Q.(ii) The degree of the differential equation $\sqrt{1+\left(\frac{dy}{dx}\right)^2}=x$ is ______.

Showing the 12 most recent of 26 on this concept.

1. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.The order and degree of the differential equation $\sqrt{\dfrac{dy}{dx}} - 4\dfrac{dy}{dx} - 7x = 0$ are respectively ______ (A) $1$ & $\dfrac{1}{2}$ (B) $2$ & $1$ (C) $1$ & $1$ (D) $1$ & $2$

   ›Reveal solutionSolution

   Order is the highest derivative present; degree is the power of the highest derivative once the equation is rationalized (radical-free) in the derivatives.

   Concept and Intuition

   The order of a differential equation is simply the order of the highest derivative appearing in it. The degree is only defined once the equation is a polynomial in its derivatives (no radicals or fractional powers involving a derivative) — so before reading off the degree, we must first clear any such radical by squaring or otherwise rationalizing.

   Step-by-Step Solution

   1. The equation is $\sqrt{\dfrac{dy}{dx}} - 4\dfrac{dy}{dx} - 7x = 0$.
   2. Only $dy/dx$ (first derivative) appears — no $d^2y/dx^2$ or higher — so the order is 1.
   3. To find the degree, isolate the radical: $\sqrt{\dfrac{dy}{dx}} = 4\dfrac{dy}{dx}+7x$.
   4. Square both sides to remove the square root: $\dfrac{dy}{dx} = \left(4\dfrac{dy}{dx}+7x\right)^2$. …

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2. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
   Q.The degree of the differential equation $\log\left(\frac{dy}{dx}\right)=\left(2x+3\frac{dy}{dx}\right)^2$ is (A) $1$ (B) $2$ (C) $3$ (D) not defined

   ›Reveal solutionSolution

   Because the derivative sits inside a transcendental function (a logarithm) rather than as a pure power, this differential equation has no polynomial form in $dy/dx$, so its degree is not defined.

   Concept and Intuition

   The "degree" of a differential equation is defined as the power of the highest-order derivative, but ONLY after the equation has been made free of radicals, fractional powers, and any derivative sitting inside a non-polynomial (transcendental) function such as $\log$, $\sin$, $e^{(\cdot)}$, etc. If the derivative cannot be isolated as a polynomial term, degree is simply not defined — order is still defined (here order $=1$), but degree is not.

   Step-by-Step Solution

   1. The equation is $\log\!\left(\dfrac{dy}{dx}\right)=\left(2x+3\dfrac{dy}{dx}\right)^2$.
   2. The highest derivative present is $\dfrac{dy}{dx}$ (order 1), but it appears inside a $\log(\cdot)$ on the left side. …

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3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.The order and degree of the differential equation $\left\{1+\left(\dfrac{dy}{dx}\right)^2\right\}^{3/2} = \dfrac{d^2y}{dx^2}$ are respectively (A) $\dfrac{3}{2},2$ (B) $2,3$ (C) $2,2$ (D) $3,4$

   ›Reveal solutionSolution

   The order is the highest derivative present ($y''$, order $2$); the degree requires first clearing the fractional power by squaring, after which the highest derivative appears to the power $2$ — so order and degree are both $2$.

   Concept and Intuition

   The order of a differential equation is simply the order of the highest derivative appearing. The degree is the power of the highest-order derivative after the equation has been made a polynomial in all the derivatives (no fractional or negative powers of any derivative allowed) — so if a fractional power like $3/2$ appears on an expression containing lower derivatives, you must algebraically clear it before reading off the degree.

   Step-by-Step Solution

   1. The equation is $\left\{1+\left(\dfrac{dy}{dx}\right)^2\right\}^{3/2}=\dfrac{d^2y}{dx^2}$.
   2. The highest derivative present is $\dfrac{d^2y}{dx^2}$ — so the order is 2.
   3. The left side has a fractional exponent $3/2$ on an expression involving $dy/dx$ (not the highest derivative), so we cannot read the degree directly; we must eliminate the fractional power.
   4. Square both sides: $\left\{1+\left(\dfrac{dy}{dx}\right)^2\right\}^{3}=\left(\dfrac{d^2y}{dx^2}\right)^{2}$. …

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4. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If the order and degree of the differential equation $x\dfrac{d^2y}{dx^2} = \left[1 + \left(\dfrac{d^2y}{dx^2}\right)^2\right]^{-1/2}$ are k and l respectively, then k, l are the roots of (A) $x^2 - 5x + 6 = 0$ (B) $x^2 - 3x + 2 = 0$ (C) $x^2 - 7x + 12 = 0$ (D) $x^2 - 6x + 8 = 0$

   ›Reveal solutionSolution

   This tests the rule that order/degree are only defined after the differential equation is made polynomial (free of fractional/negative powers) in its derivatives. Order $=2$, degree $=4$, whose roots satisfy $x^2-6x+8=0$, option (D).

