Q.The differential equation for which $y=a\cos x+b\sin x$ is a solution, is:
(A) $\frac{d^2y}{dx^2}+y=0$
(B) $\frac{d^2y}{dx^2}-y=0$
(C) $\frac{d^2y}{dx^2}+(a+b)y=0$
(D) $\frac{d^2y}{dx^2}+(a-b)y=0$

Showing the 12 most recent of 21 on this concept.

1. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.The differential equation formed by eliminating arbitrary constants $A, B$ from the equation $y = A\cos 3x + B\sin 3x$ is (A) $\dfrac{d^2y}{dx^2} + y = 0$ (B) $\dfrac{d^2y}{dx^2} + 9y = 0$ (C) $\dfrac{d^2y}{dx^2} - 9y = 0$ (D) $\dfrac{d^2y}{dx^2} - y = 0$

   ›Reveal solutionSolution

   Since $y=A\cos3x+B\sin3x$, differentiating twice reproduces $-9y$, giving $y''+9y=0$.

   Concept and Intuition

   For $y=A\cos(kx)+B\sin(kx)$, each differentiation brings down a factor of $k$ and eventually $y''=-k^2y$ — the signature ODE of simple harmonic motion with angular frequency $k$.

   Step-by-Step Solution

   1. $y=A\cos3x+B\sin3x$.
   2. $y'=-3A\sin3x+3B\cos3x$.
   3. $y''=-9A\cos3x-9B\sin3x=-9(A\cos3x+B\sin3x)=-9y$.
   4. So $y''+9y=0$. …

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2. 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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3. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.a, b, c, d are arbitrary constants. Then the corresponding differential equation to $y=ae^x+be^{-x}+c\cos x+d\sin x$ is (A) $y^{(4)}=y$ (B) $y^{(4)}+y=0$ (C) $y^{(4)}-y^{(2)}+1=0$ (D) $y^{(4)}+2y^{(2)}+1=0$

   ›Reveal solutionSolution

   Split the solution into an "exponential part" satisfying $y''=y$ and a "trigonometric part" satisfying $y''=-y$; the operator that annihilates both simultaneously is $(D^2-1)(D^2+1)=D^4-1$.

   Concept and Intuition

   When a general solution is built from two independent families of functions, each satisfying its own lower-order ODE, the differential equation with no arbitrary constants is obtained by applying the product of the two annihilating operators — this is the standard trick for building higher-order constant-coefficient ODEs from known solution families.

   Step-by-Step Solution

   1. Let $y_1=ae^x+be^{-x}$. Then $y_1'=ae^x-be^{-x}$, $y_1''=ae^x+be^{-x}=y_1$. So $(D^2-1)y_1=0$.
   2. Let $y_2=c\cos x+d\sin x$. Then $y_2''=-c\cos x-d\sin x=-y_2$. So $(D^2+1)y_2=0$.
   3. Apply $(D^2+1)$ to the whole equation $(D^2-1)y_1=0$: still $0$ (since it commutes and $(D^2+1)\cdot 0=0$), and similarly $(D^2-1)$ applied to $(D^2+1)y_2=0$ stays $0$. …

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4. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.The differential equation formed by eliminating $a$ and $b$ from the equation $y = ae^{2x} + bxe^{2x}$ is (A) $y'' - 4y' - 4y = 0$ (B) $y'' + 4y' - 4y = 0$ (C) $y'' + 4y' + 4y = 0$ (D) $y'' - 4y' + 4y = 0$

   ›Reveal solutionSolution

   The form $y=(a+bx)e^{2x}$ is the textbook general solution for a repeated characteristic root $r=2$, so the corresponding differential equation is read off directly from $(r-2)^2=0$.

   Concept and Intuition

   Whenever a family of functions has the shape $(a+bx)e^{rx}$, it is exactly the general solution of a second-order linear ODE with constant coefficients whose characteristic equation has a repeated root at $r$. Recognizing this shape avoids needing to differentiate twice and eliminate constants by hand.

   Step-by-Step Solution

   1. $y=ae^{2x}+bxe^{2x}=(a+bx)e^{2x}$ — this is the standard form for a repeated root $r=2$ of the auxiliary equation.
   2. The auxiliary equation with repeated root $2$ is $(r-2)^2=0 \Rightarrow r^2-4r+4=0$.
   3. This corresponds to the ODE $y''-4y'+4y=0$. …

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5. 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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6. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
   Q.If the differential equation obtained by eliminating A, B from $y = (\sin^{-1}x)^2 + A\cos^{-1}x + B$ is $(a - x^2)y'' - xy' = b$, then $\dfrac{b+a}{b-a} =$ (A) $2$ (B) $-2$ (C) $3$ (D) $-3$

   ›Reveal solutionSolution

   Differentiating the given function twice eliminates both arbitrary constants and produces $(1-x^2)y''-xy'=2$; matching gives $a=1,b=2$ and $\frac{b+a}{b-a}=3$.

   Concept and Intuition

   A function with two arbitrary constants ($A,B$) satisfies a second-order differential equation obtained by differentiating twice (removing both constants). We just need to carry out that differentiation carefully, using $\dfrac{d}{dx}\sin^{-1}x = \dfrac{1}{\sqrt{1-x^2}}$ and $\dfrac{d}{dx}\cos^{-1}x = \dfrac{-1}{\sqrt{1-x^2}}$.

