Find the order and degree (if defined) of the differential equation $\dfrac{d^2y}{dx^2} + x\left(\dfrac{dy}{dx}\right)^2 = 2x^2 \log\left(\dfrac{d^2y}{dx^2}\right)$.

Find the order and degree (if defined) of the differential equation (d²y)/(dx²) + x((dy)/(dx))² = 2x² log((d²y

Question

\dfrac{d^2y}{dx^2} + x\left(\dfrac{dy}{dx}\right)^2 = 2x^2 \log\left(\dfrac{d^2y}{dx^2}\right)

CBSECBSE Class XII Board 2019Subjective· 1mImportance★★★★★

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The differential equation is not a polynomial in its highest-order derivative $\frac{d^2y}{dx^2}$ because of the logarithm term, so its degree is not defined. The order is 2 (the highest derivative present).

Concept first: Order and degree — what they really mean

The order of a differential equation is simply the highest derivative that appears. That part is straightforward: you look for the largest $n$ such that $\frac{d^n y}{dx^n}$ shows up.

The degree is trickier. It is defined only when the differential equation is a polynomial in all the derivatives that appear. That means every term involving $y$ or its derivatives must be a non-negative integer power of those derivatives — no square roots, no logarithms, no sines, no fractional powers of a derivative. If the equation contains something like $\log(y'')$ or $\sqrt{y'}$ or $\sin(y')$ , then it is not a polynomial in the derivatives, and the degree is simply not defined.

Watch out

A common mistake: students try to "rearrange" a non-polynomial equation into polynomial form by, say, exponentiating both sides. But that changes the equation — you cannot force a degree where none exists. The degree is a property of the equation as given, not of some transformed version.

Step-by-step solution

1. Identify the highest-order derivative

The given equation is:

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

The derivatives present are:

$\frac{dy}{dx}$ (first derivative)
$\frac{d^2y}{dx^2}$ (second derivative)

The highest order is 2, so the order is $2$ .

2. Check if the equation is a polynomial in the derivatives …

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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.Find the order and degree (if defined) of the differential equation $\dfrac{d^2y}{dx^2} + x\left(\dfrac{dy}{dx}\right)^2 = 2x^2 \log\left(\dfrac{d^2y}{dx^2}\right)$ .

Concept understanding — Order Of Differential Equation

Order of a Differential Equation

Some examples

Why order matters …

Concept first: Order and degree — what they really mean

Step-by-step solution

Q.Find the order and degree (if defined) of the differential equation d2ydx2+x(dydx)2=2x2log⁡(d2ydx2)\dfrac{d^2y}{dx^2} + x\left(\dfrac{dy}{dx}\right)^2 = 2x^2 \log\left(\dfrac{d^2y}{dx^2}\right)dx2d2y​+x(dxdy​)2=2x2log(dx2d2y​).

Concept first: Order and degree — what they really mean

Step-by-step solution

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Q.Find the order and degree (if defined) of the differential equation $\dfrac{d^2y}{dx^2} + x\left(\dfrac{dy}{dx}\right)^2 = 2x^2 \log\left(\dfrac{d^2y}{dx^2}\right)$ .

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