Q.(iii) The number of arbitrary constants in the general solution of a differential equation of order three is ______.

Showing the 12 most recent of 26 on this concept.

1. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.Let a and b be arbitrary constants and C be a fixed constant. If $y = ae^{2x} + bxe^{2x} + C$ is the general solution of a differential equation, then the order of that differential equation is (A) $1$ (B) $2$ (C) $3$ (D) $4$

   ›Reveal solutionSolution

   The order of a differential equation equals the number of independent arbitrary constants in its general solution — here that's $2$ ($a,b$), since $C$ is fixed.

   Concept and Intuition

   A general solution of an $n$th order ODE has exactly $n$ independent arbitrary constants. It's crucial to note which "constants" in a given family are actually free parameters versus fixed values — here the problem explicitly says $C$ is fixed, so it is not counted.

   Step-by-Step Solution

   1. $y=ae^{2x}+bxe^{2x}+C$ with $a,b$ arbitrary and $C$ fixed.
   2. The number of genuinely free (arbitrary) constants is $2$: $a$ and $b$. …

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2. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.If $y = a^3 e^{b^2 x + c}$ is the general solution of a differential equation, where $a$ and $c$ are arbitrary constants and $b$ is a fixed constant, then the order of differential equation is (A) $1$ (B) $2$ (C) $3$ (D) $4$

   ›Reveal solutionSolution

   Although the solution is written using two constants $a$ and $c$, they combine into a single essential constant, so the differential equation is first order.

   Concept and Intuition

   The order of a differential equation is not decided by how many symbols appear in its general solution — it is decided by how many independent (essential) arbitrary constants the family of curves actually needs. Two constants can secretly be redundant if they always appear in a combination that behaves as one constant.

   Step-by-Step Solution

   1. Write $y = a^3 e^{b^2x+c} = a^3 e^{c}\cdot e^{b^2x}$.
   2. Since $a$ and $c$ are both arbitrary but $b$ is fixed, define $K = a^3e^{c}$. As $a,c$ range over all values, $K$ just ranges over all (nonzero) reals — it is a single essential arbitrary constant.
   3. So the general solution is really $y = Ke^{b^2x}$, a one-parameter family of curves. …

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3. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.Let $c_1,c_2,c_3,c_4$ be arbitrary constants. The order of the differential equation, corresponding to $y=c_1e^x+c_2e^{\log_e x}+c_3\sin^2 x-c_4(\cos^2 x-1)$ is (A) 1 (B) 2 (C) 3 (D) 4

   ›Reveal solutionSolution

   Simplifying the given expression shows $c_3$ and $c_4$ always appear only as the sum $c_3+c_4$, so there are really just 3 independent arbitrary constants, making the order of the corresponding DE equal to 3.

   Concept and Intuition

   The order of the differential equation satisfied by a family of curves equals the number of independent arbitrary constants in the family — not simply the number of symbols written. Redundant constants (ones that only ever appear combined) must be collapsed first.

   Step-by-Step Solution

   1. Simplify $e^{\log_e x}=x$ (for $x>0$).
   2. Simplify $-c_4(\cos^2x-1)=-c_4(-\sin^2x)=c_4\sin^2x$.
   3. So $y=c_1e^x+c_2x+c_3\sin^2x+c_4\sin^2x=c_1e^x+c_2x+(c_3+c_4)\sin^2x$.
   4. Let $C_3=c_3+c_4$ — a single arbitrary constant (since $c_3,c_4$ are both arbitrary, their sum is just another arbitrary constant, with no independent extra freedom). …

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4. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
   Q.Assertion (A): Order of the differential equations of a family of circles with constant radius is two. Reason (R): An algebraic equation having two arbitrary constants is general solution of a 2nd order differential equation. (A) (A) and (R) are true, (R) is the correct explanation to (A) (B) (A) is true, (R) is false (C) (A) and (R) are false, (R) is not the correct explanation to (A) (D) (A) is false, (R) is true

   ›Reveal solutionSolution

   This tests the link between the number of arbitrary constants in a family of curves and the order of its differential equation, applied to circles of fixed radius.

   Concept and Intuition

   The general equation of a circle is $(x-a)^2+(y-b)^2=r^2$, with 3 constants $a,b,r$ in general. If the radius is held constant (a fixed known number), only $a$ and $b$ remain arbitrary — exactly 2 constants — so eliminating them (differentiating twice) yields a 2nd-order differential equation. This is a specific instance of the general rule in Reason (R): an equation with $n$ independent arbitrary constants is the general solution of an $n$-th order DE.

