Which of the following is a second order differential equation? (A) $(y')^2+x=y^2$ (B) $y'y''+y=\sin x$ (C) $y'''+(y'')^2+y=0$ (D) $y'=y^2$

Which of the following is a second order differential equation? (A) (y')²+x=y² (B) y'y''+y=sin x (C) y'''+(y''

NCERT Exemplar · Q68

(y')^2+x=y^2

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The order of a differential equation is the highest derivative present. Only option (B) contains a second derivative ( $y''$ ) as the highest derivative, making it the second-order equation.

The order of a differential equation is defined as the order of the highest derivative that appears in it. This is a definition-based question — you don’t need to solve anything, just identify which equation has $y''$ as its highest derivative and no derivative higher than that.

Let’s check each option one by one.

Option (A): $(y')^2 + x = y^2$

The only derivative here is $y'$ (first derivative). There is no $y''$ or higher. So the order is 1, not 2.
Option (B): $y' y'' + y = \sin x$

Here we see $y''$ (second derivative) multiplied by $y'$ . The highest derivative present is $y''$ , and there is no $y'''$ or higher. So the order is 2. This matches what we need.
Option (C): $y''' + (y'')^2 + y = 0$

This contains $y'''$ (third derivative). Even though $y''$ also appears, the highest derivative is $y'''$ , so the order is 3, not 2.
Option (D): $y' = y^2$

Only $y'$ appears — order 1. …

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

COMEDK 2023Set 2023-E1 markMCQ
$\left[1-\left(\frac{d y}{d x}\right)^2\right]^{\frac{3}{2}}=k x \frac{d^2 y}{d x^2}$
›Reveal solutionSolution

Sum = order + degree = 2 + 2 = 4.

Concept: order = highest derivative present; degree = power of the highest-order derivative AFTER the equation is made free of radicals/fractional powers in the derivatives.

Equation: [1 - (dy/dx)^2]^(3/2) = k x (d^2y/dx^2).

Order: the highest derivative is d^2y/dx^2, so order = 2.

Degree: square both sides to clear the 3/2 power:

[1 - (dy/dx)^2]^3 = k^2 x^2 (d^2y/dx^2)^2. …
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COMEDK 2026Set 2026-M1 markMCQ
$\left(\frac{d y}{d x}\right)^2+\frac{d x}{d y}=x$
›Reveal solutionSolution

Clearing the fraction turns the equation into a polynomial in $\frac{dy}{dx}$ : highest derivative is order $1$ and its highest power is $3$ , so order and degree are $(1,3)$ .

Given equation.
$\left(\frac{dy}{dx}\right)^2+\frac{dx}{dy}=x.$
Express $\frac{dx}{dy}$ in terms of $\frac{dy}{dx}$ . Since $\dfrac{dx}{dy}=\dfrac{1}{\,dy/dx\,}$ , write $p=\dfrac{dy}{dx}$ :
$p^2+\frac{1}{p}=x.$
Make it a polynomial in the derivative (degree is defined only after removing radicals and fractional powers of the derivatives). Multiply through by $p$ : …
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COMEDK 2025Set 2025-M1 markMCQ
$\frac{d}{d x}\left[\left(\frac{d y}{d x}\right)^3\right]=0$
›Reveal solutionSolution
The order of a differential equation is the highest derivative present after the equation is simplified. Here, the left-hand side is a derivative of a cubic power of the first derivative, so the highest derivative that actually appears is the second derivative. The order is 2.

The key idea: Order is defined as the highest order derivative that appears in the equation after it has been cleared of radicals and fractions. We must first simplify the given expression to see what derivatives are actually present.

Let’s work through it step by step.

Rewrite the given equation The equation is

$\frac{d}{dx}\left[\left(\frac{dy}{dx}\right)^3\right] = 0.$
The expression inside the derivative is $\left(\frac{dy}{dx}\right)^3$ , which is a function of the first derivative only.

