Introduction
Differential equations form a crucial part of the JEE Main Mathematics syllabus. They are equations that involve an unknown function and its derivatives. These equations are fundamental in modeling various phenomena in physics, engineering, and other sciences. Understanding differential equations is essential for solving many practical problems.
Basic Concepts
Definition
A differential equation is an equation that relates a function with its derivatives. In general, a differential equation can be written as: $$ F(x, y, y', y'', \ldots, y^{(n)}) = 0 $$
where $y = f(x)$ is the unknown function, and $y', y'', \ldots, y^{(n)}$ are its first, second, and $n$th derivatives, respectively.
Order and Degree
- Order: The order of a differential equation is the highest order of the derivative present in the equation.
- Degree: The degree of a differential equation is the power of the highest order derivative, provided the equation is polynomial in derivatives.
Consider the differential equation: $$ \left( \frac{d^2y}{dx^2} \right)^3 + \left( \frac{dy}{dx} \right)^2 - y = 0 $$
- The order is 2 (highest derivative is $\frac{d^2y}{dx^2}$).
- The degree is 3 (the power of $\frac{d^2y}{dx^2}$).
Types of Differential Equations
Ordinary Differential Equations (ODEs)
An ordinary differential equation involves derivatives with respect to a single variable. For example: $$ \frac{d^2y}{dx^2} + 3\frac{dy}{dx} + 2y = 0 $$
Partial Differential Equations (PDEs)
A partial differential equation involves partial derivatives with respect to multiple variables. For example: $$ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 $$
NoteFor JEE Main, the focus is primarily on ordinary differential equations (ODEs).
Solutions of Differential Equations
General Solution
The general solution of a differential equation contains all possible solutions and includes arbitrary constants equal to the order of the differential equation.
Particular Solution
A particular solution is obtained by giving specific values to the arbitrary constants in the general solution.
Initial Value Problems (IVP)
An initial value problem is a differential equation along with specified values of the unknown function and its derivatives at a particular point.
ExampleSolve the initial value problem: $$ \frac{dy}{dx} = 3x^2, \quad y(0) = 4 $$
Solution:
- Integrate the differential equation: $$ y = \int 3x^2 , dx = x^3 + C $$
- Use the initial condition $y(0) = 4$: $$ 4 = 0 + C \implies C = 4 $$
- The particular solution is: $$ y = x^3 + 4 $$
Methods of Solving First Order Differential Equations
Variable Separable Method
In this method, we separate the variables $x$ and $y$ on opposite sides of the equation and then integrate.
ExampleSolve the differential equation: $$ \frac{dy}{dx} = \frac{y}{x} $$
Solution:
- Separate the variables: $$ \frac{dy}{y} = \frac{dx}{x} $$
- Integrate both sides: $$ \int \frac{1}{y} , dy = \int \frac{1}{x} , dx $$ $$ \ln|y| = \ln|x| + C $$
- Simplify: $$ y = Cx $$
Homogeneous Differential Equations
A first-order differential equation is called homogeneous if it can be written in the form: $$ \frac{dy}{dx} = f\left(\frac{y}{x}\right) $$
ExampleSolve the homogeneous differential equation: $$ \frac{dy}{dx} = \frac{x + y}{x - y} $$
Solution:
- Substitute $y = vx$ (where $v = \frac{y}{x}$): $$ \frac{dy}{dx} = x \frac{dv}{dx} + v $$
- Substitute into the differential equation: $$ x \frac{dv}{dx} + v = \frac{x + vx}{x - vx} $$ $$ x \frac{dv}{dx} + v = \frac{x(1 + v)}{x(1 - v)} $$ $$ x \frac{dv}{dx} + v = \frac{1 + v}{1 - v} $$
- Separate variables and integrate: $$ x \frac{dv}{dx} = \frac{1 + v}{1 - v} - v $$ $$ x \frac{dv}{dx} = \frac{1 + v - v + v^2}{1 - v} $$ $$ x \frac{dv}{dx} = \frac{1 + v^2}{1 - v} $$
- Integrate: $$ \int \frac{1 - v}{1 + v^2} , dv = \int \frac{1}{x} , dx $$ $$ \int \frac{1}{1 + v^2} , dv - \int \frac{v}{1 + v^2} , dv = \ln|x| + C $$ $$ \tan^{-1}(v) - \frac{1}{2} \ln|1 + v^2| = \ln|x| + C $$
- Substitute back $v = \frac{y}{x}$: $$ \tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \ln\left(1 + \left(\frac{y}{x}\right)^2\right) = \ln|x| + C $$
Linear Differential Equations
A first-order linear differential equation has the form: $$ \frac{dy}{dx} + P(x)y = Q(x) $$
The solution involves finding an integrating factor $\mu(x)$: $$ \mu(x) = e^{\int P(x) , dx} $$
ExampleSolve the linear differential equation: $$ \frac{dy}{dx} + 2y = e^x $$
Solution:
- Identify $P(x) = 2$ and $Q(x) = e^x$.
- Find the integrating factor $\mu(x)$: $$ \mu(x) = e^{\int 2 , dx} = e^{2x} $$
- Multiply through by the integrating factor: $$ e^{2x} \frac{dy}{dx} + 2e^{2x} y = e^{x} e^{2x} $$ $$ \frac{d}{dx} \left( e^{2x} y \right) = e^{3x} $$
- Integrate both sides: $$ e^{2x} y = \int e^{3x} , dx $$ $$ e^{2x} y = \frac{e^{3x}}{3} + C $$
- Solve for $y$: $$ y = \frac{e^{x}}{3} + Ce^{-2x} $$
Applications of Differential Equations
Growth and Decay
The rate of change of a quantity is proportional to the quantity itself: $$ \frac{dy}{dt} = ky $$
Newton's Law of Cooling
The rate of change of temperature is proportional to the difference between the temperature of the object and the ambient temperature: $$ \frac{dT}{dt} = -k(T - T_{\text{ambient}}) $$
Conclusion
Understanding differential equations is essential for solving many practical problems. By mastering the methods of solving first-order differential equations, you can tackle a wide range of problems in JEE Main Mathematics. Practice is key to becoming proficient in this topic.
TipAlways check your solutions by differentiating and substituting back into the original differential equation.
Common MistakeAvoid assuming that the integrating factor is always simple. Carefully compute it for each problem.
NoteRemember that the general solution includes arbitrary constants, which are crucial for forming particular solutions based on initial conditions.
