Linear First Order ODE

A linear first order ODE is a differential equation that is express as a linear combination of a function and it’s derivatives. It can be reduced into a Separable ODEs using an integrating factor.

The reason linear ODEs are so annoying is that they normally cannot be separated, and appear as a sum of functions instead of a product:

$$\frac{dy}{dx}+f(x)y+g(x)=0$$

Solving §

To solve these ODEs, the left hand side needs to be represented as a product, like $\frac{d}{dx}(A(x,y))=B(x)$, then it is separable. In order to do that, we use the Product rule to find an integrating factor ($I(x)$).

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

Why?

We need $I(x)$ such that: $\frac{d}{dx}(I\times y)=\frac{dy}{dx}I(x)+f(x)I(x)y$

Using Product rule $\frac{dI}{dx}y+\frac{dy}{dx}I=\frac{dy}{dx}I+f(x)I(x)y$ $⟹\frac{dI}{dx}\cancel{y}+\cancel{\frac{dy}{dx}I}=\cancel{\frac{dy}{dx}I}+f(x)I(x)\cancel{y}$ $⟹\frac{dI}{dx}=f(x)I(x)$ [[Separable First Order ODEs]([Separable First Order ODEs)] $⟹\int \frac{1}{I}dI=\int f(x)dx$ $⟹I=e^{\int f(x)dx}$

Since we only need one factor, we can let the constant of integration, c, equal to zero.

Then, simply multiply the entire ODE by $I(x)$:

$\frac{dy}{dx}I(x)+f(x)I(x)y=-g(x)I(x)$ $⟹\frac{d}{dx}(y\times I(x))=-g(x)I(x)$ $⟹y=\frac{1}{I(x)}\int -g(x)I(x)dx$