Apply the chain rule to differentiate Let $p = \frac{dy}{dx}$ . Then the equation becomes

$\frac{d}{dx}(p^3) = 0.$
Using the chain rule:
$\frac{d}{dx}(p^3) = 3p^2 \cdot \frac{dp}{dx} = 3\left(\frac{dy}{dx}\right)^2 \cdot \frac{d^2y}{dx^2}.$
So the equation simplifies to
$3\left(\frac{dy}{dx}\right)^2 \frac{d^2y}{dx^2} = 0.$

Identify the highest derivative

In the simplified form, we see $\frac{d^2y}{dx^2}$ (the second derivative) explicitly. There is no third derivative or higher. Therefore, the highest order derivative present is 2.

Check for any hidden higher derivatives …
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COMEDK 2025Set 2025-A1 markMCQ
$\left[1+\left(\frac{d y}{d x}\right)^2\right]^{\frac{3}{4}}=\left(\frac{d^2 y}{d x^2}\right)^{\frac{1}{3}}$
›Reveal solutionSolution
The degree of a differential equation is the power of the highest-order derivative after the equation is made polynomial in derivatives. Here, raising both sides to the 12th power clears fractional exponents, giving the highest derivative $\frac{d^2y}{dx^2}$ raised to the power 4, so the degree is 4. The correct option is (A).

The key idea: Degree is defined only when the differential equation is a polynomial in the derivatives. Fractional exponents must be eliminated by raising both sides to a suitable power, but we must be careful not to introduce extraneous roots or change the equation’s essential nature.

Why this approach works

The given equation is:
$\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/4} = \left(\frac{d^2y}{dx^2}\right)^{1/3}$
Both sides have fractional exponents. To find the degree, we need the highest-order derivative (here $\frac{d^2y}{dx^2}$ ) to appear with an integer exponent, and the equation must be a polynomial in all derivatives. The trick: find the least common multiple of the denominators of the exponents (4 and 3) and raise both sides to that power.

Step-by-step solution

Identify the highest-order derivative

The highest derivative present is $\frac{d^2y}{dx^2}$ (order 2). The equation also contains $\frac{dy}{dx}$ (order 1), but the degree is determined by the highest-order derivative.

Clear the fractional exponents

The exponents are $\frac{3}{4}$ on the left and $\frac{1}{3}$ on the right. The LCM of 4 and 3 is 12. Raise both sides to the 12th power:

$\left[ \left(1 + \left(\frac{dy}{dx}\right)^2\right)^{3/4} \right]^{12} = \left[ \left(\frac{d^2y}{dx^2}\right)^{1/3} \right]^{12}$
This simplifies to:
$\left(1 + \left(\frac{dy}{dx}\right)^2\right)^{9} = \left(\frac{d^2y}{dx^2}\right)^{4}$
because $\frac{3}{4} \times 12 = 9$ and $\frac{1}{3} \times 12 = 4$ .

Check polynomial form …
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COMEDK 2026Set 2026-A1 markMCQ
$\text { The degree of the differential equation } \sqrt{1+\left(\frac{d y}{d x}\right)^{1 / 3}}=\frac{d^2 y}{d x^2} \text { is: }$
›Reveal solutionSolution
Clear the radicals so the equation is polynomial in the derivatives: the highest-order term $\frac{d^2y}{dx^2}$ ends up raised to the power $6$ , so the degree is $\mathbf{6}$ — option (A).

Concept & Intuition

The degree of a differential equation is the power of the highest-order derivative after the equation has been made a polynomial in all its derivatives (all radicals and fractional powers removed).

Step-by-step solution

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

Square both sides to remove the square root:

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

Isolate the cube-root term and cube both sides to remove the $1/3$ power: $\left(\frac{dy}{dx}\right)^{1/3}=\left(\frac{d^2y}{dx^2}\right)^{2}-1,$ …
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COMEDK 2024Set 2024-A1 markMCQ
$\log \left(\frac{d y}{d x}\right)^{\frac{1}{2}}=5 x+4 y$
›Reveal solutionSolution

After removing the logarithm the equation becomes $\frac{dy}{dx}=e^{10x+8y}$ ; the highest derivative appears to the first power, so the degree is $1$ .

Start from $\log\left(\frac{dy}{dx}\right)^{1/2}=5x+4y$ , i.e. $\frac{1}{2}\log\frac{dy}{dx}=5x+4y$ .