   Step-by-Step Solution

   1. $y = (\sin^{-1}x)^2 + A\cos^{-1}x + B$.
   2. First derivative: $y' = 2\sin^{-1}x\cdot\dfrac{1}{\sqrt{1-x^2}} - \dfrac{A}{\sqrt{1-x^2}} = \dfrac{2\sin^{-1}x - A}{\sqrt{1-x^2}}$.
   3. Rearrange: $y'\sqrt{1-x^2} = 2\sin^{-1}x - A$ — this has eliminated $B$ and still contains $A$ (as an additive constant), but note the combination is now clean.
   4. Differentiate again: $y''\sqrt{1-x^2} + y'\cdot\dfrac{-x}{\sqrt{1-x^2}} = \dfrac{2}{\sqrt{1-x^2}}$. …

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7. 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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8. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.The differential equation for which $ax + by =1$ is general solution is (A) $\dfrac{dy}{dx}=x+c$ (B) $y\dfrac{d^2y}{dx^2}+x=1$ (C) $\dfrac{d^2y}{dx^2}=0$ (D) $\dfrac{d^3y}{dx^3}=0$

   ›Reveal solutionSolution

   Two arbitrary constants means two differentiations to eliminate them; doing so on $ax+by=1$ leaves the trivial equation $y''=0$.

   Concept and Intuition

   To find the differential equation whose general solution is a given family, differentiate the family's equation as many times as there are arbitrary constants, then eliminate those constants using the resulting equations. Here $a,b$ are 2 independent constants, so 2 differentiations (and eliminating $a$ and $b$) should produce a second-order DE free of both constants.

   Step-by-Step Solution

   1. Start with $ax+by=1$.
   2. Differentiate w.r.t. $x$: $a+by'=0$ — (i)
   3. Differentiate (i) again w.r.t. $x$: since $a$ is a constant, $\dfrac{d}{dx}(a)=0$, so $by''=0$.
   4. Since $b$ is (generically) nonzero for a genuine two-parameter family, this forces $y''=0$, i.e. $\dfrac{d^2y}{dx^2}=0$. …

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9. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.The differential equation having $y = (a+b)e^{cx+d}$ as its general solution, where a, b, c, d are arbitrary constants, is (A) $y^{(4)} + 3yy^{(3)} + 6y^{(2)}y^2 + y = 0$ (B) $y^{(3)} + 4yy^{(2)} + 6y^2 y^{(1)} + 12y = 0$ (C) $y^{(1)} - y = 0$ (D) $yy^{(2)} - \left(y^{(1)}\right)^2 = 0$

   ›Reveal solutionSolution

   Although the solution is written with four letters $a,b,c,d$, only two of them are independent; eliminating them shows the family satisfies the second-order equation $yy''=(y')^2$.

   Concept and Intuition

   The number of arbitrary constants that genuinely matter tells you the order of the corresponding differential equation. Here $a$ and $b$ only ever appear as the sum $a+b$, and $d$ only ever appears multiplying through $e^d$ — a constant factor. So the family $y=(a+b)e^{cx+d}$ is really just $y = Be^{cx}$ with two truly independent constants $B$ and $c$, meaning we should expect (and can derive) a second-order differential equation.

   Step-by-Step Solution

   1. Simplify: $y = (a+b)e^{cx+d} = \left[(a+b)e^d\right]e^{cx} = Be^{cx}$, where $B$ is one constant.
   2. Differentiate: $y' = Bc\,e^{cx} = c\left(Be^{cx}\right) = cy$, so $c = \dfrac{y'}{y}$.
   3. Differentiate again: $y'' = c\,y'$ (since $c$ is constant, differentiating $y'=cy$ gives $y''=cy'$). …

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10. 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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11. 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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12. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
    Q.If c and d are arbitrary constants, then $y=e^{2x}\left(c\cosh\sqrt{2}x+d\sinh\sqrt{2}x\right)$ is the general solution of the differential equation (A) $y''+4y'+2y=0$ (B) $y''-4y'+2y=0$ (C) $y''-4y'+4y=0$ (D) $y''-2\sqrt{2}y'+2y=0$

    ›Reveal solutionSolution

    Reading off the characteristic roots from the given exponential/hyperbolic form gives the ODE $y''-4y'+2y=0$.

    Concept and Intuition

    A general solution of the form $e^{ax}(c\cosh(bx)+d\sinh(bx))$ is exactly the solution you get from two real, distinct characteristic roots $a+b$ and $a-b$ (since $\cosh,\sinh$ are combinations of $e^{bx}$ and $e^{-bx}$). Recognizing this instantly gives the roots without needing to expand anything.

    Step-by-Step Solution

    1. Write $e^{2x}\cosh(\sqrt2x)=\dfrac{e^{(2+\sqrt2)x}+e^{(2-\sqrt2)x}}{2}$ and $e^{2x}\sinh(\sqrt2x)=\dfrac{e^{(2+\sqrt2)x}-e^{(2-\sqrt2)x}}{2}$.
    2. So the general solution is a linear combination of $e^{(2+\sqrt2)x}$ and $e^{(2-\sqrt2)x}$, meaning the characteristic roots are $r_1=2+\sqrt2$, $r_2=2-\sqrt2$.
    3. Sum of roots: $r_1+r_2=4$. Product of roots: $r_1r_2=(2)^2-(\sqrt2)^2=4-2=2$. …

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