   Concept and Intuition

   Order of a differential equation is the order of the highest derivative appearing in it. Degree is the power of the highest-order derivative, but ONLY once the equation has been rewritten as a polynomial in all the derivatives (no fractional powers, no derivatives inside roots or negative exponents). Here the right-hand side has a $-1/2$ power, so we must first algebraically clear that before reading off the degree.

   Step-by-Step Solution

   1. Given: $x\dfrac{d^2y}{dx^2}=\left[1+\left(\dfrac{d^2y}{dx^2}\right)^2\right]^{-1/2}$. Let $p=\dfrac{d^2y}{dx^2}$.
   2. So $xp=(1+p^2)^{-1/2}$. Multiply both sides by $(1+p^2)^{1/2}$: $xp(1+p^2)^{1/2}=1$.
   3. Square both sides to remove the remaining square root: $x^2p^2(1+p^2)=1$.
   4. Expand: $x^2p^2+x^2p^4=1$, i.e. $x^2p^4+x^2p^2-1=0$ — now a genuine polynomial equation in the derivative $p$.
   5. The highest derivative present is $p=y''$ — a second-order derivative, so order $k=2$. No first or third derivative appears, so order stays 2. …

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5. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.The order and degree of the differential equation whose solution is $Ax^2+By^2=1$, $A$ and $B$ are arbitrary constants, are respectively (A) 2, 2 (B) 2, 1 (C) 1, 2 (D) 1, 1

   ›Reveal solutionSolution

   Two arbitrary constants require differentiating twice, giving order 2; the resulting equation is linear in the highest derivative $y''$, giving degree 1.

   Concept and Intuition

   The order of the differential equation whose general solution has $n$ independent arbitrary constants is (generically) $n$, since eliminating $n$ constants requires $n$ differentiations. The degree is the power of the highest-order derivative once the equation is written as a polynomial in derivatives.

   Step-by-Step Solution

   1. Given: $Ax^2+By^2=1$ ... (i), with 2 arbitrary constants $A,B$ — so we expect to differentiate twice.
   2. Differentiate (i) once: $2Ax+2Byy'=0 \Rightarrow Ax+Byy'=0$ ... (ii).
   3. Differentiate (ii) again: $A + B(y'\cdot y' + y\cdot y'') = 0 \Rightarrow A + B(y'^2+yy'') = 0$ ... (iii).
   4. From (ii): $A = -\dfrac{Byy'}{x}$ (for $x\neq0$). Substitute into (iii): $-\dfrac{Byy'}{x} + B(y'^2+yy'')=0$.
   5. Factor out $B$ (nonzero generically): $-\dfrac{yy'}{x}+y'^2+yy''=0$. Multiply through by $x$: $xyy'' + xy'^2 - yy' = 0$. …

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6. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.The sum of the order and degree of the differential equation $x\left(\dfrac{d^2y}{dx^2}\right)^{1/2} = \left(1 + \dfrac{dy}{dx}\right)^{4/3}$ is (A) $5$ (B) $8$ (C) $12$ (D) $10$

   ›Reveal solutionSolution

   Order and degree are only meaningful after the equation is made polynomial in its derivatives — that requires clearing the fractional exponents first. Answer: $5$.

   Concept and Intuition

   The order of a differential equation is the order of the highest derivative present. The degree is the power of the highest-order derivative, but only after the equation has been rewritten as a polynomial in the derivatives (no fractional or negative powers of any derivative term). So before reading off the degree, any radicals or fractional exponents on derivative terms must be cleared.

   Step-by-Step Solution

   1. Given: $x\left(\dfrac{d^2y}{dx^2}\right)^{1/2} = \left(1+\dfrac{dy}{dx}\right)^{4/3}$.
   2. The highest derivative present is $y''=\dfrac{d^2y}{dx^2}$, so the order is $2$.
   3. To clear the fractional powers $\tfrac12$ and $\tfrac43$, raise both sides to the power $6$ (the LCM of denominators $2$ and $3$): $x^6(y'')^3 = (1+y')^8.$ …

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7. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.Order and degree of the differential equation $\dfrac{d^3y}{dx^3} = \left[1+\left(\dfrac{dy}{dx}\right)^2\right]^{5/2}$ respectively are (A) $5, 2$ (B) $3, 5$ (C) $3, 2$ (D) $2, 3$

   ›Reveal solutionSolution

   Order is the highest derivative present ($3$); degree requires clearing the fractional exponent first, which makes the highest derivative appear squared — giving order $3$, degree $2$: (C).

