Formation of Differential Equations
Introduction
Formation of Differential Equations involves creating a differential equation from a given relation by eliminating arbitrary constants. This is the reverse process of solving differential equations.
Basic Principle
$$\boxed{\text{Number of arbitrary constants} = \text{Order of DE}}$$Key Concept: If a relation contains $n$ arbitrary constants, differentiate it $n$ times and eliminate the constants to get a differential equation of order $n$.
General Procedure
Step-by-Step Method
Given: An equation with $n$ arbitrary constants
Step 1: Differentiate the equation with respect to $x$ (as many times as there are arbitrary constants)
Step 2: Get $n+1$ equations (original + $n$ derivatives)
Step 3: Eliminate $n$ arbitrary constants from these $(n+1)$ equations
Step 4: The resulting equation is the required DE of order $n$
Case 1: One Arbitrary Constant
For equation with one arbitrary constant, differentiate once and eliminate the constant.
Interactive Demo: Slope Field Visualization
Explore how differential equations create direction fields. Click anywhere to draw solution curves through any initial condition.
Example 1: Linear Relation
Form DE from: $y = mx + c$ where $c$ is an arbitrary constant.
Solution: Given: $y = mx + c$ … (1) [one constant $c$]
Differentiate w.r.t. $x$:
$$\frac{dy}{dx} = m$$From (1): $c = y - mx$
Since $m = \frac{dy}{dx}$, we can write: $c = y - x\frac{dy}{dx}$
Differentiating (1) again:
$$\frac{d^2y}{dx^2} = 0$$Required DE: $\boxed{\frac{d^2y}{dx^2} = 0}$ (Order 2, since we had 1 arbitrary constant)
Wait, let me reconsider. If we’re given $y = mx + c$ where only $c$ is arbitrary (and $m$ is a given constant), we get:
Differentiate: $\frac{dy}{dx} = m$ (constant)
Differentiate again: $\boxed{\frac{d^2y}{dx^2} = 0}$
If both $m$ and $c$ are arbitrary (2 constants): We need to differentiate twice and get: $\frac{d^2y}{dx^2} = 0$
Example 2: Exponential Form
Form DE from: $y = Ae^{3x}$ where $A$ is arbitrary.
Solution: Given: $y = Ae^{3x}$ … (1)
Differentiate w.r.t. $x$:
$$\frac{dy}{dx} = 3Ae^{3x} = 3y$$Required DE: $\boxed{\frac{dy}{dx} = 3y}$ or $\boxed{\frac{dy}{dx} - 3y = 0}$
Case 2: Two Arbitrary Constants
For equation with two arbitrary constants, differentiate twice and eliminate both constants.
Example 3: Circle
Form DE from: $x^2 + y^2 = a^2$ where $a$ is arbitrary.
Solution: Given: $x^2 + y^2 = a^2$ … (1) [one constant]
Differentiate w.r.t. $x$:
$$2x + 2y\frac{dy}{dx} = 0$$ $$\boxed{x + y\frac{dy}{dx} = 0}$$or
$$\boxed{xy' + x = 0}$$This is the required DE.
Example 4: General Second-Order
Form DE from: $y = Ae^x + Be^{-x}$ where $A, B$ are arbitrary.
Solution: Given: $y = Ae^x + Be^{-x}$ … (1) [two constants]
Differentiate once:
$$\frac{dy}{dx} = Ae^x - Be^{-x}$$… (2)
Differentiate again:
$$\frac{d^2y}{dx^2} = Ae^x + Be^{-x}$$… (3)
From (1) and (3):
$$\boxed{\frac{d^2y}{dx^2} = y}$$or
$$\boxed{\frac{d^2y}{dx^2} - y = 0}$$Case 3: $n$ Arbitrary Constants
Example 5: Polynomial
Form DE from: $y = c_1e^x + c_2e^{2x} + c_3e^{3x}$ where $c_1, c_2, c_3$ are arbitrary.
Solution: [3 constants] → Need 3rd order DE
$y = c_1e^x + c_2e^{2x} + c_3e^{3x}$ … (1)
$y' = c_1e^x + 2c_2e^{2x} + 3c_3e^{3x}$ … (2)
$y'' = c_1e^x + 4c_2e^{2x} + 9c_3e^{3x}$ … (3)
$y''' = c_1e^x + 8c_2e^{2x} + 27c_3e^{3x}$ … (4)
Eliminate $c_1, c_2, c_3$ from these four equations (requires systematic elimination or determinant method).
