Solving First-Order Differential Equations

Introduction

Key Concepts

Definition of First-Order Differential Equations

A first-order differential equation is an equation involving the first derivative of a function with respect to one independent variable. It can be generally expressed as:

$$ \frac{dy}{dx} + P(x)y = Q(x) $$

Here, \( y \) is the dependent variable, \( x \) is the independent variable, and \( P(x) \) and \( Q(x) \) are functions of \( x \). The order of a differential equation is determined by the highest derivative present, making this a first-order equation.

Types of First-Order Differential Equations

First-order differential equations can be categorized into several types, each with distinct characteristics and methods of solution:

• Separable Equations: These can be written as the product of a function of \( x \) and a function of \( y \).
• Linear Equations: These fit the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \).
• Exact Equations: These satisfy the condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \) for an equation of the form \( M(x,y)dx + N(x,y)dy = 0 \).
• Integrating Factor Method: Used primarily for linear equations to facilitate integration.

Separable Differential Equations

A differential equation is separable if it can be expressed as:

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

By rearranging, we obtain:

$$ \frac{1}{h(y)}dy = g(x)dx $$

Both sides can then be integrated independently:

$$ \int \frac{1}{h(y)}dy = \int g(x)dx + C $$

Where \( C \) is the constant of integration.

Example:

Solve the equation \( \frac{dy}{dx} = xy \).

Rewrite as:

$$ \frac{1}{y}dy = x dx $$

Integrate both sides:

$$ \ln|y| = \frac{x^2}{2} + C $$

Exponentiating both sides gives:

$$ y = Ce^{\frac{x^2}{2}} $$

Linear Differential Equations

A linear first-order differential equation has the form:

$$ \frac{dy}{dx} + P(x)y = Q(x) $$

To solve, an integrating factor \( \mu(x) \) is used:

$$ \mu(x) = e^{\int P(x)dx} $$

Multiplying both sides of the equation by \( \mu(x) \) transforms it into:

$$ \frac{d}{dx} [\mu(x)y] = \mu(x)Q(x) $$

Integrating both sides yields:

$$ y = \frac{1}{\mu(x)} \left( \int \mu(x)Q(x)dx + C \right) $$>

Example:

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

Here, \( P(x) = 2 \) and \( Q(x) = e^{-x} \).

Compute the integrating factor:

$$ \mu(x) = e^{\int 2 dx} = e^{2x} $$>

Multiply through by \( \mu(x) \):

$$ e^{2x}\frac{dy}{dx} + 2e^{2x}y = 1 $$>

This simplifies to:

$$ \frac{d}{dx} [e^{2x}y] = 1 $$>

Integrate both sides:

$$ e^{2x}y = x + C $$>

Solving for \( y \):

$$ y = e^{-2x}(x + C) $$>

Exact Differential Equations

An equation of the form \( M(x,y)dx + N(x,y)dy = 0 \) is exact if:

$$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$>

If exact, there exists a function \( \Psi(x,y) \) such that:

$$ \frac{\partial \Psi}{\partial x} = M \quad \text{and} \quad \frac{\partial \Psi}{\partial y} = N $$>

Example:

Consider \( (2xy + y^2)dx + (x^2 + 2xy)dy = 0 \).

Check exactness:

$$ \frac{\partial (2xy + y^2)}{\partial y} = 2x + 2y $$> $$ \frac{\partial (x^2 + 2xy)}{\partial x} = 2x + 2y $$>

Since both partial derivatives are equal, the equation is exact.

Find \( \Psi(x,y) \) by integrating \( M \) with respect to \( x \):

$$ \Psi(x,y) = \int (2xy + y^2)dx = x^2y + y^2x + h(y) $$>

Differentiate \( \Psi \) with respect to \( y \):

$$ \frac{\partial \Psi}{\partial y} = x^2 + 2xy + h'(y) $$>

Set equal to \( N \):

$$ x^2 + 2xy + h'(y) = x^2 + 2xy $$>

Thus, \( h'(y) = 0 \), implying \( h(y) = C \).

The general solution is:

$$ x^2y + y^2x = C $$>

Integrating Factors

When a differential equation is not exact, an integrating factor can sometimes be found to make it exact. The integrating factor \( \mu(x) \) depends on \( x \) or \( y \) and is determined by:

$$ \mu(x) = e^{\int P(x)dx} \quad \text{for linear equations} $$>

For non-linear equations, the integrating factor might be more complex or may not exist.

Example:

Consider \( (y + \sin x)dx + x dy = 0 \).

Check exactness:

$$ \frac{\partial M}{\partial y} = 1, \quad \frac{\partial N}{\partial x} = 1 $$>

Since they are equal, the equation is exact, and no integrating factor is needed.

Applications of First-Order Differential Equations

First-order differential equations model a wide range of phenomena:

• Population Dynamics: Describing how populations grow or decline over time.
• Radioactive Decay: Modeling the rate at which unstable nuclei decay.
• Newton’s Law of Cooling: Predicting the cooling rate of objects.
• Electrical Circuits: Analyzing current and voltage changes over time.

Example - Newton’s Law of Cooling:

The law states that the rate of change of temperature \( T \) of an object is proportional to the difference between its temperature and the ambient temperature \( T_a \).

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

Solving this linear differential equation involves finding the integrating factor and integrating both sides to determine \( T(t) \).

Solution Techniques Summary

To effectively solve first-order differential equations, students should:

• Identify the type of differential equation (separable, linear, exact).
• Apply the appropriate method (separation of variables, integrating factor, exactness condition).
• Perform accurate integration, keeping track of constants of integration.
• Verify solutions by differentiating and substituting back into the original equation.

Comparison Table

Type of Equation	Form	Solution Method
Separable	\(\frac{dy}{dx} = g(x)h(y)\)	Separation of variables followed by integration
Linear	\(\frac{dy}{dx} + P(x)y = Q(x)\)	Integrating factor method
Exact	M(x,y)dx + N(x,y)dy = 0	Find potential function \( \Psi(x,y) \) such that \( \Psi_x = M \) and \( \Psi_y = N \)
Non-Exact	Various forms	Find an integrating factor to make the equation exact

Summary and Key Takeaways