Differential Equations

Master ordinary differential equations, order, degree, and solution methods for JEE Mathematics.

Differential equations involve derivatives and are essential for modeling physical phenomena.

Overview

graph TD
    A[Differential Equations] --> B[Classification]
    A --> C[Formation]
    A --> D[Solution Methods]
    D --> D1[Variable Separable]
    D --> D2[Homogeneous]
    D --> D3[Linear]

Definitions

Order: Highest derivative in the equation

Degree: Power of highest derivative (when polynomial in derivatives)

Examples

EquationOrderDegree
$\frac{dy}{dx} = x^2$11
$\frac{d^2y}{dx^2} + y = 0$21
$(\frac{dy}{dx})^2 = x$12

Formation of DE

From a family of curves with n arbitrary constants, eliminate constants using n differentiations.

Example: $y = ae^x + be^{-x}$ (2 constants → order 2)

Solution of DE

General Solution

Contains arbitrary constants equal to order.

Particular Solution

Obtained by giving specific values to constants.

Methods of Solution

1. Variable Separable

$$\frac{dy}{dx} = f(x)g(y)$$ $$\int \frac{dy}{g(y)} = \int f(x) dx$$

2. Homogeneous Equations

$$\frac{dy}{dx} = F\left(\frac{y}{x}\right)$$

Substitute: $y = vx$, then $\frac{dy}{dx} = v + x\frac{dv}{dx}$

3. Linear First-Order

$$\frac{dy}{dx} + Py = Q$$

where P and Q are functions of x.

Integrating Factor: $IF = e^{\int P dx}$

Solution: $y \cdot IF = \int Q \cdot IF \, dx$

JEE Tip
If the equation is in the form $\frac{dx}{dy} + Px = Q$ (with P, Q functions of y), treat y as independent variable.

4. Bernoulli’s Equation

$$\frac{dy}{dx} + Py = Qy^n$$

Substitute: $v = y^{1-n}$, reduces to linear form.

5. Exact Equations

$$M dx + N dy = 0$$

is exact if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$

Applications

Growth/Decay

$$\frac{dy}{dt} = ky$$

Solution: $y = y_0 e^{kt}$

  • k > 0: Exponential growth
  • k < 0: Exponential decay

Newton’s Law of Cooling

$$\frac{dT}{dt} = -k(T - T_0)$$

Solution: $T - T_0 = (T_i - T_0)e^{-kt}$

Orthogonal Trajectories

Curves that intersect given family at right angles.

Method: Replace $\frac{dy}{dx}$ by $-\frac{dx}{dy}$

Practice Problems

  1. Solve: $\frac{dy}{dx} = \frac{y}{x}$

  2. Solve: $\frac{dy}{dx} + y = e^{-x}$

  3. Form the DE of $y = ax^2 + bx + c$

  4. Solve: $(x^2 + y^2)dx - 2xy \, dy = 0$

Quick Check
Why do we need initial conditions to find a particular solution?

Further Reading