Method of Elimination
Method 1: Direct Substitution
Solve for constants in terms of $y, y', y'', \ldots$ and substitute.
Method 2: Successive Differentiation
Keep differentiating until you can eliminate all constants.
Method 3: Determinant Method
For linear combinations, use determinants to eliminate constants.
Example: From $y = c_1e^x + c_2e^{2x}$:
Write in matrix form:
$$\begin{vmatrix} y & e^x & e^{2x} \\ y' & e^x & 2e^{2x} \\ y'' & e^x & 4e^{2x} \end{vmatrix} = 0$$Expanding gives the DE.
Special Forms
Form 1: $y = e^{ax}(A\cos bx + B\sin bx)$
This gives second-order DE:
$$\boxed{\frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2 + b^2)y = 0}$$Form 2: $y = (c_1 + c_2x)e^{ax}$
Differentiating twice and eliminating:
$$\boxed{\frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + a^2y = 0}$$Form 3: Conic Sections
Circle: $x^2 + y^2 = r^2$ → $x + yy' = 0$
Ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ → $\frac{x}{a^2} + \frac{y}{b^2}y' = 0$
Parabola: $y^2 = 4ax$ → $yy' = 2a$
Memory Tricks
🎯 Number of Constants Rule
“Constants Count = Order Mount”:
- 1 constant → 1st order DE
- 2 constants → 2nd order DE
- $n$ constants → $n$-th order DE
🎯 Differentiation Count
“Differentiate as many times as constants”:
- If 2 arbitrary constants → Differentiate 2 times
🎯 Elimination Strategy
“D-E-R”:
- Differentiate
- Eliminate constants
- Rearrange to standard form
Common Mistakes to Avoid
❌ Mistake 1: Wrong Number of Differentiations
Wrong: For $y = Ae^x + Bxe^x$ (2 constants), differentiate once ✗
Correct: Differentiate twice for 2 constants ✓
❌ Mistake 2: Not Eliminating All Constants
Wrong: Leaving $A$ or $B$ in final DE ✗
Correct: Final DE should have only $x, y, y', y'', \ldots$ ✓
❌ Mistake 3: Treating Parameters as Constants
Given: $y^2 = 4ax$ where $a$ is a parameter (not arbitrary)
Wrong: Treat $a$ as arbitrary constant ✗
Correct:
- If $a$ is given parameter: $2yy' = 4a \Rightarrow yy' = 2a$
- If $a$ is arbitrary: Differentiate again to eliminate ✓
❌ Mistake 4: Incorrect Derivative Calculation
Wrong: From $y = Ae^{2x}$, $\frac{dy}{dx} = Ae^{2x}$ ✗
Correct: $\frac{dy}{dx} = 2Ae^{2x}$ ✓
Solved Examples
Example 1: Single Constant (JEE Main)
Form differential equation from: $y = ae^{-x}$ where $a$ is arbitrary constant.
Solution: Given: $y = ae^{-x}$ … (1)
Differentiate w.r.t. $x$:
$$\frac{dy}{dx} = -ae^{-x} = -y$$Answer: $\boxed{\frac{dy}{dx} + y = 0}$
Example 2: Two Constants (JEE Main)
Eliminate $a$ and $b$ from: $y = ae^{2x} + be^{-2x}$
Solution: $y = ae^{2x} + be^{-2x}$ … (1)
$\frac{dy}{dx} = 2ae^{2x} - 2be^{-2x}$ … (2)
$\frac{d^2y}{dx^2} = 4ae^{2x} + 4be^{-2x}$ … (3)
From (1): $4ae^{2x} + 4be^{-2x} = 4y$
From (3): $\frac{d^2y}{dx^2} = 4ae^{2x} + 4be^{-2x}$
Therefore: $\boxed{\frac{d^2y}{dx^2} = 4y}$ or $\boxed{\frac{d^2y}{dx^2} - 4y = 0}$
Example 3: Trigonometric (JEE Advanced)
Form DE by eliminating $A$ and $B$ from: $y = A\cos 2x + B\sin 2x$
Solution: $y = A\cos 2x + B\sin 2x$ … (1)
$y' = -2A\sin 2x + 2B\cos 2x$ … (2)
$y'' = -4A\cos 2x - 4B\sin 2x$ … (3)
From (1) and (3):
$$y'' = -4(A\cos 2x + B\sin 2x) = -4y$$Answer: $\boxed{y'' + 4y = 0}$ or $\boxed{\frac{d^2y}{dx^2} + 4y = 0}$
Example 4: Product Form (JEE Advanced)
Eliminate constants from: $y = (A + Bx)e^x$
Solution: $y = (A + Bx)e^x$ … (1)
$y' = Be^x + (A + Bx)e^x = (A + B + Bx)e^x$ … (2)
$y'' = Be^x + (A + B + Bx)e^x = (A + 2B + Bx)e^x$ … (3)
From (1): $(A + Bx) = ye^{-x}$
From (2): $y' = Be^x + ye^{-x} \cdot e^x = Be^x + y$
So: $Be^x = y' - y$ … (4)
From (3): $y'' = Be^x + (A + B + Bx)e^x = (y' - y) + Be^x + ye^{-x} \cdot e^x$
$y'' = (y' - y) + (y' - y) + y = 2y' - y$
Answer: $\boxed{y'' - 2y' + y = 0}$ or $\boxed{\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0}$
Example 5: Geometric Application (JEE Advanced)
Form DE of all circles passing through origin with center on $x$-axis.
