Q.For each of the differential equations given below, indicate its order and degree (if defined).

Showing the 12 most recent of 93 on this concept.

1. CBSE 2019Set 65/2/11 mark
   Q.Find the order and degree (if defined) of the differential equation $\dfrac{d^2y}{dx^2} + x\left(\dfrac{dy}{dx}\right)^2 = 2x^2 \log\left(\dfrac{d^2y}{dx^2}\right)$.

   ›Reveal solutionSolution

   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

   Look at the term $\log\left(\frac{d^2y}{dx^2}\right)$. This is a logarithm of the second derivative. No amount of algebraic manipulation (adding, multiplying, raising to powers) will turn $\log(y'')$ into a polynomial in $y''$. The equation is therefore not a polynomial in $\frac{d^2y}{dx^2}$.

   3. Conclude about the degree

   Since the equation is not a polynomial in the derivatives, the degree is not defined.

   Tip

   If you ever see $\sin(y')$, $\cos(y'')$, $\log(y')$, $e^{y'}$, or $\sqrt{y''}$ in a differential equation, the degree is automatically not defined — no need to check further. The degree exists only when every derivative appears only in non-negative integer powers.

   ---

   ✓Final answer

   The order is 2 and the degree is not defined.

2. CBSE 2023Set 65/1/11 markMCQ
   Q.The sum of the order and the degree of the differential equation $\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^3 = \sin y$ is : (A) 5 (B) 2 (C) 3 (D) 4

   ›Reveal solutionSolution

   The order is the highest derivative (2), and the degree is the power of that derivative after removing radicals/fractions (1). Their sum is $2 + 1 = 3$.

   The key to this problem is understanding two separate definitions: order and degree of a differential equation. They are often confused, but once you separate them, the question becomes straightforward.

   Order is simply the highest derivative present. Look at the equation and find the derivative with the most number of primes (or the highest $n$ in $\frac{d^n y}{dx^n}$). Here we have $\frac{d^2y}{dx^2}$ (second derivative) and $\frac{dy}{dx}$ (first derivative). The highest is the second derivative, so the order is $2$.

   Degree is trickier. It is defined as the power of the highest derivative term, but only after the equation has been made free of radicals and fractions in the derivatives. That means: no square roots, cube roots, or fractional powers involving any derivative. Also, no derivative should appear inside a denominator (like $\frac{1}{dy/dx}$). In this equation, every derivative term is already a polynomial in the derivatives — there are no roots, no fractions, and no transcendental functions applied to derivatives. The highest derivative $\frac{d^2y}{dx^2}$ appears to the first power (exponent $1$). So the degree is $1$.

   Watch out

   A common mistake is to think the degree is $3$ because of the $(\frac{dy}{dx})^3$ term. But degree is defined only with respect to the highest derivative, not any lower derivative. The cube on the first derivative does not affect the degree.

   Now we simply add: order $2$ + degree $1$ = $3$.

   Tip

   If the equation had something like $\sqrt{\frac{d^2y}{dx^2}} + \frac{dy}{dx} = 0$, you would first square both sides to get $\frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2$, and then the degree of the highest derivative would be $1$ (since after squaring, the second derivative appears to the first power). Always check for radicals first.

   ✓Final answer

   The sum of the order and degree is $3$, which corresponds to option (C).

3. CBSE 2025Set 65/4/11 markMCQ
   Q.The order and degree of the differential equation $\left(\frac{d^2y}{dx^2}\right)^2 + \left(\frac{dy}{dx}\right)^2 = x \sin\left(\frac{dy}{dx}\right)$ are : (A) order 2, degree 2 (B) order 2, degree 1 (C) order 2, degree not defined (D) order 1, degree not defined

   ›Reveal solutionSolution

   The highest derivative is $\frac{d^2y}{dx^2}$, so the order is 2. The equation cannot be written as a polynomial in derivatives because of $\sin\left(\frac{dy}{dx}\right)$, so the degree is not defined. Answer: (C).

   The order of a differential equation is straightforward: it's the highest derivative that appears. The degree, however, requires more care. Degree is defined only when the equation can be expressed as a polynomial in all its derivatives (after clearing radicals and fractions). If transcendental functions like sine, cosine, exponential, or logarithm are applied to derivatives, the degree doesn't exist.

   Let me identify what we have in this equation.

