Q.(xi) The differential equation of all non-horizontal lines in a plane is $\frac{d^2x}{dy^2}=0$. (State True or False.)

Showing the 12 most recent of 26 on this concept.

1. 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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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 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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4. 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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5. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.If $l$ and $m$ are order and degree of a differential equation of all the straight lines at constant distance of P units from the origin, then $lm^2 + l^2m =$ (A) 2 (B) 6 (C) 12 (D) 30

   ›Reveal solutionSolution

   Eliminating the parameter from the family of lines at fixed distance $p$ from the origin gives a first-order, second-degree differential equation, so $l=1,m=2$ and $lm^2+l^2m=6$.

   Concept and Intuition

   Every line at a fixed perpendicular distance $p$ from the origin can be written as $x\cos\alpha+y\sin\alpha=p$, where $\alpha$ (the angle the perpendicular from the origin makes with the $x$-axis) is the single parameter distinguishing different lines in the family. Differentiating once and eliminating $\alpha$ produces the differential equation satisfied by every member of this family.

   Step-by-Step Solution

   1. Family: $x\cos\alpha + y\sin\alpha = p$ (one parameter $\alpha$, with $p$ a fixed constant).
   2. Differentiate w.r.t. $x$: $\cos\alpha + y'\sin\alpha = 0 \Rightarrow \cos\alpha = -y'\sin\alpha$.
   3. Using $\sin^2\alpha+\cos^2\alpha=1$: $y'^2\sin^2\alpha+\sin^2\alpha=1 \Rightarrow \sin\alpha = \pm\dfrac1{\sqrt{1+y'^2}}$, and correspondingly $\cos\alpha=\mp\dfrac{y'}{\sqrt{1+y'^2}}$.
   4. Substitute back into the original family equation: $x\left(\mp\dfrac{y'}{\sqrt{1+y'^2}}\right)+y\left(\pm\dfrac1{\sqrt{1+y'^2}}\right)=p \Rightarrow \pm\dfrac{y-xy'}{\sqrt{1+y'^2}}=p$.
   5. Squaring to remove the $\pm$ ambiguity: $(y-xy')^2 = p^2(1+y'^2)$.
   6. This equation contains only the first derivative $y'$ (no higher derivatives), so the order $l=1$. …

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6. 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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7. 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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8. 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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9. 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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10. 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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11. 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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12. 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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