Determine the order and degree: $\dfrac{d^4y}{dx^4}+\sin\!\left(\dfrac{dy}{dx}\right)=0$

Question

Accepted Answer

$\dfrac{d^4y}{dx^4}+\sin\!\left(\dfrac{dy}{dx}\right)=0$ . Highest derivative $\dfrac{d^4y}{dx^4}$ : order $4$ . But $\dfrac{dy}{dx}$ (a lower-order derivative) sits inside $\sin(\cdot)$ , a transcendental function, and no algebraic step removes it — the equation can never be wri …

Start your 14-day free trial to unlock the full solution →

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 24 on this concept.

CBSE 2026Set ANNUAL1 mark
$(y''')^2+3(y'')^3+3xy'+5y=0$
›Reveal solutionSolution

The degree is the power of the highest-order derivative, once the equation is a polynomial in derivatives (no radicals/fractions of derivatives).

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

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. State whether the following statement is true or false: Order and degree of a differential equation are always positive integers.$
(a) True
(b) False
›Reveal solutionSolution

Order = highest derivative order and degree = power of the highest-order derivative (equation cleared of radicals/fractions in the derivatives); each is a positive integer, so the statement is True.

By definition, the order of a differential equation is the order of the highest derivative it contains — this is always a whole counting number, i.e. a positive integer ( $1, 2, 3, \dots$ ).

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}=\left(\dfrac{d^2y}{dx^2}\right)^{3/2}$
›Reveal solutionSolution

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

The given differential equation is

$\sqrt{1+\left(\frac{dy}{dx}\right)^2}=\left(\frac{d^2y}{dx^2}\right)^{3/2}$

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$\sqrt{\dfrac{dy}{dx}}-4\dfrac{dy}{dx}-7x=0$
(a) $1, 2$
(b) $2, 1$
(c) $2, 2$
(d) $1, 1$
›Reveal solutionSolution
Isolating the radical and squaring turns the equation into a polynomial in $dy/dx$ ; the power of the highest derivative after clearing the radical gives the degree.

Given $\sqrt{\dfrac{dy}{dx}}-4\dfrac{dy}{dx}-7x=0$ . Let $p=\dfrac{dy}{dx}$ .

Isolate the radical: $\sqrt p=4p+7x$ .

Square both sides to remove the radical (a fractional power of a derivative is not allowed for degree to be defined): $p=(4p+7x)^2=16p^2+56px+49x^2$ .

Rearranged: $16p^2+(56x-1)p+49x^2=0$ — a polynomial equation in $p=\dfrac{dy}{dx}$ . …
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$\left(\frac{d^2y}{dx^2}\right)^2 + \left(\frac{dy}{dx}\right)^2 = a^x$
(a) 1, 1
(b) 1, 2
(c) 2, 2
(d) 2, 1
›Reveal solutionSolution

The highest derivative is $\frac{d^2y}{dx^2}$ (order 2), and it appears raised to power 2 (degree 2).

The equation is

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

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

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set MARCH1 markMCQ
$\left(\dfrac{dx}{dy}\right)^{3}+2y^{1/2}=x$
(a) of order 1 and degree 6
(b) of order 2 and degree 1
(c) of order 1 and degree 2
(d) of order 1 and degree 3
›Reveal solutionSolution

The only derivative present is $\dfrac{dx}{dy}$ (order $1$ ), raised to power $3$ in a polynomial form, so the degree is $3$ ; option (d).

Order. The equation $\left(\dfrac{dx}{dy}\right)^{3}+2y^{1/2}=x$ contains only the first derivative $\dfrac{dx}{dy}$ , so the order is $1$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. Order and degree of the differential equation y = dy/dx + c/(dy/dx) are$
(a) 1, 2
(b) 2, 2
(c) 1, 1
(d) 2, 1
›Reveal solutionSolution

Rewrite the equation as a polynomial in $dy/dx$ ; the highest power of the highest-order derivative gives the degree.

