Find the differential equation of the family of curves $y = \text{Ae}^{2x} + \text{Be}^{-2x}$, where A and B are arbitrary constants.

Find the differential equation of the family of curves y = Ae2x + Be-2x, where A and B are arbitrary constants

Question

y = \text{Ae}^{2x} + \text{Be}^{-2x}

CBSECBSE Class XII Board 2019Subjective· 2mImportance★★★★★

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Concept understanding — Eliminating Arbitrary Constant

Eliminating Arbitrary Constants – The Core Idea

Consider a family of curves — say all circles centred at the origin: $x^2 + y^2 = r^2$ . Here $r$ is an arbitrary constant: each value gives one specific circle, but the whole family shares the same shape.

Suppose you're asked: "What differential equation does every member satisfy?" You need to eliminate the arbitrary constant $r$ to get an equation that holds for all such circles, regardless of $r$ .

That is the goal: start with an equation containing arbitrary constants, differentiate enough times to remove them, and obtain a differential equation representing the whole family.

Why Differentiate?

Arbitrary constants are constants — their derivative is zero. So differentiating either makes the constant disappear or lets you eliminate it by combining the original equation with its derivatives.

Differentiate $x^2 + y^2 = r^2$ with respect to $x$ : $\;2x + 2y \frac{dy}{dx} = 0$ . Divide by 2:

x + y \frac{dy}{dx} = 0

The constant $r$ is gone. The differential equation $x + y y' = 0$ is satisfied by every circle centred at the origin.

Note

We needed one differentiation because there was one arbitrary constant. In general, $n$ arbitrary constants require $n$ differentiations.

The Precise Statement

Eliminating arbitrary constants means: given an equation involving $x$ , $y$ , and $n$ arbitrary constants, differentiate it $n$ times (with respect to the independent variable) and then algebraically eliminate the $n$ constants from the system of $n+1$ equations (the original plus the $n$ derivatives). The result is an $n$ -th order ordinary differential equation representing the entire family of curves.

A Second Example (Two Constants)

Take all curves $y = A x + B x^2$ , where $A$ and $B$ are arbitrary constants.

Differentiate once: $\;y' = A + 2B x$

Differentiate again: $\;y'' = 2B$

Now we have three equations:

$y = A x + B x^2$
$y' = A + 2B x$
$y'' = 2B$

From (3), $B = \frac{y''}{2}$ . Substitute into (2): $y' = A + x y''$ , so $A = y' - x y''$ .

Substitute $A$ and $B$ into (1):

y = (y' - x y'') x + \left(\frac{y''}{2}\right) x^2 = x y' - \frac{x^2 y''}{2}

Multiply through by 2 and rearrange:

x^2 y'' - 2x y' + 2y = 0

This second-order ODE is satisfied by every curve $y = A x + B x^2$ , no matter what $A$ and $B$ are.

Watch out

A common mistake: trying to eliminate constants by substitution before differentiating enough times. You must differentiate first, then eliminate — substituting too early loses information.

Why This Matters …

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To find the differential equation, we differentiate the given family of curves twice to eliminate the two arbitrary constants, A and B, resulting in the second-order differential equation $\frac{d^2y}{dx^2} - 4y = 0$ .

When we are asked to find the differential equation of a family of curves, the core idea is to eliminate the arbitrary constants present in the equation of the family. A differential equation describes a relationship between a function and its derivatives, without any specific constants that define a particular member of the family.

The number of arbitrary constants in the equation of the family of curves dictates the order of the differential equation we will obtain. If there are $n$ arbitrary constants, we will need to differentiate the equation $n$ times to generate enough equations to eliminate all constants. In this problem, we have two arbitrary constants, A and B, so we expect to arrive at a second-order differential equation.

Let's proceed step-by-step:

Start with the given equation of the family of curves. The given equation is:

y = \text{Ae}^{2x} + \text{Be}^{-2x} \quad \text{(Equation 1)}

Here, A and B are the arbitrary constants we need to eliminate.

2. Differentiate the equation once with respect to $x$ .

This step introduces the first derivative, $\frac{dy}{dx}$ , into our system of equations.

\frac{dy}{dx} = \frac{d}{dx}(\text{Ae}^{2x} + \text{Be}^{-2x})

Using the chain rule, $\frac{d}{dx}(e^{kx}) = ke^{kx}$:

\frac{dy}{dx} = \text{A}(2\text{e}^{2x}) + \text{B}(-2\text{e}^{-2x})

\frac{dy}{dx} = 2\text{Ae}^{2x} - 2\text{Be}^{-2x} \quad \text{(Equation 2)}

Notice that A and B are still present. Since we have two constants, we need another differentiation.

3. Differentiate the first derivative again with respect to $x$ .

This step introduces the second derivative, $\frac{d^2y}{dx^2}$ , which is necessary for a second-order differential equation.

\frac{d^2y}{dx^2} = \frac{d}{dx}(2\text{Ae}^{2x} - 2\text{Be}^{-2x})

Again, applying the chain rule:

\frac{d^2y}{dx^2} = 2\text{A}(2\text{e}^{2x}) - 2\text{B}(-2\text{e}^{-2x})

\frac{d^2y}{dx^2} = 4\text{Ae}^{2x} + 4\text{Be}^{-2x} \quad \text{(Equation 3)}

Now we have three equations involving $y$, $\frac{dy}{dx}$, $\frac{d^2y}{dx^2}$, and the constants A and B.

4. Eliminate the arbitrary constants A and B. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set ANNUAL1 mark
$Q. What is the differential equation of the family of lines passing through the origin?$
›Reveal solutionSolution

Eliminate the single arbitrary slope constant $m$ from $y=mx$ by differentiating once.

