NCERT Class 12 Maths: Differential Equations Exercise 9.3

Formation of Differential Equations: Theoretical Foundation

To form a differential equation representing a given family of curves, our primary objective is to eliminate the arbitrary constants. Suppose we are given an equation representing a family of curves:

$F(x, y, c_1, c_2, ..., c_n) = 0$

where $c_1, c_2, ..., c_n$ are $n$ independent arbitrary constants. To eliminate these $n$ constants, we must differentiate the equation successively $n$ times with respect to $x$. This gives us $n$ additional equations containing derivatives up to order $n$:

1. First derivative: $g_1(x, y, y', c_1, ..., c_n) = 0$
2. Second derivative: $g_2(x, y, y', y'', c_1, ..., c_n) = 0$

...

1. $n$-th derivative: $g_n(x, y, y', ..., y^{(n)}, c_1, ..., c_n) = 0$

By utilizing these $n+1$ equations (the original curve equation plus the $n$ derivative equations), we algebraically eliminate the $n$ constants $c_1, c_2, ..., c_n$. The resulting relation is of the form:

$\Phi(x, y, y', y'', ..., y^{(n)}) = 0$

This final equation is the required differential equation. Notice that the order of the resulting differential equation is precisely equal to the number of independent arbitrary constants present in the original family of curves.

Key Terminology & Rules

Arbitrary Constant

A parameter in the equation of a family of curves that can take any real value, defining a specific member of that family (e.g., 'a' and 'b' in the equation of an ellipse).

Family of Curves

A set of curves represented by a single algebraic equation containing one or more parameters (arbitrary constants).

Order Rule of Formation

The order of a differential equation formed by eliminating arbitrary constants is equal to the number of independent arbitrary constants in the equation of the family of curves.

Step-by-Step Methodology to Form a Differential Equation

1. Identify and Count — Analyze the given equation of the family of curves and identify the number of independent arbitrary constants (let's say $n$).
2. Successive Differentiation — Differentiate the given equation with respect to $x$ successively $n$ times to obtain $n$ derivative equations.
3. Eliminate Constants — Use algebraic manipulation, substitution, or determinants to eliminate the arbitrary constants from the system of $n+1$ equations.
4. Write Final Equation — Express the final equation in terms of $x, y, y', y'', ..., y^{(n)}$, ensuring it is free of all arbitrary constants.

Deeply Worked Board-Level Examples

• Example 1: Form the differential equation representing the family of curves $y = a e^{3x} + b e^{-2x}$ by eliminating the arbitrary constants $a$ and $b$. Step 1: Count the arbitrary constants. There are two constants: $a$ and $b$. Thus, the resulting differential equation must be of order 2. Step 2: First differentiation. $y' = 3a e^{3x} - 2b e^{-2x} \quad \text{--- (Equation 1)}$ Step 3: Second differentiation. $y'' = 9a e^{3x} + 4b e^{-2x} \quad \text{--- (Equation 2)}$ Step 4: Eliminate constants. We can express $y''$ in terms of $y$ and $y'$. Let us write: $y'' = 9a e^{3x} + 4b e^{-2x} = 3(3a e^{3x} - 2b e^{-2x}) + 10b e^{-2x} = 3y' + 10b e^{-2x}$ Alternatively, a standard linear algebra approach or direct substitution works best. Let's solve Equation 1 and the original equation for $a e^{3x}$ and $b e^{-2x}$: From original equation: $y = a e^{3x} + b e^{-2x} \implies b e^{-2x} = y - a e^{3x}$. Substitute this into Equation 1: $y' = 3a e^{3x} - 2(y - a e^{3x}) = 5a e^{3x} - 2y \implies a e^{3x} = \frac{y' + 2y}{5}$ Similarly, we can find: $b e^{-2x} = y - \left(\frac{y' + 2y}{5}\right) = \frac{3y - y'}{5}$ Now, substitute these expressions into Equation 2: $y'' = 9\left(\frac{y' + 2y}{5}\right) + 4\left(\frac{3y - y'}{5}\right)$ $5y'' = 9y' + 18y + 12y - 4y'$ $5y'' = 5y' + 30y \implies y'' - y' - 6y = 0$ Final Answer: The required differential equation is $y'' - y' - 6y = 0$.
• Example 2: Form the differential equation of the family of circles touching the y-axis at the origin. Step 1: Write the equation of the family of curves. A circle touching the y-axis at the origin has its center on the x-axis at $(a, 0)$ and radius equal to $a$. Equation of circle: $(x - a)^2 + y^2 = a^2$ $x^2 - 2ax + a^2 + y^2 = a^2 \implies x^2 + y^2 = 2ax \quad \text{--- (Equation 1)}$ There is only one arbitrary constant, $a$. Hence, the differential equation will be of order 1. Step 2: Differentiate Equation 1 with respect to $x$: $2x + 2y \frac{dy}{dx} = 2a \quad \text{--- (Equation 2)}$ Step 3: Substitute the value of $2a$ from Equation 2 back into Equation 1 to eliminate $a$: $x^2 + y^2 = \left(2x + 2y \frac{dy}{dx}\right)x$ $x^2 + y^2 = 2x^2 + 2xy \frac{dy}{dx}$ $y^2 - x^2 = 2xy \frac{dy}{dx}$ Final Answer: The required differential equation is $2xy \frac{dy}{dx} + x^2 - y^2 = 0$.