   1. Finding the order

      The derivatives present are $\frac{dy}{dx}$ (first derivative) and $\frac{d^2y}{dx^2}$ (second derivative). The highest derivative is the second derivative.

      Therefore, the order is 2.

   2. Checking if the equation is a polynomial in derivatives

      For degree to be defined, we need the equation in the form of a polynomial in $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$. Let's examine each term:

      • $\left(\frac{d^2y}{dx^2}\right)^2$ is a polynomial term (power 2 in the second derivative)
      • $\left(\frac{dy}{dx}\right)^2$ is a polynomial term (power 2 in the first derivative)
      • $x \sin\left(\frac{dy}{dx}\right)$ contains a transcendental function applied to the derivative

   3. Why the degree is not defined

      The term $\sin\left(\frac{dy}{dx}\right)$ is the problem. This is not a polynomial expression in $\frac{dy}{dx}$. You cannot expand sine into a finite polynomial.

      Even if you tried to manipulate the equation, there's no algebraic way to eliminate the sine function and express everything as a polynomial in the derivatives.

   Watch out

   A common mistake is to think degree is always defined. Remember: degree exists only when the equation is (or can be rewritten as) a polynomial in all derivatives. Transcendental functions of derivatives immediately disqualify the equation from having a defined degree.

   Tip

   Quick check for degree: scan for $\sin$, $\cos$, $e^{(\cdot)}$, $\log$, etc. applied to any derivative. If present, degree is not defined.

   ✓Final answer

   The correct option is (C): order 2, degree not defined.

4. CBSE 2024Set 65/2/11 markMCQ
   Q.The degree of the differential equation $(y'')^2 + (y')^3 = x \sin(y')$ is: (A) $1$ (B) $2$ (C) $3$ (D) not defined

   ›Reveal solutionSolution

   The degree of a differential equation is defined only when the equation is a polynomial in the derivatives. Here, the term $\sin(y')$ is non-polynomial, so the degree is not defined. The correct option is (D).

   The degree of a differential equation is a precise, formal property — it is not just the highest power of the highest derivative you see. For the degree to exist, the equation must be a polynomial in all the derivatives that appear. That means every derivative term (like $y'$, $y''$, etc.) must be raised only to a non-negative integer power, and no transcendental functions (sine, cosine, exponential, log) can wrap around any derivative.

   Here, the equation is:

   $(y'')^2 + (y')^3 = x \sin(y')$

   The left-hand side is fine: $(y'')^2$ and $(y')^3$ are polynomial in $y''$ and $y'$. But the right-hand side contains $\sin(y')$ — the sine of the first derivative. That is not a polynomial in $y'$; it is a transcendental function of $y'$. So the equation as a whole is not a polynomial in the derivatives.

   Because the definition of degree requires a polynomial form, the degree simply does not exist here.

   Let’s walk through the reasoning step by step.

   1. Recall the definition of degree.

      The degree of a differential equation is the power of the highest-order derivative, provided the equation is a polynomial in all the derivatives. If any derivative appears inside a non-polynomial function (like $\sin$, $\cos$, $e$, $\log$, etc.), the degree is not defined.

   2. Identify the highest-order derivative.

      The highest derivative present is $y''$ (second order). It appears as $(y'')^2$, which is polynomial. So the order is $2$, but that is not what we are asked.

   3. Check the condition for degree.

      Look at the term $\sin(y')$. Here $y'$ is inside a sine function. No matter how you rewrite it, you cannot express $\sin(y')$ as a finite sum of powers of $y'$ — it is inherently non-polynomial. Therefore the equation is not a polynomial in $y'$.

   4. Conclude about degree.

      Since the equation fails the polynomial condition, the degree is not defined. Among the given options, this corresponds to (D).

   Watch out

   A common mistake is to see $(y'')^2$ and think “the highest power is 2, so degree is 2”. But that ignores the $\sin(y')$ term. The degree is not simply the exponent on the highest derivative — the equation must be polynomial in all derivatives first.

   Tip

   If the equation had been $(y'')^2 + (y')^3 = x \cdot y'$ (replacing $\sin(y')$ with $y'$), then it would be polynomial and the degree would be $2$. The presence of $\sin$, $\cos$, $\exp$, $\log$, or any non-polynomial function of a derivative always makes the degree undefined.

   ✓Final answer

   The degree is not defined, so the correct option is (D).

5. CBSE 2023Set 65/3/11 markMCQ
   Q.What is the product of the order and degree of the differential equation $\frac{d^2 y}{dx^2} \sin y + \left(\frac{dy}{dx}\right)^3 \cos y = \sqrt{y}$ ?