Multiply by 2 and exponentiate:

$\log\frac{dy}{dx}=10x+8y \;\Rightarrow\; \frac{dy}{dx}=e^{10x+8y}.$ …
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COMEDK 2024Set 2024-E1 markMCQ
$\left(\frac{d^2 y}{d x^2}\right)^5+\frac{4\left(\frac{d^2 y}{d x^2}\right)^3}{\left(\frac{d^3 y}{d x^3}\right)}+\frac{d^3 y}{d x^3}=x^2-1$
›Reveal solutionSolution
The key is to rewrite the equation so that every derivative appears with a positive integer exponent, then identify the highest order (3) and the degree (3) — their sum is 6.

We are given the differential equation:
$\left(\frac{d^2 y}{d x^2}\right)^5+\frac{4\left(\frac{d^2 y}{d x^2}\right)^3}{\left(\frac{d^3 y}{d x^3}\right)}+\frac{d^3 y}{d x^3}=x^2-1$
Concept and intuition:

The order of a differential equation is the highest derivative present. The degree is the power of that highest derivative after the equation has been made polynomial in all derivatives (i.e., no fractional powers, no derivatives in denominators). Here, the term $\frac{4(d^2y/dx^2)^3}{d^3y/dx^3}$ has a derivative in the denominator, so we must first clear that denominator to find the true degree.

Identify the highest-order derivative.

The derivatives appearing are $\frac{d^2y}{dx^2}$ (order 2) and $\frac{d^3y}{dx^3}$ (order 3). The highest is order 3. So the order is $3$ .

Rewrite to remove the denominator.

Multiply the entire equation by $\frac{d^3y}{dx^3}$ to eliminate the fraction:

$\left(\frac{d^2y}{dx^2}\right)^5 \cdot \frac{d^3y}{dx^3} + 4\left(\frac{d^2y}{dx^2}\right)^3 + \left(\frac{d^3y}{dx^3}\right)^2 = (x^2-1)\frac{d^3y}{dx^3}$

Bring all terms to one side. Rearranging:

$\left(\frac{d^2y}{dx^2}\right)^5 \cdot \frac{d^3y}{dx^3} + 4\left(\frac{d^2y}{dx^2}\right)^3 + \left(\frac{d^3y}{dx^3}\right)^2 - (x^2-1)\frac{d^3y}{dx^3} = 0$

Determine the degree. The degree is the exponent of the highest-order derivative ( $\frac{d^3y}{dx^3}$ ) when the equation is a polynomial in derivatives. Look at each term involving $\frac{d^3y}{dx^3}$ : …
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KCET 2026Set UNKNOWN1 markMCQ
$2y' + (y'')^2 = \sqrt{y'' - 3}$
›Reveal solutionSolution

Identify the order (highest derivative present) and degree (its power once the radical is cleared), then add their squares.

Step 1 — Identify the order

$2y' + (y'')^2 = \sqrt{y'' - 3}$ . The highest-order derivative present is $y''$ , so the order is $2$ .

Step 2 — Clear the radical to find the degree

The degree of a differential equation is defined only after it is written as a polynomial in the derivatives, free of radicals/fractional powers. Since $y''$ sits inside a square root on the right, square both sides:
$\left(2y' + (y'')^2\right)^2 = y'' - 3$
Step 3 — Read off the degree …
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Q.Which of the following is a second order differential equation?
(A) $(y')^2+x=y^2$
(B) $y'y''+y=\sin x$
(C) $y'''+(y'')^2+y=0$
(D) $y'=y^2$

Concept understanding — Order Of Differential Equation

Order of a Differential Equation

Some examples

Why order matters …

Method: Identifying the Order of a Differential Equation

Steps

Common Mistakes

Why this approach works

Step-by-step solution

Step 1 — Identify the order

Step 2 — Clear the radical to find the degree

Step 3 — Read off the degree …

Q.Which of the following is a second order differential equation? (A) (y′)2+x=y2(y')^2+x=y^2(y′)2+x=y2 (B) y′y′′+y=sin⁡xy'y''+y=\sin xy′y′′+y=sinx (C) y′′′+(y′′)2+y=0y'''+(y'')^2+y=0y′′′+(y′′)2+y=0 (D) y′=y2y'=y^2y′=y2

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Q.Which of the following is a second order differential equation?
(A) $(y')^2+x=y^2$
(B) $y'y''+y=\sin x$
(C) $y'''+(y'')^2+y=0$
(D) $y'=y^2$

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