Q.Find the order and degree of the differential equation $\left(\dfrac{d^3y}{dx^3}\right)^{2/3} = \dfrac{dy}{dx} + y$.

Showing the 12 most recent of 24 on this concept.

1. CBSE 2026Set ANNUAL1 mark
   Q.Write the degree of the differential equation $(y''')^2+3(y'')^3+3xy'+5y=0$

   ›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: $(y''')^2+3(y'')^3+3xy'+5y=0$.

   The highest-order derivative present is $y'''$ (third order), and it already appears as a polynomial term.

   The power to which $y'''$ (the highest order derivative) is raised is $2$.

   ✓Final answer

   Degree $=2$

2. 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 ($1, 2, 3, \dots$).

   The degree is the power (exponent) of the highest-order derivative after the equation has been expressed as a polynomial in its derivatives (free of radicals and fractional powers of the derivatives). When it exists in this standard polynomial form, the degree is likewise a positive integer.

   For example, $\left(\dfrac{d^2y}{dx^2}\right)^3 + \dfrac{dy}{dx} = 0$ has order $2$ and degree $3$ — both positive integers.

   ✓Final answer

   The statement is True — option (i).

3. CBSE 2025Set ANNUAL1 mark
   Q.Write the order of the differential equation $\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}=\left(\dfrac{d^2y}{dx^2}\right)^{3/2}$

   ›Reveal solutionSolution

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

   The given differential equation is

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

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

   ✓Final answer

   Order $=2$.

4. CBSE 2025Set ANNUAL1 markMCQ
   Q.The order and degree of the differential equation $\sqrt{\dfrac{dy}{dx}}-4\dfrac{dy}{dx}-7x=0$ are respectively :

   (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.

   1. Given $\sqrt{\dfrac{dy}{dx}}-4\dfrac{dy}{dx}-7x=0$. Let $p=\dfrac{dy}{dx}$.
   2. Isolate the radical: $\sqrt p=4p+7x$.
   3. Square both sides to remove the radical (a fractional power of a derivative is not allowed for degree to be defined): $p=(4p+7x)^2=16p^2+56px+49x^2$.
   4. Rearranged: $16p^2+(56x-1)p+49x^2=0$ — a polynomial equation in $p=\dfrac{dy}{dx}$.
   5. Order = highest order of derivative present = $1$ (only $\dfrac{dy}{dx}$ appears, no higher derivative).
   6. Degree = power of the highest-order derivative once the equation is a polynomial free of radicals/fractional powers of derivatives = $2$ (from the $p^2$ term).

   ✓Final answer

   (a) $1, 2$

5. CBSE 2025Set ANNUAL1 markMCQ
   Q.The order and degree of the differential equation $\left(\frac{d^2y}{dx^2}\right)^2 + \left(\frac{dy}{dx}\right)^2 = a^x$ are ______ respectively.

   (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.

   The equation is already polynomial in the derivatives (the right side $a^x$ contains no derivatives). The power to which the highest-order derivative $\dfrac{d^2y}{dx^2}$ is raised is $2$, so the degree is 2.

   ✓Final answer

   The order and degree are $2$ and $2$ respectively — the third option.

6. CBSE 2025Set MARCH1 markMCQ
   Q.The differential equation $\left(\dfrac{dx}{dy}\right)^{3}+2y^{1/2}=x$ is :

   (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 $1$.

   Degree. Degree is the highest power of the highest-order derivative when the equation is polynomial in its derivatives. Here that derivative appears as $\left(\dfrac{dx}{dy}\right)^{3}$.

   The radical $y^{1/2}$ involves the dependent variable, not a derivative, so it does not change the degree. Hence the degree is $3$.

   ✓Final answer

   Option (d) order $1$ and degree $3$.

7. 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 $p$:

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

   This is a polynomial equation in $p = dy/dx$. The highest-order derivative present is the first derivative, so the order is 1. The highest power to which that derivative is raised (after clearing the fraction) is $2$, so the degree is 2.

   ✓Final answer

   Order $1$, degree $2$ — option (a).

8. 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 $1, 2, 3, \dots$
   • Degree = the power of the highest-order derivative once the equation is written as a polynomial in the derivatives (free of radicals and fractions in the derivatives). When it is defined, this power is a positive integer.

   Both quantities are therefore positive integers, matching the standard textbook statement.

   ✓Final answer

   True. (Note: the degree must first be defined — i.e. the equation is a polynomial in its derivatives; where it exists it is a positive integer.)

9. CBSE 2024Set MARCH1 markMCQ
   Q.The order and degree of the differential equation $\sqrt{\frac{d^2y}{dx^2}}=\sqrt{\frac{dy}{dx}+5}$ are respectively :

   (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:

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

   The highest-order derivative present is $\dfrac{d^2y}{dx^2}$ (order $2$), and after the equation is made rational and integral its power is $1$ (degree $1$).

   ✓Final answer

   Option (a) 2 and 1.

10. CBSE 2023Set ANNUAL1 mark
    Q.Write the degree of the differential equation $e^{\frac{dy}{dx}} + \dfrac{dy}{dx} = x$

    ›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 form), so this cannot be reduced to polynomial form. Hence the degree of this differential equation is not defined.

    ✓Final answer

    Not defined

11. 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.

    Step 3. Hence the order is $2$ (its degree, incidentally, is $2$ since it appears squared after clearing radicals/fractions — but the question asks only for order).

    ✓Final answer

    Order $=2$ (option c).

12. CBSE 2023Set ANNUAL1 markMCQ
    Q.The degree of the differential equation $\left(\frac{d^2y}{dx^2}\right)^2 + \left(\frac{dy}{dx}\right)^3 = a^x$ is 3.

    (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 $2$.

    So the degree is $2$. The exponent $3$ belongs to the lower-order term $\dfrac{dy}{dx}$ and does not decide the degree. Hence the statement that the degree is $3$ is wrong.

    ✓Final answer

    False — the order is $2$ and the degree is $2$.