   Concept and Intuition

   "Order" of a differential equation is simply the order of the highest derivative appearing. "Degree" is the power of the highest-order derivative after the equation has been made a polynomial in derivatives (i.e., all fractional/negative powers of derivatives must first be cleared by algebraic manipulation such as squaring).

   Step-by-Step Solution

   1. The highest derivative present is $\dfrac{d^3y}{dx^3}$ — a third-order derivative. So order $=3$.
   2. The RHS, $\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^{5/2}$, has a fractional exponent, so the equation is not yet in polynomial form.
   3. Square both sides to remove the fractional power: $\left(\dfrac{d^3y}{dx^3}\right)^2=\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^5$. …

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8. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.The order and degree of the differential equation $\left(\dfrac{d^3y}{dx^3}\right)^{1/2} - 2\left(\dfrac{dy}{dx}\right)^{1/4} + xy = 0$ are respectively (A) 3 and 12 (B) 3 and 2 (C) 3 and 4 (D) 3 and 6

   ›Reveal solutionSolution

   The order is 3 (from the third derivative); after systematically squaring twice to clear all fractional exponents (the $1/2$ power on $y'''$ and the $1/4$ power on $y'$), the equation becomes polynomial with $y'''$ appearing to the 4th power — so degree $=4$.

   Concept and Intuition

   Order is simply the highest derivative present (here, the third derivative, so order 3). Degree requires the equation to first be made polynomial in all the derivatives (no fractional or negative powers, no derivatives inside radicals) — only then is the degree the power of the highest-order derivative. Since we have two different fractional powers ($1/2$ on $y'''$ and $1/4$ on $y'$), we must clear both, which can require squaring more than once and tracks a growing power on $y'''$.

   Step-by-Step Solution

   1. Order: the highest derivative is $\dfrac{d^3y}{dx^3}$, so order $=3$.

   2. Clearing the fractional powers — write $p=y'''$, $q=y'$, $a=xy$:

   $p^{1/2} - 2q^{1/4} + a = 0 \implies p^{1/2} = 2q^{1/4}-a$

   3. Square once to remove the $1/2$ power on $p$:

   $p = 4q^{1/2} - 4a\,q^{1/4} + a^2$

   This still carries fractional powers of $q$ (namely $q^{1/2}$ and $q^{1/4}$), which must also be cleared since all derivative terms must end up with integer powers, not just the highest one.

   4. Isolate the term with the highest remaining fractional power ($q^{1/4}$), letting $w=q^{1/4}$ (so $q^{1/2}=w^2$):

   $p - a^2 = 4w^2 - 4a\,w$

   This is a quadratic in $w$:

   $4w^2 - 4a\,w - (p-a^2) = 0 \implies w = \frac{a \pm \sqrt{p}}{2}$

   (after simplifying the quadratic formula, using $a^2+p-a^2=p$ inside the root). …

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9. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.If a and b are respectively the order and degree of a differential equation $y^2(y'')^2 + 3x(y')^{1/3} + x^2y^2 = \sin x$, then (A) $b = a$ (B) $a = 3b$ (C) $b = 3a$ (D) $ab = 6$

   ›Reveal solutionSolution

   Order is fixed by the highest derivative ($y''$, so $a=2$); clearing the fractional power on $y'$ by cubing the whole equation raises the power of $y''$ to $6=3a$, giving $b=3a$.

   Concept and Intuition

   The order of a differential equation is simply the order of the highest derivative appearing. The degree is only defined once the equation is written as a polynomial in all the derivatives (integer, non-negative powers) — so any fractional or negative power on a derivative must first be cleared by an appropriate algebraic operation (like raising both sides to a power), and this operation can also change the power of other derivative terms present in the equation.

