Q.Determine the order and degree, if defined, of the differential equation: $(y''')^2 + (y'')^3 + (y')^4 + y^5 = 0$

Showing the 12 most recent of 26 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If the degree of the differential equation $\left(\dfrac{d^2y}{dx^2}\right)^{3/2} + 5\left(\dfrac{d^2y}{dx^2}\right)^{5/2} = 7y$ is $m$ and its order is $n$, then $y = Ae^{mx} + Be^{nx}$ is solution of the differential equation (A) $\dfrac{d^2y}{dx^2} - 12\dfrac{dy}{dx} + 20y = 0$ (B) $\dfrac{d^2y}{dx^2} - 7\dfrac{dy}{dx} + 14y = 0$ (C) $\dfrac{d^2y}{dx^2} - 10\dfrac{dy}{dx} + 16y = 0$ (D) $\dfrac{d^2y}{dx^2} - 8\dfrac{dy}{dx} + 12y = 0$

   ›Reveal solutionSolution

   Clearing both fractional powers of $y''$ in turn (not at once) gives degree $10$, order $2$; the ODE with roots $10$ and $2$ is $y''-12y'+20y=0$.

   Concept and Intuition

   Order is simply which derivative is the highest one present — here only $y''$ (and $y$) appear, so order $n=2$, independent of any algebra. Degree requires the equation to be written as a polynomial in the derivatives (no fractional/negative powers); when there is more than one differently-fractional-powered term of the same derivative, each radical is removed in turn: isolate one radical term, raise both sides to clear it, and repeat for whatever fractional power still remains, until the equation is fully polynomial in $y''$.

   Step-by-Step Solution

   Let $t=\dfrac{d^2y}{dx^2}$, so the equation is $t^{3/2}+5t^{5/2}=7y$.

   1. Isolate the higher-power term and square:

   $5t^{5/2}=7y-t^{3/2}\ \Rightarrow\ 25t^5 = 49y^2-14y\,t^{3/2}+t^3$

   2. A fractional power, $t^{3/2}$, still remains — isolate it:

   $14y\,t^{3/2}=49y^2+t^3-25t^5$

   3. Square again to finish clearing it:

   $196\,y^2 t^3=\big(49y^2+t^3-25t^5\big)^2$

   4. This is now a genuine polynomial in $t=y''$. Expanding the right side, the single highest-power contribution is $(-25t^5)^2=625\,t^{10}$, and no other term reaches power $10$, so the highest power of $y''$ present is $10$ $\Rightarrow$ degree $m=10$. The order (highest derivative appearing anywhere) is $n=2$. …

   🔒 Start your 14-day free trial to see the full solution →
2. 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: $\left\{1+\left(\dfrac{dy}{dx}\right)^2\right\}^{3}=\left(\dfrac{d^2y}{dx^2}\right)^{2}$. …

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

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

   🔒 Start your 14-day free trial to see the full solution →
5. 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: $\dfrac{dy}{y}=\dfrac{dx}{2x}$. …

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

   🔒 Start your 14-day free trial to see the full solution →
7. 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 →
8. 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 →
9. 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: $\left(\dfrac{d^3y}{dx^3}\right)^2=\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^5$. …

   🔒 Start your 14-day free trial to see the full solution →
10. 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 →
11. 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 →
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. …

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