Q.Family $y=Ax+A^3$ of curves will correspond to a differential equation of order:
(A) 3
(B) 2
(C) 1
(D) not defined

Showing the 12 most recent of 26 on this concept.

1. 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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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 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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4. 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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5. 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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6. 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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7. 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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8. 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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9. 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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10. 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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11. 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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12. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
    Q.Among the options given below, from which option a differential equation of order two can be formed? (A) All circles passing through origin (B) All parabolas passing through origin and having focus on x-axis (C) All the lines passing through the origin (D) All hyperbolas of the form $x^2 - y^2=k^2$

    ›Reveal solutionSolution

    Count the surviving arbitrary constants in each family after applying the stated conditions; only "circles through the origin" keeps 2 independent constants, so only it needs a second-order differential equation.

    Concept and Intuition

    The ORDER of the differential equation of a family of curves equals the number of essential arbitrary constants in the family's equation. A condition like "passes through the origin" or "passes through origin with focus on the x-axis" often uses up one of the constants, reducing the order by one. So the real task is to write each family's general equation, apply the given condition, and see how many constants remain.

    Step-by-Step Solution

    1. Circles through origin: general circle $x^2+y^2+2gx+2fy+c=0$ has 3 constants $(g,f,c)$. Passing through $(0,0)$ forces $c=0$, leaving $x^2+y^2+2gx+2fy=0$ with 2 free constants $g,f$ → needs a 2nd-order DE.
    2. Parabolas through origin with focus on x-axis: since the focus lies on the x-axis, the axis of the parabola is the x-axis, so the vertex also lies on it: $y^2=4a(x-h)$. Passing through origin gives $0=4a(0-h)\Rightarrow h=0$ (for $a\ne0$), leaving just $y^2=4ax$ — only 1 constant → 1st-order DE.
    3. Lines through origin: $y=mx$, 1 constant $m$ → 1st-order DE. …

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