Hello it's a me again drifter1! Today we continue with **Mathematics **to talk about the **Laplace method** that let's us **solve **(mostly) any **Differential equation**! We will get into the **method **and **theory **behind it today, and next time into examples/exercises.

So, without further do, let's get straight into it!

## Laplace Transformation

Before getting into today's solving method we first need to talk about the Laplace tranformation in general!

For an function f: [0, +∞) -> R we define:

This is the **Laplace transformation of f(t)** or f(x) if you want to use x.

The Laplace tranformation result is **a function of s**!

Supposing there is an **inverse Laplace transformation** we have:

This means that we can get our function f(x) or f(t) back by using some inverse Laplace tranformation on the Laplace transformation result L(f(x)) or L(f(t)).

The **most useful property **of the transformation is that it is **linear **and so:

*L[kf(x) +- λg(x)] = kL[f(x)] +- λL[g(x)]*

Because the **integral goes to infinity** we calculate the Laplace transformation using:

L(f(x)) =

Here (on wikipedia) you can read more about the Laplace transformation **theorems **and can also find a **table/list **of some transformations!

**Transformations **that you will need very often are:

**f(x) = c**->**F(s) = c/s****f(x) = x^n**->**F(s) = n! / s^(n+1)****f(x) = sin(ax)**->**F(s) = a / (s^2 + a^2)****f(x) = cos(ax)**->**F(s) = s / (s^2 + a^2)****f(x) = e^(ax)**->**F(s) = 1 / (s-a)****f(x) = e^(bx)*sin(ax)**->**F(s) = a / [(s-b)^2 + a^2]****f(x) = e^(bx)*cos(ax)**->**F(s) = (s-b) / [(s-b)^2 + a^2]****f(x) = x*sin(ax)**->**F(s) = 2as / [s^2 + a^2]^2****f(x) = x*cos(ax)**->**F(s) = (s^2 - a^2) / [s^2 + a^2]^2**

We also use these in the inverse Laplace transformation!

For **example**:

f(x) = e^(-2x) * sin5x => 5 / [(s+2)^2 + 25] = F(s) [Laplace transformation]

F(s) = s / (s-1) => cos(-x) = f(x) [Inverse Laplace transformation]

The **Laplace transformation of an function f(x)=y **is:

*L(y) = Y*

*L(y') = s*Y - y(0)*

*L(y") = s^2*Y - s*y(0) - y'(0)*

*...*

*L(y^(n)) = s^n*Y - s^(n-1)*y(0) - s^(n-2)*y'(0) - ... - s* y^(n-2)(0) - y^(n-1)(0)*

Which means that we need a **starting condition y(0)** and so y at x = 0.

## Solving differential equations

Suppose a constant coefficient linear 2nd-order ODE:

ay" + by" + cy = R(x)

Let's **find the Laplace transformation for each side**:

L(ay" + by" + cy) = L(R(x)) =>

aL(y") + bL(y') + cL(y) = L(R(x)) =>

... do some **calculations**

That way we end up with Y(s) = ...

Using the **inverse Laplace transformation** we can then find y(x).

These method can of course be **applied in n-order Differential equations** **of any type** and it doesn't have to be any of the cases we covered previously!

The **difficulty **comes in the **calculations **and in the way that we **distinguish the different cases** **and transformations **from each other.

So, to sum up the **steps **are:

- Use the Laplace transformation on each side of the equation
- Get to the form: Y(s) = ..., by solving for Y(s)
- Use the inverse Laplace transformation to find y(x)

## Solving (differential) systems

Using the same method we can of course **solve systems that contain derivatives** now!

When having y(x) and z(x) be functions of x in a linear system of two equations (order 2) then:

- we
**find**the**Laplace transformation for each equation independently**. - we
**substitute**each one in the other equation**to end up with Y(s) = ... and Z(s) = ...**, where each one doesn't contain the other. - we
**calculate**the**inverse Laplace transformation**to get our y(x) and z(x) functions.

If the system is not so easy then we sometimes have to use some Linear algebra to solve the system of Laplace transformations (find Y and Z). I will get into a Linear algebra method of solving ODE's in general in a post later on!

Also, in my Systems and Signals series of Mathematics we will cover Laplace transformations more in depth and even more, but this will come much later on cause first I have to cover some more Mathematical Analysis and also Complex numbers!

Next time we will get into Exercises!

### Image sources:

### Previous posts of the series:

Introduction -> Definition and Applications

First-order part(1) -> Separable, homogeneous and exact 1st-order ODE's

First-order part(2) -> Linear, Bernoulli and Riccati first-order ODE's

First-order exercises -> Exercises for all the 1st-order ODE types

Second-order linear with const coeffs -> Constant coefficient linear 2nd-order ODE's

Special second-order forms -> Linear Euler ODE, Wronsky and Lagrange (Canonical) methods

Second-order exercises -> Exercises for 2nd-order ODE's

And this is actually it for today and I hope that you enjoyed it!

Next time we will apply this method in some exercises!

Bye!