Certain ODEs can be simplified into basic linear ODEs or separable ODEs via substitution.
Homogenous Type First Order ODEs
A first order homogenous ODE is one of the form: $$\frac{dy}{dx} =f( \frac{y}{x} )
First order homogenous ODEs can be simplified by using a substitution of $u=\frac{y}x$ . Note that since a change of variable has occurred, we need to derive $\frac{du}{dx}$ (via the [Product rule](Product%20rule.md)).
>[!warning]
>Note that this is **not the same as** as a [homogenous ODE](Second%20Order%20Linear%20ODE.md) in higher power ODEs.
>[!example]- 1: Homogenous ODE
>
>**Solve the following ODE: $\frac{dy}{dx} = \frac{y}{x} + \cos^2({\frac{y}{x}})$**
>
>Let $u = \frac{y}{x} \implies y = ux$
>Then $\frac{dy}{dx} = \frac{d}{dx}(ux) = \frac{du}{dx}x + u$ (Product Rule)
>Subbing back into original equation: $\frac{du}{dx}x + \cancel{u} = \cancel{u} + \cos^2(u)$
>$\frac{du}{dx}x = \cos^2(u) \implies \frac{1}{\cos^2(u)} \frac{du}{dx}= x$
>$\int sec(u)^2 \, du = \int x \, dx$
>$\tan(u) = \frac{x^2}{2} + c$
>$\frac{y}{x} = \arctan(\frac{x^2}{2} + c)$
>$y = x \arctan(\frac{x^2}{2} + c)$
### Bernoulli's Equation
#TODO Which equation exactly?
The equation has the form: $$\frac{dy}{dx} + P(x)y = Q(x)y^n
By substituting , the equation can be reduced to a linear ODE