An ordinary differential equation is one with only one independent variable. For example:

\(\dfrac{dy}{dx}=f(x)\)

The order of a differential equation is the number of differentials of \(y\) included. For example one with the second derivative of \(y\) is of order \(2\).

Ordinary equations can can either implicit or explicit. An explicit function shows the highest order derivative as a function of other terms.

An implicit function is one which is not explicit.

A linear ODE is an explicit ODE where the derivative terms of \(y\) do not multiply together, that is, in the form:

\(y^{(n)}=\sum_ia_i(x)y^{(i)}+r(x)\)

We have an evolution:

\(\dfrac{dy}{dt}=f(t,y)\)

And a starting condition:

\(y_0=f(t_0)\)

We now discuss various ways to solve these.

An ordinary differential equation is one with only one independent variable. For example:

\(\dfrac{dy}{dx}=f(x)\)

The order of a differential equation is the number of differentials of \(y\) included. For example one with the second derivative of \(y\) is of order \(2\).

Ordinary equations can can either implicit or explicit. An explicit function shows the highest order derivative as a function of other terms.

An implicit function is one which is not explicit.

A linear ODE is an explicit ODE where the derivative terms of \(y\) do not multiply together, that is, in the form:

\(y^{(n)}=\sum_ia_i(x)y^{(i)}+r(x)\)

We have an evolution:

\(\dfrac{dy}{dt}=f(t,y)\)

And a starting condition:

\(y_0=f(t_0)\)

We now discuss various ways to solve these.

For some we can write:

\(\dfrac{dy}{dt}=f(t,y)\)

\(\dfrac{dy}{dt}=q(t)-p(t)y\)

This can be solved by multiplying by an unknown function \(\mu (t)\):

\(\dfrac{dy}{dt}+p(t)y=q(t)\)

\(\mu (t)[\dfrac{dy}{dt}+p(t)y]=\mu (t)q(t)\)

We can then set \(\mu(t)=e^{\int p(t)dt}\). This means that \(\dfrac{d\mu }{dt}=p(t)u(t)\)

\(\dfrac{d}{dt}[\mu(t)y]=\mu (t)q(t)\)

\(\mu(t)y=\int \mu (t)q(t)dt + C\)

In some cases, this can then be solved.

\(\dfrac{\delta y}{\delta x}=cy\)

\(y=Ae^{c(y+a)}\)

\(\dfrac{\delta^2 y}{\delta x^2}=cy\)

\(y=Ae^{\sqrt c (y+a)}\)

For some we can write:

\(\dfrac{dy}{dt}=f(t,y)\)

\(\dfrac{dy}{dt}=\dfrac{g(t)}{h(y)}\)

We can then do the following:

\(h(y)\dfrac{dy}{dt}=g(t)\)

\(\int h(y)\dfrac{dy}{dt}dt=\int g(t)dt + C\)

\(\int h(y)dy=\int g(t)dt + C\)

In some cases, these functions can then be integrated and solved.

These are of the form

\(\dfrac{d^2y}{dt^2}+p(t)\dfrac{dy}{dt}+q(t)y=g(t)\)

There are two types. Homogenous equations are where \(g(t)=0\). Otherwise they are heterogenous.

We explore the case with constants:

\(a\dfrac{d^2y}{dt^2}+b\dfrac{dy}{dt}+cy=0\)