Q.(vi) The differential equation representing the family of circles $x^2+(y-a)^2=a^2$ will be of order two. (State True or False.)

Showing the 12 most recent of 26 on this concept.

1. 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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2. 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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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 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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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 2026Set eng-2026-05-15-FN1 markMCQ
   Q.Among the following the differential equations, the equation having order 2 and degree 3 is (A) $\dfrac{dy}{dx} - \sin y = \dfrac{d^2y}{dx^2}\left(\sqrt{\dfrac{d^2y}{dx^2} - 1}\right)$ (B) $\left(\dfrac{d^2y}{dx^2}\right)^3 = \dfrac{dy}{dx} + y^2\left(\dfrac{d^3y}{dx^3}\right)^2$ (C) $\left(\dfrac{d^2y}{dx^2}\right)^3 = \left(\dfrac{d^2y}{dx^2}\right)^{3/2} + x^2$ (D) $\dfrac{dy}{dx} - \sin y = \left(\dfrac{d^2y}{dx^2}\right)^3\left(\sqrt{\dfrac{d^2y}{dx^2}} - 1\right)$

   ›Reveal solutionSolution

   Order is the highest derivative present; degree is the power of that highest derivative once the equation is made free of radicals/fractional powers involving derivatives. Only option (A) reduces to order 2, degree 3. Answer: (A).

   Concept and Intuition

   To find the degree of a differential equation, first make sure it is written as a polynomial in derivatives — any square roots, fractional powers, or derivatives inside denominators must be cleared first (by squaring, cubing, etc., as needed) before reading off the exponent of the highest-order derivative term.

   Step-by-Step Solution

   1. Option (A): $\dfrac{dy}{dx}-\sin y=y''\sqrt{y''-1}$. Highest derivative: $y''$ (order 2). Isolate the radical (already isolated) and square both sides: $\left(\dfrac{dy}{dx}-\sin y\right)^2=(y'')^2(y''-1)=(y'')^3-(y'')^2$. This is polynomial in $y''$ with highest power $3$ — order 2, degree 3. ✓ Matches what's asked.
   2. Option (B): $(y'')^3=y'+y^2(y''')^2$. Highest derivative is $y'''$ (order 3), appearing squared — order 3, degree 2. Does not match.
   3. Option (C): $(y'')^3=(y'')^{3/2}+x^2$. Isolate the fractional power: $(y'')^3-x^2=(y'')^{3/2}$; squaring: $\left[(y'')^3-x^2\right]^2=(y'')^3$, giving highest power $(y'')^6$ — order 2, degree 6. Does not match. …

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10. 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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11. 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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12. 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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