Ordinary Differential Equations are equations expressed in terms of ONE independent variable and AT LEAST one of the derivatives is a function of this variable. (lolwut)

The order of an ODE is the order of the highest derivative present.

Essentially any solution to an nth order ODE is a function which when differentiated n times creates the ODT.

example:

(1)# Initial value problems

An initial value problem (IVP) is an nth order ODE with the extra information of a set of values of the solution and its first (n-1) derivatives at some fixed point.

These values are called the initial conditions of the IVP.

Example:

Equation, y(0) = 2, dy/dx is 7 for x = 0

# First order

## Separable

A separable ODE is a differential equation where the two variables can be separated to have all of x on one side, all of y on the other.

These can usually be written as

(2)We solve these by writing it as $h(y)dy = g(x)dx$, then integrating to find the implicit solution $H(y) = G(x) + C$

## 1st Order linear

These are called linear since there are no non-linear terms of y or y’ (i.e. no sin(y), powers of y, etc)

To solve them:

- Write them in the form:

- Let h(x) (the integrating factor) be:

- Multiply the equation by h(x):

- Use the product rule of differentiation

(i.e. remembering that $h(x) = e^{\int f(x)dx}$ then h(x)y differentiates to h(x)f(x)y + h(x)d/dx (as in, uv -> u'v + uv'))

to write it as:

This step seems confusing and like it doesn't make sense, so I'm going to analyse it more in depth:

(7)What's happening here is that as $h(x) = e^{\int f(x)dx}$, it differentiates to $\int f(x)dx \cdot e^{\int f(x)dx}$. Essentially, deriving $h(x)y(x) = h'(x)y(x) + h(x)y'(x)$. Which leads to $f(x)h(x)y(x) + h(x)\frac{dy}{dx}$

- Integrate both sides and rearrange to get y on one side

## Exact ODEs

Exact ODEs deal with functions of two variables.

(8)If $\frac{\partial F}{\partial y} = \frac{\partial G}{\partial x}$ (i.e they have the same derivative) then this is an **exact** function

If the function is exact then there is a solution $H(x,y) = C$, where $F(x,y) = \frac{\partial H}{\partial x} \text{and} G(x, y) = \frac{\partial H}{\partial y}$

Or approaching from the other side, and starting , with the equation: $H(x, y) = C$

If the second order mixed partial derivatives of H are equal then $H(x, y) = C$ has solutions $F(x,y) \text{and} G(x,y)$, where they are respectively equal to $\frac{\partial H}{\partial x} \text{and} \frac{\partial H}{\partial y}$

# Second Order - Linear ODEs with Constant Coefficients

Second order equations have the form of:

(9)## Homogeneous

A 2nd Order Equation is Homogeonous if f(x) = 0. i.e.

(10)or more simple

(11)If $y_1$ and $y_2$ are solutions to the above equation then any linear combination ($Ay_1 + By_2$) is also a solution to the equation.

The proof for the above is simple - if $y_1$ goes to 0, and $y_2$ goes to 0, then multiplying them by numbers and adding them will still end up with 0.

### Characteristic Equation

The characteristic equation is:

(12)Essentially what we've done so far, (but using m instead of $\lambda$) is let m^{2} = y'', m = y' and 1 = y. The equation $y'' + ay' + by = 0$ can now be written as $m^2 + am + b = 0$. We work out m in the next section

$\delta = b^2 - 4ac$

- If $\delta > 0$

There are two solutions to the characteristic equation.

We take the results of m and set them as $m_1 \text{ and } m_2$, as well as setting A and B to be the factorisation of the characteristic equation (e.g. $(m - A)(m - B)$.

Hence:

- If $\delta < 0$

There are two **complex** solutions, of the form:

We solve these by setting $y = e^{ax}(Acos(bx) + Bsin(bx))$

If we do not know A and/or B, we work them out with initial conditions, or leave them as variables.

E.g. $y'' + 2y' + 2y = 0, y(0) = 1, y'(0) = 0$

(15)- If $\delta = 0$

There is one solution to the equation, such that $m = \frac{-b}{2}$

Hence

## Inhomogeneous

Inhomogeneous solutions are defined as those where f(x) != 0.

We solve these by setting $f(x) = h_x + p_x$ (homogeneous solution + polynomial solution).

### Polynomial Solution

We solve this by making educated guesses:

(where p and q polynomials of the same degree)

If the guess is a solution multiply the guess by x.

If it's still a solution multiply it again by x (i.e. by x^{2})