Q.Find the differential equation representing the family of curves $y = ae^{bx+5}$, where $a$ and $b$ are arbitrary constants.

Showing the 12 most recent of 21 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.The differential equation having $y = \dfrac{A}{x^5} + \dfrac{\log x}{5} - \dfrac{1}{25}$ as its general solution is (A) $x\dfrac{dy}{dx} - 5y = x$ (B) $x\dfrac{d^2y}{dx^2} + y = \log x$ (C) $x\dfrac{dy}{dx} + 5y = \log x$ (D) $x^2\dfrac{dy}{dx} + 5y = \log x$

   ›Reveal solutionSolution

   Differentiating the given general solution once and combining terms to eliminate the arbitrary constant $A$ directly produces $x\,dy/dx+5y=\log x$.

   Concept and Intuition

   To recover the differential equation behind a stated general solution with one arbitrary constant, differentiate once and algebraically eliminate that constant between the original equation and its derivative.

   Step-by-Step Solution

   1. $y=Ax^{-5}+\dfrac{\log x}{5}-\dfrac{1}{25}$.
   2. Differentiate: $y'=-5Ax^{-6}+\dfrac{1}{5x}$.
   3. Multiply by $x$: $xy'=-5Ax^{-5}+\dfrac15$.
   4. Multiply the original equation by 5: $5y=5Ax^{-5}+\log x-\dfrac15$.
   5. Add the two: $xy'+5y=\big(-5Ax^{-5}+5Ax^{-5}\big)+\dfrac15-\dfrac15+\log x=\log x$. …

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2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.The differential equation of the family of curves given by $y = e^{3x}(Ax + B)$ where A, B are arbitrary constants, is (A) $\dfrac{d^2y}{dx^2} + 6\dfrac{dy}{dx} + 9y = 0$ (B) $\dfrac{d^2y}{dx^2} + 6\dfrac{dy}{dx} - 9y = 0$ (C) $\dfrac{d^2y}{dx^2} - 6\dfrac{dy}{dx} - 9y = 0$ (D) $\dfrac{d^2y}{dx^2} - 6\dfrac{dy}{dx} + 9y = 0$

   ›Reveal solutionSolution

   This tests recognizing that $y=(Ax+B)e^{kx}$ is the solution family for a repeated root $k,k$, so the ODE is $y''-2ky'+k^2y=0$; here $k=3$ gives $y''-6y'+9y=0$.

   Concept and Intuition

   When a linear second-order ODE with constant coefficients has a repeated characteristic root $m=k$, its general solution is $y=(Ax+B)e^{kx}$ — the extra factor of $x$ compensates for the root being repeated. Recognizing this pattern means we don't need to eliminate constants by differentiating twice; we can go straight from the solution form to the characteristic equation.

   Step-by-Step Solution

   1. The given family $y=e^{3x}(Ax+B)$ matches the repeated-root pattern with $k=3$.
   2. The characteristic equation for a repeated root $m=3$ is $(m-3)^2=0$, i.e. $m^2-6m+9=0$.
   3. Translating back to the differential equation: $\dfrac{d^2y}{dx^2}-6\dfrac{dy}{dx}+9y=0$. …

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3. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If $a$ and $h$ are arbitrary constants, then the differential equation corresponding to the family of curves $ax^2 + 2hxy = 1$ is (A) $x^2\dfrac{d^2y}{dx^2} + x\dfrac{dy}{dx} + y = 0$ (B) $x^2\dfrac{d^2y}{dx^2} - x\dfrac{dy}{dx} + y = 0$ (C) $x^2\dfrac{d^2y}{dx^2} + x\dfrac{dy}{dx} - y = 0$ (D) $x^2\dfrac{d^2y}{dx^2} - x\dfrac{dy}{dx} - y = 0$

   ›Reveal solutionSolution

   With two arbitrary constants $a,h$, we differentiate the family twice and eliminate both constants between the resulting equations, landing on $x^2y''+xy'-y=0$.