Exam Trap: Counting 'Disguised' Arbitrary Constants

A very common Board Exam trap is presenting equations with "disguised" arbitrary constants. The order of the differential equation is determined ONLY by the number of independent arbitrary constants, not the total number of constant symbols written.

Example Trap:
Consider the family of curves: $y = A e^{x + B}$
It looks like there are two constants: $A$ and $B$. However, we can rewrite the equation as:
$y = A \cdot e^x \cdot e^B = (A e^B) e^x$
Since $A$ and $e^B$ are both constants, their product $(A e^B)$ can be replaced by a single arbitrary constant, say $C$:
$y = C e^x$
This equation actually contains only one independent arbitrary constant. Therefore, its differential equation will be of order 1, not order 2!
Always simplify exponential, logarithmic, and trigonometric constant expressions before counting the order.

Practice Questions with Solutions

• Q: Form the differential equation representing the family of curves $y = c_1 e^{2x} + c_2 e^{-2x}$ where $c_1$ and $c_2$ are arbitrary constants. A: Step 1: Identify arbitrary constants. Here, there are two independent arbitrary constants ($c_1$ and $c_2$). The resulting differential equation must be of order 2. Step 2: Differentiate the given equation with respect to $x$: $y' = 2c_1 e^{2x} - 2c_2 e^{-2x} \quad \text{--- (Equation 1)}$ Step 3: Differentiate with respect to $x$ once more: $y'' = 4c_1 e^{2x} + 4c_2 e^{-2x}$ Step 4: Factor out the common term to relate it back to the original equation: $y'' = 4(c_1 e^{2x} + c_2 e^{-2x})$ Since $y = c_1 e^{2x} + c_2 e^{-2x}$, substitute $y$ back into the second derivative expression: $y'' = 4y \implies y'' - 4y = 0$ Final answer: The required differential equation is $\frac{d^2y}{dx^2} - 4y = 0$.
• Q: Form the differential equation of the family of ellipses having foci on the x-axis and center at the origin. A: Step 1: Write the standard equation of the ellipse. The equation of an ellipse centered at the origin with foci on the x-axis is: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad \text{--- (Equation 1)}$ Here, $a$ and $b$ are two arbitrary constants. Thus, the order of the differential equation will be 2. Step 2: Differentiate Equation 1 with respect to $x$: $\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 \implies \frac{x}{a^2} + \frac{y}{b^2} y' = 0$ Rearranging the terms, we get: $\frac{y y'}{x} = -\frac{b^2}{a^2} \quad \text{--- (Equation 2)}$ Step 3: Differentiate Equation 2 with respect to $x$ using the quotient rule on the LHS (noting that the RHS is a pure constant $-b^2/a^2$ which becomes $0$): $\frac{d}{dx}\left(\frac{y y'}{x}\right) = 0$ $\frac{x \cdot \frac{d}{dx}(y y') - (y y') \cdot 1}{x^2} = 0$ $\frac{x(y y'' + (y')^2) - y y'}{x^2} = 0$ Step 4: Simplify the numerator: $x y y'' + x (y')^2 - y y' = 0$ Final answer: The required differential equation is $x y \frac{d^2y}{dx^2} + x \left(\frac{dy}{dx}\right)^2 - y \frac{dy}{dx} = 0$.
• Q: Form the differential equation representing the family of curves $y^2 = a(b^2 - x^2)$ where $a$ and $b$ are arbitrary constants. A: Step 1: Write down the equation: $y^2 = ab^2 - ax^2$. There are two arbitrary constants, $a$ and $b$. Thus, we need a second-order differential equation. Step 2: Differentiate with respect to $x$: $2y y' = -2ax \implies y y' = -ax \quad \text{--- (Equation 1)}$ Step 3: Differentiate Equation 1 again with respect to $x$ using the product rule on the LHS: $y y'' + (y')^2 = -a \quad \text{--- (Equation 2)}$ Step 4: We can now eliminate the constant $a$. From Equation 1, we have $-a = \frac{y y'}{x}$. Substitute this value of $-a$ into Equation 2: $y y'' + (y')^2 = \frac{y y'}{x}$ Multiply the entire equation by $x$ to clear the fraction: $x y y'' + x (y')^2 - y y' = 0$ Final answer: The required differential equation is $x y \frac{d^2y}{dx^2} + x \left(\frac{dy}{dx}\right)^2 - y \frac{dy}{dx} = 0$.
• Q: Form the differential equation representing the family of parabolas having their vertices at the origin and axes along the positive y-axis. A: Step 1: The standard equation of parabolas with vertices at $(0,0)$ and axes along the positive y-axis is: $x^2 = 4ay$ Here, $a$ is the single arbitrary constant. The differential equation must be of order 1. Step 2: Differentiate with respect to $x$: $2x = 4a y' \implies 4a = \frac{2x}{y'}$ Step 3: Substitute the value of $4a$ back into the original equation: $x^2 = \left(\frac{2x}{y'}\right)y$ $x^2 y' = 2xy$ Divide both sides by $x$ (since $x \neq 0$ for a general point on the curve): $x y' = 2y \implies x \frac{dy}{dx} - 2y = 0$ Final answer: The required differential equation is $x \frac{dy}{dx} - 2y = 0$.

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