# 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 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.