   Concept and Intuition

   A family of curves with $n$ independent arbitrary constants satisfies a differential equation of order $n$ (generically), obtained by differentiating the family $n$ times and eliminating the constants among the original equation and its derivatives. Here there are two constants ($a$ and $h$), so we expect (and need) a second-order ODE.

   Step-by-Step Solution

   1. Start with $ax^2 + 2hxy = 1$.
   2. Differentiate once w.r.t. $x$:

   $2ax + 2h(xy'+y) = 0 \;\Longrightarrow\; ax + h(xy'+y) = 0 \quad \text{...(I)}$

   3. Differentiate (I) again w.r.t. $x$:

   $a + h\big(xy''+y'+y'\big) = 0 \;\Longrightarrow\; a + h(xy''+2y') = 0 \quad \text{...(II)}$

   4. From (I): $a = -\dfrac{h(xy'+y)}{x}$ (for $x\neq0$).
   5. Substitute into (II):

   $-\frac{h(xy'+y)}{x} + h(xy''+2y') = 0$

   6. Divide through by $h$ (assuming $h\neq0$) and multiply by $x$:

   $-(xy'+y) + x^2y''+2xy' = 0$

   7. Simplify: …

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4. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.The differential equation obtained by eliminating $A$ and $B$ from the equation $y = A(x+B)^2$ which represents a family of curves (A) $2yy'' = (y')^2$ (B) $yy'' = 2y'$ (C) $2yy'' = y' + y$ (D) $2yy'' = y' - y$

   ›Reveal solutionSolution

   Differentiating the two-parameter family $y=A(x+B)^2$ twice and eliminating $A,B$ algebraically yields the differential equation $2yy''=(y')^2$.

   Concept and Intuition

   A family of curves with $n$ arbitrary constants generally satisfies an $n$th-order differential equation obtained by differentiating $n$ times and eliminating the constants. Here there are two constants ($A$ and $B$), so we differentiate twice and use the original equation plus its first and second derivatives to eliminate both.

   Step-by-Step Solution

   1. Given $y=A(x+B)^2$.
   2. Differentiate once: $y'=2A(x+B)$.
   3. Differentiate again: $y''=2A$, so $A=\dfrac{y''}{2}$.
   4. From step 2, solve for $(x+B)$: $(x+B)=\dfrac{y'}{2A}=\dfrac{y'}{y''}$ (substituting $2A=y''$). …

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5. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.If the differential equation of the family of curves given by the equation $y = ae^x + b\cos x$, where $a$ and $b$ are arbitrary constants is $y_2(\cos x + \sin x) + y(\cos x - \sin x) = 2y_1 f(x)$, then $f(x) =$ (A) $\sin x$ (B) $\cos x$ (C) $-\cos x$ (D) $-\sin x$

   ›Reveal solutionSolution

   Eliminating the two arbitrary constants $a,b$ from $y=ae^x+b\cos x$ by differentiating twice and combining terms shows $f(x)=\cos x$.

   Concept and Intuition

   A family of curves with $n$ arbitrary constants satisfies an $n$th-order ODE obtained by differentiating $n$ times and eliminating the constants. Here we have two constants ($a,b$), so we differentiate twice and combine $y,y_1,y_2$ algebraically to isolate them.

