crucial to remember that functions of the dependent variable $y$ (like $\sin y$ or $\sqrt{y}$ ) do not affect whether the equation is a polynomial in its derivatives.

Let's apply these concepts to the given problem.

Identify the differential equation: The given differential equation is:

\frac{d^2 y}{dx^2} \sin y + \left(\frac{dy}{dx}\right)^3 \cos y = \sqrt{y}

Determine the Order:

We need to identify all the derivatives present in the equation and find the highest order among them.
- The first term contains $\frac{d^2 y}{dx^2}$ , which is a second-order derivative.
- The second term contains $\frac{dy}{dx}$ , which is a first-order derivative. The highest order derivative present in the equation is $\frac{d^2 y}{dx^2}$ . Therefore, the order of the differential equation is $\mathbf{2}$ .
Determine the Degree:

Before finding the degree, we must ensure that the differential equation is a polynomial in its derivatives. This means that the derivatives themselves (like $\frac{dy}{dx}$ or $\frac{d^2 y}{dx^2}$ ) should not be inside radicals, fractional powers, or transcendental functions.
- In our equation, the derivatives $\frac{d^2 y}{dx^2}$ and $\frac{dy}{dx}$ appear with integer powers (1 and 3, respectively). …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 93 on this concept.

CBSE 2026Set 65/1/11 markMCQ
$Q. The order and degree of the differential equation d dx(ey) = 0 respectively are 1 (A) 0, 1 (B) 1, 1 (C) 2, 1 (D) 1, not defined$
›Reveal solutionSolution
The given equation $\frac{d}{dx}(e^y)=0$ simplifies to $e^y \frac{dy}{dx}=0$ , which is a first-order differential equation. Since $e^y \neq 0$ , the highest derivative is $\frac{dy}{dx}$ raised to the power 1, so the degree is 1. The correct option is (B).

The order of a differential equation is the highest order derivative present. The degree is the power of the highest order derivative, provided the equation is polynomial in derivatives. Here, the equation looks deceptively simple — but we must first expand it properly.

Expand the derivative. The given equation is $\frac{d}{dx}(e^y) = 0$ . Using the chain rule:

$\frac{d}{dx}(e^y) = e^y \cdot \frac{dy}{dx}.$
So the equation becomes:
$e^y \frac{dy}{dx} = 0.$

Identify the highest derivative.

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

Determine the degree.

The degree is defined only when the differential equation is a polynomial in the derivatives. Here, the term $e^y$ is not a polynomial in $y$ or its derivatives — it's an exponential function of $y$ . However, the derivative $\frac{dy}{dx}$ itself appears with power 1, and the equation is already in the form $e^y \cdot \frac{dy}{dx} = 0$ .

Since $e^y$ is never zero for any real $y$ , we can divide both sides by $e^y$ to get:

$\frac{dy}{dx} = 0.$
This is a polynomial in $\frac{dy}{dx}$ (specifically, it is $\left(\frac{dy}{dx}\right)^1 = 0$ ). So the degree is $1$ . …
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CBSE 2026Set V11 mark
$[0, 3, -1, 2, -2, 1]$
›Reveal solutionSolution
Order $m=2$ and degree $n=1$ give $m+n=3$ .

In $2x^2\dfrac{d^2y}{dx^2}-3\dfrac{dy}{dx}+y=0$ :

The highest-order derivative is $\dfrac{d^2y}{dx^2}$ , so the order is $m=2$ . …
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CBSE 2026Set A1 markMCQ
$\left(\frac{d^2 s}{dt^2}\right)^2 + \left(\frac{ds}{dt}\right)^3 + 4 = 0$
(a) order = 2, degree = 1
(b) order = 2, degree = 2
(c) order = 1, degree = 2
(d) order = 1, degree = 1
›Reveal solutionSolution

Highest derivative is second order; it appears squared → order 2, degree 2.

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

…

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CBSE 2026Set ANNUAL1 markMCQ
$\frac{d^4y}{dx^4}-\sin\left(\frac{d^3y}{dx^3}\right)=0$
(a) 4
(b) 3
(c) 0
(d) Undefined
›Reveal solutionSolution

The degree of a differential equation is defined only when it can be written as a polynomial in derivatives; here a $\sin(\cdot)$ of a derivative appears, so the degree is undefined.

The degree of a differential equation is the highest power of the highest-order derivative, provided the equation is a polynomial in all its derivatives (after clearing any fractional powers/roots and radicals).

Here the equation is

$\frac{d^4y}{dx^4}-\sin\left(\frac{d^3y}{dx^3}\right)=0$

…

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CBSE 2026Set ANNUAL1 markMCQ
$2x^2 \frac{d^2y}{dx^2} - 3\frac{dy}{dx} + y = 0$
(a) $2$
(b) $1$
(c) $0$
(d) Not defined
›Reveal solutionSolution

The order of a differential equation is the order of the highest-order derivative appearing in it.