   (a) $3$

   (b) $2$

   (c) $6$

   (d) not defined

   ›Reveal solutionSolution

   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.

   1. Identify the differential equation: The given differential equation is:

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

   2. 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 $\mathbf{2}$.

   3. 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).
      • The terms $\sin y$, $\cos y$, and $\sqrt{y}$ involve the dependent variable $y$, not its derivatives. These terms do not prevent the equation from being a polynomial in its derivatives. Since the equation is already in a form where it is a polynomial in its derivatives, we can proceed to find the degree. The degree is the power of the highest order derivative. The highest order derivative is $\frac{d^2 y}{dx^2}$. Its power in the equation is 1. Therefore, the degree of the differential equation is $\mathbf{1}$.
      Watch out

      A common mistake is to confuse functions of the dependent variable $y$ (like $\sin y$, $\cos y$, $\sqrt{y}$) with functions of the derivatives (like $\sin(\frac{dy}{dx})$ or $\sqrt{\frac{d^2 y}{dx^2}}$). Only the latter would make the degree undefined or require manipulation to define it.

   4. Calculate the product of the order and degree:

      Product = Order $\times$ Degree

      Product = $2 \times 1 = 2$

   The product of the order and degree of the given differential equation is 2.

   ✓Final answer

   The product of the order and degree of the differential equation is $\boxed{2}$.

6. CBSE 2019Set 65/3/11 mark
   Q.Write the order and degree of the differential equation $\left(\dfrac{d^4y}{dx^4}\right)^2 = \left[x + \left(\dfrac{dy}{dx}\right)^2\right]^3$.

   ›Reveal solutionSolution

   The order of a differential equation is the highest derivative present; the degree is the power of that highest derivative when the equation is polynomial in all derivatives. Here, order is $4$ and degree is $2$.

   Understanding Order and Degree

   The order of a differential equation tells you the highest number of times you've differentiated the dependent variable. It's the "deepest" derivative that appears.

   The degree is subtler: once you've cleared radicals and fractions involving derivatives (making the equation polynomial in all its derivatives), the degree is the exponent on the highest-order derivative. Think of it as the algebraic degree of the "leading term" when you view the equation as a polynomial in derivatives.

   Finding the Order

   1. Identify all derivatives present.

      The equation is:

   $\left(\frac{d^4y}{dx^4}\right)^2 = \left[x + \left(\frac{dy}{dx}\right)^2\right]^3$

   We see $\dfrac{d^4y}{dx^4}$ (the fourth derivative) and $\dfrac{dy}{dx}$ (the first derivative).

   2. Pick the highest order.

      The highest derivative is $\dfrac{d^4y}{dx^4}$, which is a fourth-order derivative.

   Order = $4$.

   Finding the Degree

   3. Check if the equation is polynomial in all derivatives.

      The equation is already in a form where no derivative appears under a radical or in a denominator. Both sides are polynomial expressions in the derivatives:

      • Left side: $\left(\dfrac{d^4y}{dx^4}\right)^2$ is the fourth derivative raised to power $2$.
      • Right side: $\left[x + \left(\dfrac{dy}{dx}\right)^2\right]^3$ expands to terms involving powers of $\dfrac{dy}{dx}$, but the highest-order derivative $\dfrac{d^4y}{dx^4}$ only appears on the left.

   4. Determine the power of the highest-order derivative.

      The highest-order derivative $\dfrac{d^4y}{dx^4}$ appears raised to the power $2$ on the left-hand side. It does not appear on the right-hand side at all.

   Degree = $2$.

   Watch out

   Don't confuse the power of a lower-order derivative with the degree. The degree is determined solely by the exponent on the highest-order derivative. Here, $\left(\dfrac{dy}{dx}\right)^2$ is squared, but that's irrelevant for degree—only the power of $\dfrac{d^4y}{dx^4}$ matters.

   Tip

   If the equation had been $\sqrt{\dfrac{d^4y}{dx^4}} = \text{something}$, you'd first square both sides to eliminate the radical before determining the degree. The degree is defined only when the equation is polynomial in all derivatives.

   ✓Final answer

   The order is $\boxed{4}$ and the degree is $\boxed{2}$.

