Differential Equations

1. First Order Linear Differential Equations and Integrating Factors

What is a First Order Linear Differential Equation?

A first order linear differential equation is an equation involving a function y, its first derivative dy/dx, and possibly x. It has the standard form:

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

where P(x) and Q(x) are functions of x (or constants).

Examples:

• $\frac{dy}{dx} - 2y = x^2$ (here P(x) = -2, Q(x) = x²)
• $x\frac{dy}{dx} - y = x^4$ (needs to be rearranged first)
• $\frac{dy}{dx} + y\coth x = \cosh x$ (here P(x) = coth x, Q(x) = cosh x)

What is an Integrating Factor?

An integrating factor is a special function that we multiply through the entire differential equation to make it easier to solve. It's like a mathematical "trick" that transforms the equation into something we can integrate directly.

The integrating factor is denoted by I(x) or just I, and is calculated using:

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

where P(x) is the coefficient of y in the standard form.

How to Solve Using an Integrating Factor

Step-by-step method:

Step 1: Rearrange the equation into standard form $\frac{dy}{dx} + P(x)y = Q(x)$

Step 2: Identify P(x) and calculate the integrating factor $I = e^{\int P(x)\,dx}$

Step 3: Multiply the entire equation by the integrating factor

Step 4: The left side becomes $\frac{d}{dx}(Iy)$, so write: $\frac{d}{dx}(Iy) = I \cdot Q(x)$

Step 5: Integrate both sides: $Iy = \int I \cdot Q(x)\,dx$

Step 6: Solve for y: $y = \frac{1}{I}\int I \cdot Q(x)\,dx$

Worked Example 1: $\frac{dy}{dx} - 2y = x^2$

This is already in standard form with P(x) = -2 and Q(x) = x².

Calculate integrating factor: $I = e^{\int -2\,dx} = e^{-2x}$

Multiply through by I: $e^{-2x}\frac{dy}{dx} - 2e^{-2x}y = x^2e^{-2x}$

The left side is $\frac{d}{dx}(ye^{-2x})$, so: $\frac{d}{dx}(ye^{-2x}) = x^2e^{-2x}$

Integrate both sides: $ye^{-2x} = \int x^2e^{-2x}\,dx$

To integrate the right side, use integration by parts twice: $\int x^2e^{-2x}\,dx = -\frac{1}{2}x^2e^{-2x} + \frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} + c$

Therefore: $ye^{-2x} = -\frac{1}{2}x^2e^{-2x} + \frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} + c$

Multiply through by $e^{2x}$: $y = -\frac{1}{2}x^2 + \frac{1}{2}x - \frac{1}{4} + ce^{2x}$

This is the general solution.

Worked Example 2: $x\frac{dy}{dx} - y = x^4$

First, rearrange to standard form by dividing through by x: $\frac{dy}{dx} - \frac{1}{x}y = x^3$

Now P(x) = -1/x and Q(x) = x³.

Calculate integrating factor: $I = e^{\int -\frac{1}{x}\,dx} = e^{-\ln x} = e^{\ln x^{-1}} = x^{-1} = \frac{1}{x}$

Multiply through by I = 1/x: $\frac{1}{x}\frac{dy}{dx} - \frac{1}{x^2}y = x^2$

This gives: $\frac{d}{dx}\left(\frac{y}{x}\right) = x^2$

Integrate: $\frac{y}{x} = \frac{x^3}{3} + c$

Therefore: $y = \frac{x^4}{3} + cx$

Worked Example 3: $\frac{dy}{dx} + y\coth x = \cosh x$

This is in standard form with P(x) = coth x and Q(x) = cosh x.

Calculate integrating factor: $I = e^{\int \coth x\,dx} = e^{\ln(\sinh x)} = \sinh x$

Multiply through: $\sinh x\frac{dy}{dx} + y\sinh x\coth x = \sinh x\cosh x$

Since $\coth x = \frac{\cosh x}{\sinh x}$: $\sinh x\frac{dy}{dx} + y\cosh x = \sinh x\cosh x$

This is: $\frac{d}{dx}(y\sinh x) = \sinh x\cosh x$

Integrate: $y\sinh x = \int \sinh x\cosh x\,dx$

Using the substitution u = sinh x, du = cosh x dx: $y\sinh x = \frac{\sinh^2 x}{2} + c$

Therefore: $y = \frac{\sinh x}{2} + \frac{c}{\sinh x}$