The family of lines through the origin is $y = mx$ , where $m$ is an arbitrary constant. Differentiating with respect to $x$ : $\dfrac{dy}{dx}=m$ . Substituting back into $y=mx$ : …

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CBSE 2024Set ANNUAL1 mark
$y = A\sin x + B\cos x$
›Reveal solutionSolution

Differentiate twice and observe the second derivative is $-y$ .

Given $y = A\sin x + B\cos x$ .

Step 1: Differentiate once:

$\frac{dy}{dx} = A\cos x - B\sin x$

Step 2: Differentiate again:

$\frac{d^2y}{dx^2} = -A\sin x - B\cos x = -(A\sin x + B\cos x) = -y$

…

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CBSE 2023Set AX1 mark
$y=a\sin(x+b)$
›Reveal solutionSolution

Two arbitrary constants need two differentiations; eliminating $a,b$ gives $\dfrac{d^2y}{dx^2}+y=0$ .

Concept. A family with two arbitrary constants ( $a$ and $b$ ) yields a second-order differential equation, obtained by differentiating twice and eliminating the constants.

Solution. Given $y=a\sin(x+b)$ . …

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CBSE 2023Set ANNUAL1 markMCQ
$y = a\sin(x+b)$
(a) $\dfrac{d^2y}{dx^2} = y$
(b) $a\dfrac{d^2y}{dx^2} = by$
(c) $b\dfrac{d^2y}{dx^2} = ay$
(d) $\dfrac{d^2y}{dx^2} + y = 0$
›Reveal solutionSolution

Differentiate twice to eliminate the two arbitrary constants $a,b$ .

$y=a\sin(x+b)$

$\dfrac{dy}{dx}=a\cos(x+b)$

$\dfrac{d^2y}{dx^2}=-a\sin(x+b)=-y$

…

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CBSE 2022Set ANNUAL1 mark
$y=Ae^{2x}$
›Reveal solutionSolution

Differentiate the family and eliminate the arbitrary constant $A$ using the original equation.

$y=Ae^{2x} \Rightarrow \dfrac{dy}{dx}=2Ae^{2x}=2(Ae^{2x})=2y$ .

…

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CBSE 2020Set ANNUAL1 mark
$Q. Write the differential equation of all non-horizontal lines in a plane.$
›Reveal solutionSolution

Writing every non-horizontal line as $x=my+c$ and eliminating the arbitrary constants $m,c$ by differentiating twice gives $\dfrac{d^2x}{dy^2}=0$ .

A horizontal line cannot be written as $x=$ (function of $y$ ) in a one-to-one way, but every non-horizontal line (including vertical ones) can be written as

$x=my+c$

where $m,c$ are arbitrary constants (this family covers all non-horizontal lines in the plane).

…

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CBSE 20191 mark
$y = ae^{2x} + 5$
›Reveal solutionSolution

Differentiate once to eliminate the single arbitrary constant $a$ ; the family $y = ae^{2x} + 5$ yields the differential equation $\frac{dy}{dx} = 2(y - 5)$ .

Why we differentiate to eliminate constants

A family of curves with one arbitrary constant describes infinitely many curves—one for each value of the constant. The differential equation captures the relationship between $y$ , $x$ , and the slope $\frac{dy}{dx}$ that every member of the family satisfies, without reference to the specific constant.

The rule is simple: differentiate as many times as there are arbitrary constants. Here we have one constant $a$ , so one differentiation will let us express everything in terms of $y$ , $x$ , and $\frac{dy}{dx}$ alone.

Step-by-step elimination

1. Start with the given family:
$y = ae^{2x} + 5$
The constant $a$ multiplies $e^{2x}$ , and we have a vertical shift by $5$ .

2. Differentiate both sides with respect to $x$ :

The derivative of $ae^{2x}$ uses the chain rule: the derivative of $e^{2x}$ is $2e^{2x}$ , and the constant $5$ vanishes.
$\frac{dy}{dx} = a \cdot 2e^{2x} = 2ae^{2x}$
3. Eliminate $a$ by expressing $ae^{2x}$ in terms of $y$ :

From the original equation, rearrange to isolate the term containing $a$ :
$ae^{2x} = y - 5$
4. Substitute into the derivative:

Replace $ae^{2x}$ in the expression for $\frac{dy}{dx}$ :

$\frac{dy}{dx} = 2(y - 5)$ …
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CBSE 2017Set ANNUAL1 markMCQ
$y = c_1 e^x + c_2 e^{-x}$
(a) $\frac{d^2y}{dx^2}+1=0$
(b) $\frac{d^2y}{dx^2}-1=0$
(c) $\frac{d^2y}{dx^2}-y=0$
(d) $\frac{d^2y}{dx^2}+y=0$
›Reveal solutionSolution

Differentiating $y=c_1e^x+c_2e^{-x}$ twice reproduces $y$ itself, giving $y''-y=0$ .

…

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CBSE 2016Set ANNUAL1 mark
$c$
›Reveal solutionSolution

eliminate the arbitrary constant by differentiating

Given $xy=c^2$ . Differentiate both sides with respect to $x$ (the right side is a constant, so its derivative is $0$ ):

$x\dfrac{dy}{dx}+y\cdot 1=0$

…

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Q.Find the differential equation of the family of curves y=Ae2x+Be−2xy = \text{Ae}^{2x} + \text{Be}^{-2x}y=Ae2x+Be−2x, where A and B are arbitrary constants.

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Q.Find the differential equation of the family of curves $y = \text{Ae}^{2x} + \text{Be}^{-2x}$ , where A and B are arbitrary constants.

Unlock everything free for 14 days