# List of Notes - Category: First-Order ODE

## Separable Differential Equations

Consider differential equation `y'=f(t,y)` or `(dy)/(dt)=f(t,y)`. If function `f(t,y)` can be written as product of function `g(t)` (function that depends only on t) and function `u(y)` (function that depends only on y) than such differential equation is called separable.

## Homogeneous Equations

If in differential equation y'=f(t,y), function f(t,y) has the property that f(at,ay)=f(t,y) then such differential equation is called homogeneous.

It can be transformed into separable equation using substitution ` y=ut` along with the corresponding derivative `(dy)/(dt)=(du)/(dt)t+u`. Resulting equation is solved as separable equation and required solution is obtained by back subtitution.

## Exact Equations

Differential equation `M(x,y)dx+N(x,y)dy=0` is exact if there exist function f such that `df=M(x,y)dx+N(x,y)dy` .

In this case equation can be rewritten as `df=0` which gives solution f=C.

Test for exactness: if M(x,y) and N(x,y) are continuous functions and have continuous first partial derivatives on some rectangle of the xy-plane, then differential equation is exact if and only if `(partial M(x,y))/(partial y)=(partial N(x,y))/(partial x)`.

## Linear Differential Equations

A first-order linear differential equation has form `y'+p(t)y=q(t)` .

To solve it, rewrite it in the differential form: `(dy)/(dt)+p(t)y=q(t)` or `(p(t)y-q(t))dt+dy=0` .

Now, using the facts about exact equations we see that this equation is not exact, however since `M(t,y)=p(t)y-q(t)` and `N(t,y)=1` then `(partial M)/(partial y)=p(t)` and `(partial N)/(partial x)=0` . So, `1/N((partial M)/(partial y)-(partial N)/(partial x))=1/1 (p(t)-0)=p(t)` . Thus, integrating factor `I(t,y)=e^(int p(t) dt)` .

## Bernoulli Equations

A Bernoulli equation has form `y'+p(t)y=q(t)y^n` where n is real number.

Using substituion `z=y^(1-n)` this equation can be transformed into linear.

Example 1. Solve `y'-3/ty=t^4y^(1/3)` .

This is Bernoulli equation. Use substitution `z=y^(1-1/3)=y^(2/3)` . Then `y=z^(3/2)` and `y'=3/2 z^(1/2)z'` . Plugging these values into equation gives.