   Step-by-Step Solution

   1. $y=ae^x+b\cos x$
   2. $y_1=ae^x-b\sin x$
   3. $y_2=ae^x-b\cos x$
   4. Add (1) and (3): $y+y_2=2ae^x \Rightarrow ae^x=\dfrac{y+y_2}{2}$.
   5. Subtract (3) from (1): $y-y_2=2b\cos x \Rightarrow y_2-y=-2b\cos x$.
   6. Now evaluate $y_2(\cos x+\sin x)+y(\cos x-\sin x)=\cos x(y_2+y)+\sin x(y_2-y)$ $=\cos x\,(2ae^x)+\sin x\,(-2b\cos x)=2\cos x\,(ae^x-b\sin x)$. …

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6. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.The differential equation corresponding to the family of parabolas whose axis is along $x = 1$ is (A) $\dfrac{d^2y}{dx^2}-(x-1)\dfrac{dy}{dx}=0$ (B) $(x-1)\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx}=0$ (C) $\dfrac{d^2y}{dx^2}+(x-1)\dfrac{dy}{dx}-y=0$ (D) $(x-1)\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}=0$

   ›Reveal solutionSolution

   This tests forming a differential equation by eliminating arbitrary constants from a family of curves; the answer is (B).

   Concept and Intuition

   A family of parabolas with a fixed vertical axis $x=1$ has the general equation $y=a(x-1)^2+b$, where $a$ (controls width/orientation) and $b$ (vertical shift of vertex) are the two free parameters. Since there are 2 arbitrary constants, we need a 2nd-order ODE to eliminate them completely.

   Step-by-Step Solution

   1. Write the family: $y=a(x-1)^2+b$.
   2. Differentiate once: $\dfrac{dy}{dx}=2a(x-1)$.
   3. Differentiate again: $\dfrac{d^2y}{dx^2}=2a$, so $a=\dfrac12\dfrac{d^2y}{dx^2}$.
   4. Substitute this back into the first derivative relation: $\dfrac{dy}{dx}=\left(\dfrac{d^2y}{dx^2}\right)(x-1)$.
   5. Rearranging: $(x-1)\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx}=0$, which eliminates both $a$ and $b$ (note $b$ already dropped out automatically after the first derivative). …

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7. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.The differential equation of the family of circles passing through the origin and having centre on X-axis is (A) $(y^2+x^2)dx - 2ydy = 0$ (B) $(y^2-x^2)dx - 2xydy = 0$ (C) $(y^2-x^2)dx + 2ydy = 0$ (D) $(y^2+x^2)dx + 2ydy = 0$

   ›Reveal solutionSolution

   Eliminating the one parameter $a$ (the centre's $x$-coordinate) from the circle's equation via differentiation gives the differential equation $(y^2-x^2)dx-2xy\,dy=0$.

   Concept and Intuition

   A one-parameter family of curves satisfies a first-order differential equation obtained by differentiating the family's equation once and eliminating the parameter. Here the family is "circles through the origin with centre on the x-axis," which has exactly one parameter: the centre's $x$-coordinate $a$ (the radius must equal $a$ since the circle passes through the origin).

   Step-by-Step Solution

   1. Circle with centre $(a,0)$, radius $a$ (so it passes through the origin): $(x-a)^2+y^2=a^2\Rightarrow x^2+y^2=2ax$.
   2. Differentiate w.r.t. $x$: $2x+2yy'=2a\Rightarrow a=x+yy'$.
   3. Substitute back into $x^2+y^2=2ax$: $x^2+y^2=2x(x+yy')=2x^2+2xyy'$. …

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8. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.The differential equation for which $y^2=4a(x+a)$ (a is the parameter) is the general solution is (A) $y=2x\dfrac{dy}{dx}+y\left(\dfrac{dy}{dx}\right)^2$ (B) $y=y\dfrac{dy}{dx}-x\left(\dfrac{dy}{dx}\right)^2$ (C) $x=3\dfrac{dy}{dx}+y\left(\dfrac{dy}{dx}\right)^2$ (D) $y=3x^2\dfrac{dy}{dx}+y^2\left(\dfrac{dy}{dx}\right)^2$

   ›Reveal solutionSolution

   This tests eliminating the arbitrary constant from a one-parameter family to get its differential equation. Differentiate once, solve for $a$, substitute back. Answer: (A).