Solution: Equation of circle with center $(a, 0)$ and radius $a$ (to pass through origin):
$$(x - a)^2 + y^2 = a^2$$Expanding:
$$x^2 - 2ax + a^2 + y^2 = a^2$$ $$x^2 + y^2 = 2ax$$… (1) [one constant $a$]
Differentiate w.r.t. $x$:
$$2x + 2yy' = 2a$$… (2)
From (2): $a = x + yy'$
Substitute in (1):
$$x^2 + y^2 = 2x(x + yy')$$ $$x^2 + y^2 = 2x^2 + 2xyy'$$Answer: $\boxed{y^2 = x^2 + 2xyy'}$ or $\boxed{y^2 - x^2 = 2xy\frac{dy}{dx}}$
Practice Problems
Level 1: JEE Main Basics
Problem 1.1: Form DE from $y = cx^2$ where $c$ is arbitrary.
Solution
$y = cx^2$ … (1)
Differentiate: $y' = 2cx$
From (1): $c = \frac{y}{x^2}$
Substitute: $y' = 2 \cdot \frac{y}{x^2} \cdot x = \frac{2y}{x}$
DE: $xy' = 2y$ or $x\frac{dy}{dx} - 2y = 0$
Problem 1.2: Eliminate $a$ from $y = a\sin x + \cos x$.
Solution
$y = a\sin x + \cos x$ … (1)
$y' = a\cos x - \sin x$ … (2)
From (1): $a = \frac{y - \cos x}{\sin x}$
Substitute in (2): $y' = \frac{y - \cos x}{\sin x} \cdot \cos x - \sin x$
$y'\sin x = (y - \cos x)\cos x - \sin^2 x$
$y'\sin x = y\cos x - \cos^2 x - \sin^2 x$
$y'\sin x = y\cos x - 1$
DE: $y\sin x - y\cos x = -1$ or $\frac{dy}{dx}\sin x - y\cos x = -1$
Actually, let’s reconsider. If only $a$ is arbitrary:
$y = a\sin x + \cos x$, $y' = a\cos x - \sin x$
From first: $a = \frac{y - \cos x}{\sin x}$
Into second: $y' = \frac{y - \cos x}{\sin x}\cos x - \sin x$
Multiply by $\sin x$: $y'\sin x = y\cos x - \cos^2 x - \sin^2 x = y\cos x - 1$
DE: $y' = y\cot x - \csc x$
Problem 1.3: Form DE from $y = e^{ax}$ where $a$ is arbitrary.
Solution
$y = e^{ax}$
Taking logarithm: $\ln y = ax$
So: $a = \frac{\ln y}{x}$
Differentiate original: $y' = ae^{ax} = ay$
Substitute $a$: $y' = \frac{\ln y}{x} \cdot y = \frac{y\ln y}{x}$
DE: $xy' = y\ln y$ or $x\frac{dy}{dx} = y\ln y$
Level 2: JEE Main Advanced
Problem 2.1: Eliminate $a, b$ from $y = a\cos x + b\sin x$.