Let $p = dy/dx$ . The equation $y = p + c/p$ becomes, after multiplying through by $p$ :

$yp = p^2 + c \quad\Rightarrow\quad p^2 - yp + c = 0$

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. State whether the following statement is true or false: Order and degree of a differential equation are always positive integers.$
(a) True
(b) False
›Reveal solutionSolution
The order of a differential equation is always a positive integer, and its degree — whenever it exists — is also a positive integer, so the statement is taken as True.

Order = the order of the highest-order derivative appearing in the equation. Since we count derivatives ( $\frac{dy}{dx}$ , $\frac{d^2y}{dx^2}$ , …), the order is always a positive integer such as $1, 2, 3, \dots$ …
Start your 14-day free trial to see the full solution →
CBSE 2024Set MARCH1 markMCQ
$\sqrt{\frac{d^2y}{dx^2}}=\sqrt{\frac{dy}{dx}+5}$
(a) 2 and 1
(b) 2 and 3
(c) 2 and 2
(d) 3 and 2
›Reveal solutionSolution

Square both sides $\Rightarrow \frac{d^2y}{dx^2}=\frac{dy}{dx}+5$ ; order $2$ , degree $1$ .

Starting from $\sqrt{\dfrac{d^2y}{dx^2}}=\sqrt{\dfrac{dy}{dx}+5}$ , square both sides to clear the radicals:

$\frac{d^2y}{dx^2}=\frac{dy}{dx}+5.$

…

Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 mark
$e^{\frac{dy}{dx}} + \dfrac{dy}{dx} = x$
›Reveal solutionSolution

The derivative appears inside an exponential (transcendental) term, so the equation cannot be written as a polynomial in $\frac{dy}{dx}$ .

Degree is defined only when the differential equation can be expressed as a polynomial in the derivatives. Here $\frac{dy}{dx}$ occurs as the exponent of $e$ (a non-polynomial/transcendental …

Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 markMCQ
$Q. Order of the differential equation (d²y/dx²)² = 1 + (dy/dx)³ is$
(a) 1
(b) 3
(c) 2
(d) 4
›Reveal solutionSolution

The order of a differential equation is the order of the highest derivative that appears in it — degree (the power it's raised to) is a separate idea.

Step 1. The equation is $\left(\dfrac{d^2y}{dx^2}\right)^2 = 1+\left(\dfrac{dy}{dx}\right)^3$ .

Step 2. The highest-order derivative present is $\dfrac{d^2y}{dx^2}$ , a second derivative.

…

Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 markMCQ
$\left(\frac{d^2y}{dx^2}\right)^2 + \left(\frac{dy}{dx}\right)^3 = a^x$
(a) True
(b) False
›Reveal solutionSolution
The highest-order derivative is $\dfrac{d^2y}{dx^2}$ , raised to power $2$ , so the degree is $2$ — the claim of $3$ is False.

The degree of a differential equation is the power to which the highest-order derivative is raised, once the equation is expressed as a polynomial in its derivatives (free of radicals and fractional powers).

The equation is:

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

The highest-order derivative is $\dfrac{d^2y}{dx^2}$ (order $2$ ).

Its power in the equation is $2$ . …
Start your 14-day free trial to see the full solution →

Q.Determine the order and degree: $\dfrac{d^4y}{dx^4}+\sin\!\left(\dfrac{dy}{dx}\right)=0$

Concept understanding — Order and Degree of a Differential Equation

Q.Determine the order and degree: d4ydx4+sin⁡ ⁣(dydx)=0\dfrac{d^4y}{dx^4}+\sin\!\left(\dfrac{dy}{dx}\right)=0dx4d4y​+sin(dxdy​)=0

🎓Unlock everything free for 14 days

Q.Determine the order and degree: $\dfrac{d^4y}{dx^4}+\sin\!\left(\dfrac{dy}{dx}\right)=0$

Unlock everything free for 14 days