7. CBSE 2024Set 65/1/11 markMCQ
   Q.The order and degree of the differential equation $\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]^{3}=\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$ are respectively : (A) 1, 2 (B) 2, 3 (C) 2, 1 (D) 2, 6

   ›Reveal solutionSolution

   The order is the highest derivative present (second derivative), so order = 2. The degree is the power of the highest derivative after removing radicals and fractions; here the highest derivative $\frac{d^2y}{dx^2}$ appears to the first power, so degree = 1. The correct option is (C).

   The order of a differential equation is simply the highest derivative that appears. The degree is trickier: it is the power of the highest derivative after the equation has been made polynomial in all derivatives — meaning you must clear any radicals, fractions, or roots that involve the derivatives.

   Let’s see why this matters here.

   1. Identify the highest derivative. The equation is

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

   The left side contains only the first derivative $\frac{dy}{dx}$, but the right side has $\frac{d^2 y}{dx^2}$ — the second derivative. No derivative higher than the second appears, so the order is 2.

   2. Now check the degree.

      The degree is defined only when the equation is a polynomial in the derivatives. Here the highest derivative $\frac{d^2 y}{dx^2}$ is already isolated on one side, and it appears to the power 1. There is no root, fraction, or radical wrapping around it.

      So the degree is simply the exponent of $\frac{d^2 y}{dx^2}$, which is 1.

      Watch out

      A common mistake is to think the exponent 3 on the left side somehow affects the degree. It does not — degree is about the power of the highest derivative, not about powers on lower derivatives or on the whole expression. The left side is just a polynomial in $\frac{dy}{dx}$; it doesn’t change the exponent of $\frac{d^2 y}{dx^2}$.

   3. Double-check by rewriting in polynomial form.

      If you wanted to be absolutely formal, bring everything to one side:

   $\frac{d^2 y}{dx^2} - \left[1+\left(\frac{dy}{dx}\right)^2\right]^3 = 0.$

   This is already a polynomial in $\frac{d^2 y}{dx^2}$ (it appears to the first power) and also a polynomial in $\frac{dy}{dx}$ (to the sixth power, but that doesn’t matter). The highest derivative’s exponent is 1, so degree = 1.

   Tip

   If the equation had been something like $\sqrt{\frac{d^2 y}{dx^2}} = \frac{dy}{dx}$, you would first square both sides to get $\frac{d^2 y}{dx^2} = \left(\frac{dy}{dx}\right)^2$, and then the degree of the highest derivative would be 1. Always clear radicals before reading off the degree.

   Thus order = 2, degree = 1.

   ✓Final answer

   The order is 2 and the degree is 1, which corresponds to option (C).

8. CBSE 2019Set 65/1/11 mark
   Q.Find the order and the degree of the differential equation $x^2 \frac{d^2y}{dx^2} = \left\{ 1 + \left(\frac{dy}{dx}\right)^2 \right\}^4$.

   ›Reveal solutionSolution

   The order of a differential equation is determined by the highest derivative present, and its degree is the power of that highest derivative after ensuring the equation is a polynomial in its derivatives. For the given equation, the order is 2 and the degree is 1.

   When we work with differential equations, two fundamental characteristics help us classify and understand them: their order and their degree. These concepts are crucial because they often dictate the methods we use to solve the equations and the nature of their solutions.

   The order of a differential equation tells us the "highest level of change" or the "highest rate of change of a rate of change" that the equation describes. For instance, a first-order equation might describe velocity, while a second-order equation could describe acceleration or curvature.

   The degree of a differential equation, on the other hand, describes the "algebraic complexity" of the highest derivative. It tells us the power to which the highest order derivative is raised. This is important because equations with higher degrees can sometimes have multiple solutions or require more advanced techniques to solve. However, for the degree to be defined, the equation must be expressible as a polynomial in its derivatives. If derivatives appear inside functions like $\sin(\frac{dy}{dx})$ or $e^{\frac{d^2y}{dx^2}}$, the degree is undefined.

   Let's apply these ideas to the given differential equation:

   $x^2 \frac{d^2y}{dx^2} = \left\{ 1 + \left(\frac{dy}{dx}\right)^2 \right\}^4$

   1. Identify all derivatives present:

      The equation contains two derivatives:

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

   2. Determine the order:

      The order of a differential equation is the order of the highest derivative appearing in it. Comparing $\frac{dy}{dx}$ (order 1) and $\frac{d^2y}{dx^2}$ (order 2), the highest order derivative is $\frac{d^2y}{dx^2}$.