…

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CBSE 2026Set ANNUAL1 markMCQ
$\dfrac{d^2y}{dx^2} = \sqrt{1+\left(\dfrac{dy}{dx}\right)^2}$
(a) 1
(b) 2
(c) 3
(d) None of these
›Reveal solutionSolution

The order of a differential equation is the order of the highest derivative appearing in it.

…

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CBSE 2026Set ANNUAL1 markMCQ
$\left(\dfrac{d^2y}{dx^2}\right)^3 + \left(\dfrac{dy}{dx}\right)^2 + \sin\left(\dfrac{dy}{dx}\right) + 1 = 0$
(a) 3
(b) 2
(c) 1
(d) Not defined
›Reveal solutionSolution

Degree is defined only when the differential equation is a polynomial in derivatives; the $\sin(dy/dx)$ term breaks this.

The degree of a differential equation is the power of the highest-order derivative, but this is only defined when the equation can be written as a polynomial in the derivatives (no transcendental functions of a derivative).

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Order of differential equation d²y/dx² - 2 dy/dx + 3y = 0 is:$
(a) 3
(b) 2
(c) 1
(d) 0
›Reveal solutionSolution

The order of a differential equation is the order of the highest derivative present in it.

The given differential equation is:

$\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 3y = 0$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q.Order and degree of differential equation \left(\dfrac{d^2y}{dx^2}\right)^3 = \left[y + \left(\dfrac{dy}{dx}\right)^2\right] are:$
(a) 2, 4
(b) 2, 3
(c) 2, 2
(d) 1, 2
›Reveal solutionSolution

Order = highest-order derivative present; degree = power of that highest-order derivative, once the equation is a polynomial in derivatives.

Working: The equation is

$\left(\frac{d^2y}{dx^2}\right)^3 = y + \left(\frac{dy}{dx}\right)^2$

The highest-order derivative present is $\dfrac{d^2y}{dx^2}$ (a second derivative) $\Rightarrow$ order $=2$ . …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Order and degree of differential equation (y''')² + (y'')³ + (y')⁴ + y⁵ = 0 is$
(a) Order = 2, Degree = 3
(b) Order = 3, Degree = 2
(c) Order = 1, Degree = 4
(d) Order = 3, Degree = 3
›Reveal solutionSolution

Order is the index of the highest derivative present; degree is the power of that highest derivative once the equation is polynomial in derivatives.

Given: $(y''')^2 + (y'')^3 + (y')^4 + y^5 = 0$ .

Order: the highest-order derivative appearing is $y'''$ (the third derivative), so the order is 3.

…

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CBSE 2026Set ANNUAL1 markMCQ
$\left(\dfrac{dy}{dx}\right)^4 + 3y\dfrac{d^2y}{dx^2} = 0$
(a) 2 and 4
(b) 4 and 2
(c) 2 and 1
(d) 1 and 2
›Reveal solutionSolution

The order is the highest derivative present; the degree is the power of that highest-order derivative once the equation is written as a polynomial in the derivatives.

Given: $\left(\dfrac{dy}{dx}\right)^4 + 3y\dfrac{d^2y}{dx^2} = 0$

Order: The highest-order derivative appearing is $\dfrac{d^2y}{dx^2}$ , a second-order derivative. So order $=2$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The number of arbitrary constants in the general solution of a differential equation of third order will be:$
(a)
(i) 0
(b)
(ii) 1
(c)
(iii) 2
(d)
(iv) 3
›Reveal solutionSolution
A third-order differential equation has $3$ arbitrary constants in its general solution — option (iv).

Concept. The order of a differential equation is the order of its highest derivative, and the general solution contains exactly as many independent arbitrary constants as the order (each integration introduces one constant).

Steps.

Order of the equation $=3$ . …
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Q.What is the product of the order and degree of the differential equation $\frac{d^2 y}{dx^2} \sin y + \left(\frac{dy}{dx}\right)^3 \cos y = \sqrt{y}$ ?

Concept understanding — Order Of Differential Equation

Order of a Differential Equation

Some examples

Why order matters …

Q.What is the product of the order and degree of the differential equation d2ydx2sin⁡y+(dydx)3cos⁡y=y\frac{d^2 y}{dx^2} \sin y + \left(\frac{dy}{dx}\right)^3 \cos y = \sqrt{y}dx2d2y​siny+(dxdy​)3cosy=y​ ?

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Q.What is the product of the order and degree of the differential equation $\frac{d^2 y}{dx^2} \sin y + \left(\frac{dy}{dx}\right)^3 \cos y = \sqrt{y}$ ?

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