Solution
$y = a\cos x + b\sin x$ … (1)
$y' = -a\sin x + b\cos x$ … (2)
$y'' = -a\cos x - b\sin x$ … (3)
From (1) and (3): $y'' = -(a\cos x + b\sin x) = -y$
DE: $y'' + y = 0$
Problem 2.2: Form DE from $(x - a)^2 + (y - b)^2 = r^2$ where $a, b$ are arbitrary and $r$ is constant.
Solution
$(x - a)^2 + (y - b)^2 = r^2$ … (1) [2 constants]
Differentiate: $2(x - a) + 2(y - b)y' = 0$
$(x - a) + (y - b)y' = 0$ … (2)
Differentiate (2): $1 + (y')^2 + (y - b)y'' = 0$
From (2): $(y - b) = -\frac{x - a}{y'}$
Substitute: $1 + (y')^2 - \frac{x - a}{y'} \cdot y'' = 0$
From (2): $x - a = -(y - b)y'$
This becomes complex. Alternative:
From (2): $x - a = -(y - b)y'$
Differentiate: $1 = -(y')^2 - (y - b)y''$
So: $(y - b) = -\frac{1 + (y')^2}{y''}$
DE: $[1 + (y')^2]^{3/2} = |r \cdot y''|$ (requires more work)
Standard result: $[1 + (y')^2]^3 = r^2(y'')^2$
Problem 2.3: Eliminate $c_1, c_2$ from $y = c_1e^x + c_2e^{-x}$.
Solution
$y = c_1e^x + c_2e^{-x}$ … (1)
$y' = c_1e^x - c_2e^{-x}$ … (2)
$y'' = c_1e^x + c_2e^{-x}$ … (3)
From (1) and (3): $y'' = y$
DE: $y'' - y = 0$
Level 3: JEE Advanced
Problem 3.1: Form DE representing all parabolas with vertex at origin and axis along positive $x$-axis.
Solution
Equation: $y^2 = 4ax$ where $a$ is arbitrary.
Differentiate: $2yy' = 4a$
So: $a = \frac{yy'}{2}$
Substitute back: $y^2 = 4 \cdot \frac{yy'}{2} \cdot x = 2xyy'$
DE: $y = 2xy'$ or $y = 2x\frac{dy}{dx}$
Problem 3.2: Eliminate $a, b, c$ from $z = ax + by + c$ where $z$ is function of $x$ and $y$.
Solution
This is a partial DE problem.
$z = ax + by + c$
$\frac{\partial z}{\partial x} = a$, $\frac{\partial z}{\partial y} = b$
$\frac{\partial^2 z}{\partial x \partial y} = 0$
DE: $\frac{\partial^2 z}{\partial x \partial y} = 0$ or $z_{xy} = 0$
Problem 3.3: Form DE from $y = Ae^{mx}\cos(nx + B)$ where $A, B$ are arbitrary and $m, n$ are constants.
Solution
This requires two differentiations.
$y = Ae^{mx}\cos(nx + B)$
$y' = Ame^{mx}\cos(nx + B) - Ane^{mx}\sin(nx + B)$
$= my - Ane^{mx}\sin(nx + B)$
This leads to: $y'' - 2my' + (m^2 + n^2)y = 0$
DE: $\frac{d^2y}{dx^2} - 2m\frac{dy}{dx} + (m^2 + n^2)y = 0$
Important Results Summary
| Given Relation | Differential Equation |
|---|---|
| $y = ae^{kx}$ | $\frac{dy}{dx} = ky$ |
| $y = ae^x + be^{-x}$ | $\frac{d^2y}{dx^2} - y = 0$ |
| $y = a\cos kx + b\sin kx$ | $\frac{d^2y}{dx^2} + k^2y = 0$ |
| $x^2 + y^2 = a^2$ | $x + y\frac{dy}{dx} = 0$ |
| $y^2 = 4ax$ | $y\frac{dy}{dx} = 2a$ or $yy' = 2a$ |
| $(y - a)^2 = 4a(x - b)$ | $(y')^2 = 4ay''$ (after elimination) |
Cross-References
- Order and Degree: Understanding DE structure → Order and Degree
- Differentiation Rules: Required for formation → Differentiation
- Solving DEs: Reverse process → Variable Separable Method
Quick Revision Points
- Number of arbitrary constants = Order of DE
- Differentiate as many times as constants
- Eliminate all constants from final DE
- Final DE contains only $x, y$ and derivatives
- Parameters (given values) are not eliminated
- Check: Final DE should not have $A, B, C, \ldots$
Last updated: November 8, 2025