      Therefore, the order of the differential equation is 2.

   3. Check for polynomial form and prepare for degree:

      For the degree to be defined, the differential equation must be expressible as a polynomial in its derivatives. This means there should be no fractional powers of derivatives, no derivatives inside transcendental functions (like $\sin$, $\cos$, $e^x$, $\log$), and no radicals involving derivatives that cannot be cleared.

      Our equation is $x^2 \frac{d^2y}{dx^2} = \left\{ 1 + \left(\frac{dy}{dx}\right)^2 \right\}^4$.

      This equation is already in a polynomial form with respect to its derivatives. The right-hand side is a power of an expression involving a derivative, but it's not a fractional power of a derivative itself, nor is a derivative inside a non-polynomial function. We do not need to perform any operations to clear radicals or fractional powers.

   4. Determine the degree:

      The degree of a differential equation is the power of the highest order derivative, once the equation is expressed as a polynomial in its derivatives.

      From Step 2, we identified the highest order derivative as $\frac{d^2y}{dx^2}$.

      In the given equation, $\frac{d^2y}{dx^2}$ appears as $x^2 \left(\frac{d^2y}{dx^2}\right)^1$. Its power is 1.

      The right-hand side, $\left\{ 1 + \left(\frac{dy}{dx}\right)^2 \right\}^4$, contains only first-order derivatives. Even if we were to expand this expression, it would not introduce any higher powers of $\frac{d^2y}{dx^2}$.

      Therefore, the highest power of the highest order derivative ($\frac{d^2y}{dx^2}$) in the equation is 1.

      Watch out

      A common mistake is to look for the highest power of any derivative in the equation. For example, the term $\left(\frac{dy}{dx}\right)^2$ has a power of 2, and the entire right side is raised to the power of 4, which would lead to terms like $\left(\frac{dy}{dx}\right)^8$ if expanded. However, the degree is defined as the power of the highest order derivative, which is $\frac{d^2y}{dx^2}$ in this case, not $\frac{dy}{dx}$.

   The order is 2 and the degree is 1.

   ✓Final answer

   The order of the differential equation is 2 and its degree is 1.

9. CBSE 2023Set 65/2/11 markMCQ
   Q.Degree of the differential equation $\sin x + \cos\left(\frac{dy}{dx}\right) = y^2$ is:

   (a) 2

   (b) 1

   (c) not defined

   (d) 0

   ›Reveal solutionSolution

   The degree of a differential equation is defined only when it is a polynomial in the derivatives. Since $\cos\left(\frac{dy}{dx}\right)$ is a transcendental function of the derivative, the equation is not a polynomial in $\frac{dy}{dx}$, so its degree is not defined. The correct option is (c).

   The degree of a differential equation is a surprisingly subtle idea. Many students rush to count the highest power of the highest-order derivative, but that only works when the equation is a polynomial in the derivatives. If the equation contains terms like $\sin(y')$, $\cos(y')$, $e^{y'}$, or $\log(y')$, the very notion of "degree" breaks down — because these are not polynomial expressions.

   Let’s see why this matters here.

   ---

   1. Identify the order first. The given equation is:

   $\sin x + \cos\left(\frac{dy}{dx}\right) = y^2$

   The highest derivative present is $\frac{dy}{dx}$ (first derivative). So the order is 1. That’s straightforward.

   2. Now check if the equation is a polynomial in the derivative.

      For degree to be defined, the equation must be expressible as a polynomial in $\frac{dy}{dx}$ (after clearing radicals, if any). Here, the term $\cos\left(\frac{dy}{dx}\right)$ is a cosine of the derivative. No amount of algebraic manipulation will turn $\cos(y')$ into a polynomial in $y'$ — it’s a transcendental function.

   3. Apply the definition.

      The standard definition (NCERT, CBSE, and all major boards) states:

      The degree of a differential equation is the power of the highest-order derivative, provided the equation is a polynomial equation in derivatives.

      Since $\cos(y')$ is not a polynomial in $y'$, the condition fails. Therefore, the degree is not defined.

   Watch out

   A common mistake is to say the degree is 1 because the derivative appears only to the first power inside the cosine. But the cosine itself makes the equation non-polynomial. The degree is not the exponent of the derivative inside a trigonometric function — it’s the exponent when the derivative stands alone as a polynomial term.