   Concept and Intuition

   A family of curves with $n$ independent parameters satisfies a differential equation of order $n$. Here only $a$ is a parameter, so one differentiation should let us eliminate it completely and land back on a relation purely in $x, y, y'$.

   Step-by-Step Solution

   1. Start with $y^2 = 4a(x+a) = 4ax + 4a^2$.
   2. Differentiate both sides with respect to $x$ (treating $a$ as constant):

   $2y\frac{dy}{dx} = 4a \implies a = \frac{1}{2}y\frac{dy}{dx}.$

   3. Substitute this expression for $a$ back into the original equation to eliminate $a$:

   $y^2 = 4x\left(\frac{1}{2}yy'\right) + 4\left(\frac{1}{2}yy'\right)^2 = 2xyy' + y^2(y')^2.$

   4. Divide throughout by $y$ (valid away from $y=0$):

   $y = 2x\frac{dy}{dx} + y\left(\frac{dy}{dx}\right)^2.$

   5. This matches option (A) exactly. …

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9. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.If a and b are arbitrary constants, then the differential equation corresponding to the family of curves $y = \tan(ax+b)$ is (A) $(1+x^2)y_2 - 2yy_1 + y = 0$ (B) $(1+y^2)y_2 - 2yy_1^2 = 0$ (C) $(1+x^2)y_2 + 2yy_1^2 = 0$ (D) $(1+y^2)y_2 - 2yy_1^2 + y = 0$

   ›Reveal solutionSolution

   Differentiating $y=\tan(ax+b)$ twice and eliminating the arbitrary constant $a$ (using $\sec^2=1+\tan^2$) directly gives the second-order ODE $(1+y^2)y_2-2yy_1^2=0$.

   Concept and Intuition

   A 2-parameter family of curves satisfies a second-order differential equation obtained by differentiating twice and eliminating both arbitrary constants. Since $b$ only shifts the argument (it never appears explicitly after one differentiation of $\tan$), only $a$ needs eliminating here.

   Step-by-Step Solution

   1. $y=\tan(ax+b)$. First derivative: $y_1=a\sec^2(ax+b)=a(1+\tan^2(ax+b))=a(1+y^2)$.
   2. So $a=\dfrac{y_1}{1+y^2}$ — note $b$ has already dropped out.
   3. Differentiate $y_1=a(1+y^2)$ again with respect to $x$: $y_2=a\cdot2yy_1$ (since $a$ is constant).
   4. Substitute $a=\dfrac{y_1}{1+y^2}$: $y_2=\dfrac{y_1}{1+y^2}\cdot2yy_1=\dfrac{2yy_1^2}{1+y^2}$. …

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10. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
    Q.If $y = At^2 + \dfrac{B}{t}$ (A, B are parameters) is general solution of the differential equation $f(t)\,y''(t) + g(t)\,y'(t) + h(t)\,y = 0$ then $2f(t) + t^2 h(t) =$ (A) $g(t) - h(t)$ (B) $g(t) + f(t)$ (C) $g(t)\,f(t)$ (D) $(f(t))^{g(t)}$

    ›Reveal solutionSolution

    This tests recognizing $y=At^2+B/t$ as the solution family of a Cauchy–Euler equation, whose middle ($y'$) coefficient must vanish, forcing $2f(t)+t^2h(t)\equiv0$.

    Concept and Intuition

    A linear 2nd-order ODE with independent solutions $t^{m_1},t^{m_2}$ is (up to an overall scalar function) the Euler equation $t^2y''+pty'+qy=0$, where $m$ satisfies $m(m-1)+pm+q=0$. Since the two solutions here are $t^2$ and $t^{-1}$, we can read off $p,q$ from the roots $m=2,-1$ and hence the exact ratio $f:g:h$.