   Step-by-Step Solution

   1. Write the family: $(x-a)^2+(y-b)^2=r^2$, $r$ fixed, $a,b$ arbitrary.
   2. Two arbitrary constants ⇒ need to differentiate twice to eliminate both ⇒ resulting DE has order 2. So Assertion (A) is TRUE.
   3. Reason (R): an algebraic equation with two arbitrary constants is the general solution of a 2nd order DE — this is a true, standard fact. …

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5. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
   Q.For the differential equation $\dfrac{d^3y}{dx^3} = 0$, $y = ax^2 + bx + c$ is (A) the general solution (B) a particular solution (C) not a solution (D) a solution, but not a particular solution

   ›Reveal solutionSolution

   A 3rd-order ODE's general solution needs exactly 3 arbitrary constants; $y=ax^2+bx+c$ has exactly 3, so it's the general (not particular) solution.

   Concept and Intuition

   A particular solution has all constants fixed to specific numeric values (satisfying given initial/boundary conditions); a general solution keeps as many arbitrary constants as the order of the differential equation. Order 3 $\Rightarrow$ 3 independent arbitrary constants needed.

   Step-by-Step Solution

   1. Verify $y=ax^2+bx+c$ solves the equation: $y'=2ax+b$, $y''=2a$, $y'''=0$. Yes, it satisfies $d^3y/dx^3=0$ for any $a,b,c$.
   2. Count arbitrary constants: $a,b,c$ — three of them, matching the order of the ODE (3rd order). …

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6. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.By eliminating the arbitrary constants from $y = (a+b)\sin(x+c) - de^{x+e+f}$, the differential equation obtained is of order (A) 6 (B) 4 (C) 3 (D) 5

   ›Reveal solutionSolution

   Although $6$ symbols appear, several combine into single effective constants, leaving only $3$ independent ones — so the resulting differential equation has order $3$.

   Concept and Intuition

   The order of the differential equation obtained by eliminating arbitrary constants equals the number of independent essential constants, not the raw count of symbols written down. Here $a$ and $b$ only ever appear as the sum $a+b$, and $d,e,f$ only ever appear as the single combination $d\,e^{e+f}$ (after factoring $e^{x+e+f}=e^{e+f}\cdot e^x$) — so several "constants" are redundant.

   Step-by-Step Solution

   1. Rewrite: $y=(a+b)\sin(x+c)-d\,e^{x+e+f} = A\sin(x+c) - B\,e^{x}$, where $A=a+b$ and $B=d\,e^{e+f}$ are single combined constants, and $c$ is separate.
   2. So really only $3$ independent arbitrary constants remain: $A$, $c$, $B$ (absorbing the sign into $B$).
   3. Differentiate repeatedly: $y'=A\cos(x+c)+Be^x$ (using $B$ generically for the exponential coefficient), $y''=-A\sin(x+c)+Be^x$, $y'''=-A\cos(x+c)+Be^x$.
   4. From $y$ and $y''$: $y+y''=2Be^x$. Differentiating this relation: $y'+y'''=2Be^x$ as well (same right side). …

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7. 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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8. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
   Q.$y = ax + b$ is (A) General solution of $\dfrac{d^3y}{dx^3} = 0$ (B) General solution of $\dfrac{dy}{dx} = a + b$ (C) General solution for both $\dfrac{d^2y}{dx^2} = 0$ and $\dfrac{d^3y}{dx^3} = 0$ (D) General solution for $\dfrac{d^2y}{dx^2} = 0$

   ›Reveal solutionSolution

   The number of arbitrary constants in the solution must match the order of the ODE for it to be the general solution; $y=ax+b$ has 2 constants, matching a 2nd-order equation, $d^2y/dx^2=0$.

   Concept and Intuition

   A key rule: the general solution of an $n$th-order ODE contains exactly $n$ independent arbitrary constants. $y=ax+b$ has two constants ($a$ and $b$), so it can only be the general solution of a second-order equation. For a third-order equation like $y'''=0$, the general solution is a full quadratic $y=c_1x^2+c_2x+c_3$ (three constants) — $y=ax+b$ is merely one special case of that family (with $c_1=0$), not the general solution.

   Step-by-Step Solution

   1. Differentiate $y=ax+b$ once: $y'=a$ (constant).
   2. Differentiate again: $y''=0$. Since this holds for all choices of $a,b$ and uses up exactly the 2 constants, $y=ax+b$ is the general solution of $y''=0$.
   3. Check $y'''$: differentiating a third time, $y'''=0$ also holds — but this is now a 3rd-order ODE whose general solution should have 3 arbitrary constants (e.g. $y=c_1x^2+c_2x+c_3$). $y=ax+b$ is only a subset (particular family within) that general solution, not the general solution itself, since it's missing the free quadratic term. …

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9. 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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10. 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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11. 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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12. 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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