   Step-by-Step Solution

   1. The equation is $y^2(y'')^2+3x(y')^{1/3}+x^2y^2=\sin x$.
   2. The highest-order derivative present is $y''$ (second derivative), so the order $a=2$.
   3. The term $(y')^{1/3}$ has a fractional exponent, so the equation is not yet a polynomial in the derivatives.
   4. Isolate that term: $3x(y')^{1/3} = \sin x - y^2(y'')^2 - x^2y^2$.
   5. Cube both sides to clear the cube root: $27x^3(y') = \big[\sin x - y^2(y'')^2-x^2y^2\big]^3$. …

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10. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
    Q.If the degree of the differential equation corresponding to the family of curves $y = ax + \frac{1}{a}$ (where $a \ne 0$ is an arbitary constant) is r and it's order is m, then the solution of $\frac{dy}{dx} = \frac{y}{2x}, y(1) = \sqrt{r+m}$ is (A) $y = 3^x$ (B) $y^2 = 3x$ (C) $x^2 = 3y$ (D) $y = 3\log x$

    ›Reveal solutionSolution

    Find the order/degree of the DE for the given family, use $r+m$ as the initial condition, then solve a separable linear-in-$x$ ODE: $y^2=3x$.

    Concept and Intuition

    The family $y=ax+1/a$ has one arbitrary constant $a$, so its differential equation has order 1. But eliminating $a$ (since $a$ appears both linearly and as $1/a$) forces a quadratic in $y'$, giving degree 2. This r,m pair then feeds a separate, simple variable-separable ODE.

    Step-by-Step Solution

    1. Differentiate $y=ax+1/a$: $y'=a$.
    2. Substitute $a=y'$ back into the family equation: $y=y'x+\dfrac{1}{y'}$. Multiply through by $y'$: $yy'=x(y')^2+1$, i.e. $x(y')^2-yy'+1=0$.
    3. This equation is first order (only $y'$ appears, no higher derivative) $\Rightarrow m=1$; the highest power of $y'$ is 2 $\Rightarrow r=2$. So $r+m=3$.
    4. Now solve $\dfrac{dy}{dx}=\dfrac{y}{2x}$ with $y(1)=\sqrt{3}$. Separate: $\dfrac{dy}{y}=\dfrac{dx}{2x}$. …

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11. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
    Q.The order and degree of the differential equation $3x^2\dfrac{d^2y}{dx^2}-\sin\left(\dfrac{d^3y}{dx^3}\right)+\cos(xy)=0$ are (A) Order can't be defined and degree is 3 (B) Order is 3 and degree can't be defined (C) Order is 3 and degree is 1 (D) Order is 1 and degree is 3

    ›Reveal solutionSolution

    Because the highest derivative $y'''$ sits inside a $\sin(\cdot)$, the equation is not a polynomial in derivatives, so its degree is undefined even though its order (3) is perfectly well defined.

    Concept and Intuition

    Order of a DE = order of the highest derivative appearing. Degree = the power of the highest-order derivative, but this is only defined when the equation, after clearing radicals/fractions, is a polynomial in all the derivatives. If the highest derivative appears inside a non-polynomial function (sin, log, exponential, etc.), degree simply cannot be assigned.

    Step-by-Step Solution

    1. The DE is $3x^2y''-\sin(y''')+\cos(xy)=0$.
    2. The highest-order derivative present is $y'''$ (third derivative) — so the order is $3$.
    3. However, $y'''$ appears as the argument of $\sin(\cdot)$, not raised to an integer power — this equation is not a polynomial in $y'''$ (or any of the derivatives), even after any algebraic rearrangement, since $\sin$ cannot be turned into a finite polynomial expression in $y'''$. …

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12. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
    Q.The sum of the order and degree of the differential equation $\dfrac{d^4y}{dx^4}=\left\{c+\left(\dfrac{dy}{dx}\right)^2\right\}^{3/2}$ is (A) $4$ (B) $6$ (C) $5$ (D) $8$

    ›Reveal solutionSolution

    Squaring both sides clears the $3/2$ power, giving order $4$ and degree $2$; their sum is $6$.

    Concept and Intuition

    Degree is only meaningful once the differential equation is written as a polynomial in all the derivatives involved — any radicals (fractional powers) on a derivative term must first be removed by an algebraic operation such as squaring, cubing, etc.

    Step-by-Step Solution

    1. Given: $\dfrac{d^4y}{dx^4}=\left\{c+\left(\dfrac{dy}{dx}\right)^2\right\}^{3/2}$.
    2. Square both sides to remove the $3/2$ power: $\left(\dfrac{d^4y}{dx^4}\right)^2 =\left\{c+\left(\dfrac{dy}{dx}\right)^2\right\}^3$.
    3. This is now a genuine polynomial equation in the derivatives $y''''$ and $y'$.
    4. Highest-order derivative: $y''''=\dfrac{d^4y}{dx^4}$, so order $=4$. …

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