    Step-by-Step Solution

    1. Write $m^2+(p-1)m+q=0$ with roots $2$ and $-1$: sum $=1=-(p-1)\Rightarrow p=0$; product $=-2=q$.
    2. So the defining ODE is $t^2y''+0\cdot ty'-2y=0$, i.e. any valid $(f,g,h)$ must be of the form $f(t)=\mu(t)t^2,\ g(t)=0,\ h(t)=-2\mu(t)$ for some nonzero function $\mu(t)$ (multiplying the whole equation by $\mu$ keeps the same solution set).
    3. Compute the target: $2f(t)+t^2h(t)=2\mu t^2+t^2(-2\mu)=2\mu t^2-2\mu t^2=0$ — identically zero, independent of $\mu$. …

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11. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
    Q.If $Ax^3 + Bxy = 4$ (A and B are arbitrary constants) is the general solution of the differential equation $F(x)\dfrac{d^2y}{dx^2} + G(x)\dfrac{dy}{dx} - 2y = 0$, then $F(1) + G(1) =$ (A) $1$ (B) $0$ (C) $4$ (D) $9$

    ›Reveal solutionSolution

    Solving the given relation for $y$ shows it is $y=C_1/x+C_2x^2$, whose ODE is $x^2y''-2y=0$; so $F(x)=x^2$, $G(x)=0$, giving $F(1)+G(1)=1$.

    Concept and Intuition

    A relation with two arbitrary constants that is claimed to be the general solution of a 2nd-order linear ODE can be turned INTO that ODE by solving explicitly for $y$ (when possible) and then eliminating the two constants via two differentiations — this is often cleaner than blind implicit differentiation.

    Step-by-Step Solution

    1. From $Ax^3+Bxy=4$, solve for $y$: $Bxy=4-Ax^3 \Rightarrow y=\dfrac{4}{Bx}-\dfrac{A}{B}x^2$.
    2. Since $A,B$ are arbitrary, $C_1=4/B$ and $C_2=-A/B$ are themselves two independent arbitrary constants: $y=C_1x^{-1}+C_2x^2$.
    3. Differentiate: $y'=-C_1x^{-2}+2C_2x$, and again: $y''=2C_1x^{-3}+2C_2$.
    4. Compute $x^2y''=2C_1x^{-1}+2C_2x^2$. But $2y=2C_1x^{-1}+2C_2x^2$ too — so $x^2y''=2y$, i.e. $x^2y''-2y=0.$ …

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12. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
    Q.The differential equation representing the family of circles having their centres on Y-axis is $\left(y_1=\dfrac{dy}{dx} \text{ and } y_2=\dfrac{d^2y}{dx^2}\right)$ (A) $y_2=y(y_1^2+1)$ (B) $y_2=xy(y_1^2+1)$ (C) $xy_2=y_1(y_1^2+1)$ (D) $xy_2=y(y_1^2+1)$

    ›Reveal solutionSolution

    Eliminating the two constants $k$ (center) and $a$ (radius) from $x^2+(y-k)^2=a^2$ via two differentiations yields $xy_2=y_1(y_1^2+1)$.

    Concept and Intuition

    A family of curves with $n$ independent arbitrary constants satisfies a differential equation of order $n$ obtained by differentiating $n$ times and eliminating the constants. Circles centered anywhere on the Y-axis have two free parameters — the center's y-coordinate $k$ and the radius $a$ — so we need exactly two differentiations.

    Step-by-Step Solution

    1. General equation of a circle with center $(0,k)$ and radius $a$: $x^2+(y-k)^2=a^2$.
    2. Differentiate once w.r.t. $x$: $2x+2(y-k)y_1=0\Rightarrow x+(y-k)y_1=0\Rightarrow (y-k)=-\dfrac{x}{y_1}$.
    3. Differentiate again: $1+y_1\cdot y_1+(y-k)y_2=0$, i.e. $1+y_1^2+(y-k)y_2=0$.
    4. Substitute $(y-k)=-x/y_1$ from step 2: $1+y_1^2-\dfrac{x}{y_1}y_2=0$. …

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