Q.The number of arbitrary constants in the general solution of a differential equation of fourth order are: (A) 0 (B) 2 (C) 3 (D) 4

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 $b$. …

   🔒 Start your 14-day free trial to see the full solution →
2. 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). …

   🔒 Start your 14-day free trial to see the full solution →
3. 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. …

   🔒 Start your 14-day free trial to see the full solution →
4. 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). …

   🔒 Start your 14-day free trial to see the full solution →
5. 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. …

   🔒 Start your 14-day free trial to see the full solution →
6. 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 $=4$. …

   🔒 Start your 14-day free trial to see the full solution →
7. 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). …

   🔒 Start your 14-day free trial to see the full solution →
8. 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 $x$: $xyy'' + xy'^2 - yy' = 0$. …

   🔒 Start your 14-day free trial to see the full solution →
9. 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. …

   🔒 Start your 14-day free trial to see the full solution →
10. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
    Q.p and q are positive integers and $n < r < m$. If the order and degree of the differential equation $\left(\dfrac{d^m y}{dx^m} + \dfrac{d^n y}{dx^n}\right)^{p/q} = 5\dfrac{d^r y}{dx^r}$ are respectively 4 and 3, then (A) $n=4, q=3$ (B) $m=4, q=3$ (C) $r=4, q=3$ (D) $m=4, p=3$

    ›Reveal solutionSolution

    Order = highest derivative present = $m$; degree = power of that highest-order term after clearing the fractional exponent = $p$. Given order $4$, degree $3$: $m=4$, $p=3$.

    Concept and Intuition

    Order of an ODE is simply the order of the highest derivative appearing in it. Degree is the power of the highest-order derivative after the equation has been made polynomial in all derivatives (i.e., all fractional/negative powers of derivatives removed by clearing radicals). Here the equation has a fractional power $p/q$ on the bracket containing the highest-order derivative, so we must raise the whole equation to the power $q$ first.

    Step-by-Step Solution

    1. Among $m,n,r$ we're told $n<r<m$, so the highest-order derivative present is $\dfrac{d^my}{dx^m}$. Hence the order of the equation is $m$.
    2. We are told the order is $4$, so $m=4$.
    3. The equation is $\left(\dfrac{d^my}{dx^m}+\dfrac{d^ny}{dx^n}\right)^{p/q} = 5\dfrac{d^ry}{dx^r}$. To remove the fractional exponent $p/q$, raise both sides to the power $q$: $\left(\dfrac{d^my}{dx^m}+\dfrac{d^ny}{dx^n}\right)^{p} = 5^q\left(\dfrac{d^ry}{dx^r}\right)^{q}$. …

    🔒 Start your 14-day free trial to see the full solution →
11. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
    Q.The difference of the order and degree of the differential equation $\left(\dfrac{d^2y}{dx^2}\right)^{-7/2}\left(\dfrac{d^3y}{dx^3}\right)^2-\left(\dfrac{d^2y}{dx^2}\right)^{-5/2}\left(\dfrac{d^4y}{dx^4}\right)=0$ is (A) $5$ (B) $3$ (C) $4$ (D) $2$

    ›Reveal solutionSolution

    Order $=4$, degree $=1$, so their difference is $4-1=3$.

    Concept and Intuition

    The order of a differential equation is the order of its highest derivative. The degree is the power of the highest-order derivative once the equation is made a polynomial in all derivatives (free of fractional/negative exponents).

    Step-by-Step Solution

    1. Highest derivative is $\dfrac{d^4y}{dx^4}$ so the order is 4.
    2. Multiply the whole equation by $\left(\dfrac{d^2y}{dx^2}\right)^{7/2}$ to remove negative fractional powers.
    3. First term: $\left(\dfrac{d^3y}{dx^3}\right)^2$; second term: $\left(\dfrac{d^2y}{dx^2}\right)^{1}\dfrac{d^4y}{dx^4}$.
    4. The equation is now polynomial, with $\dfrac{d^4y}{dx^4}$ appearing to the first power, so degree $=1$.
    5. Difference $= 4-1 = 3$.

    Common Mistakes

    • Reading degree directly from the fractional exponents without first clearing them. …

    🔒 Start your 14-day free trial to see the full solution →
12. 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 $3$): $x^6(y'')^3 = (1+y')^8.$ …

    🔒 Start your 14-day free trial to see the full solution →