   4. Why not option (a), (b), or (d)?
      • (a) 2: No derivative is raised to power 2.
      • (b) 1: Tempting, but the equation isn’t polynomial in $y'$.
      • (d) 0: The equation is not free of derivatives; it contains $y'$ inside $\cos$, so degree isn’t zero either. Only (c) fits the definition.

   Tip

   A quick litmus test: if you see $\sin(y')$, $\cos(y')$, $e^{y'}$, $\log(y')$, or any non-polynomial function of a derivative, the degree is not defined — regardless of what power the derivative appears to have inside that function.

   ✓Final answer

   The degree of the given differential equation is not defined, so the correct option is (c).

10. 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.

    1. 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:

    $e^y \frac{dy}{dx} = 0.$

    2. Identify the highest derivative.

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

    3. 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 $1$.

    Watch out

    A common mistake is to think that because $e^y$ is not a polynomial, the degree is "not defined". But the degree is defined for the highest derivative term — here $\frac{dy}{dx}$ appears with power 1, and the equation can be simplified to a polynomial form. So degree is defined and equals 1.

    Tip

    If the equation had been something like $\frac{d}{dx}(\sin y) = 0$, the same logic applies: $\cos y \cdot \frac{dy}{dx} = 0$ simplifies to $\frac{dy}{dx} = 0$ (since $\cos y$ is not identically zero), giving order 1, degree 1.

    4. Match with the options. Order = 1, Degree = 1 corresponds to option (B).

    ✓Final answer

    The order is 1 and the degree is 1, so the correct option is (B).

11. CBSE 2025Set 65/2/11 markMCQ
    Q.If $p$ and $q$ are respectively the order and degree of the differential equation $\frac{d}{dx}\left(\left(\frac{dy}{dx}\right)^{3}\right) = 0$, then $(p - q)$ is: (A) 0 (B) 1 (C) 2 (D) 3

    ›Reveal solutionSolution

    Expand the derivative to find the highest derivative and the power of that derivative when the equation is polynomial in derivatives. Here $p = 2$, $q = 1$, so $(p - q) = 1$.

    The order of a differential equation is the highest derivative that appears. The degree is the exponent on that highest derivative after the equation has been written as a polynomial in all derivatives (no radicals, no derivatives in denominators, etc.).

    The trap here is reading the equation too quickly. We have a derivative of something, not just that something itself. Let's expand it properly.

    Finding the order

    1. Expand the outer derivative using the chain rule.

       We're differentiating $\left(\frac{dy}{dx}\right)^3$ with respect to $x$:

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

    2. Rewrite the differential equation.

       The equation becomes:

    $3\left(\frac{dy}{dx}\right)^2 \cdot \frac{d^2y}{dx^2} = 0$

    3. Identify the highest derivative.

       The highest derivative present is $\frac{d^2y}{dx^2}$, which is the second derivative.

       Therefore, the order $p = 2$.

    Watch out

    A common mistake is to think the order is $1$ because you see $\frac{dy}{dx}$ raised to the third power. But order counts the number of times you differentiate, not the power. The outer $\frac{d}{dx}$ operator adds one more level of differentiation.

    Finding the degree

    4. Check if the equation is polynomial in derivatives.

       Our expanded form is:

    $3\left(\frac{dy}{dx}\right)^2 \cdot \frac{d^2y}{dx^2} = 0$

    This is already polynomial in both $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ — no roots, no fractions involving derivatives.

    5. Find the power of the highest derivative.

       The highest derivative is $\frac{d^2y}{dx^2}$, and it appears to the power $1$.

       Therefore, the degree $q = 1$.

    Tip

    Degree is always a positive integer and is defined only when the equation can be written as a polynomial in derivatives. If you see $\sqrt{\frac{d^2y}{dx^2}}$, you'd first square both sides to make it polynomial before finding the degree.

    Computing $(p - q)$

    6. Subtract degree from order.

    $p - q = 2 - 1 = 1$

    ✓Final answer

    The correct option is (B) $1$.

12. CBSE 2026Set V11 mark
    Q.Choose from $[0, 3, -1, 2, -2, 1]$. If $m$ and $n$ are respectively the order and degree of the differential equation $2x^2\frac{d^2y}{dx^2}-3\frac{dy}{dx}+y=0$ then $m+n=$ ____.

    ›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 $m=2$.
    • The equation is polynomial in the derivatives and the highest derivative occurs to the first power, so the degree is $n=1$.

    Therefore $m+n=2+1=3$.

    ✓Final answer

    $3$