# Second order linear differential equations

A second order differential equation is one containing not only the the derivative, of the function we want to find, but also the derivative of this derivative: $$f''(x) + af'(x) + bf(x) = 0$$ ## Intro

Put a cat in a box with some radioactive material and a flask of poison. Add a device measuring radioactivity. Close the box. If decay happens, the flask breaks and the cat dies. This is the most famous illustration of quantum mechanics, and it was proposed by physicist Erwin Schrödinger.

The crux is: as long as we keep the box closed, for all we know,the cat is both dead and alive.

And if we'd like to know something about the cat, without opening the box? Say hello to the Schrödinger equation, the most important equation in quantum mechanics. It's a second order differential equation used to study the behavior of quantum particles.

(A small side note: it is strongly discouraged to try the cat experiment at home.)

## Concept

Differential equations are all about relating quantity to change.

A differential equation of first order contains variables and their first derivatives. In second order differential equations there are also second derivatives.

As an example, the acceleration is the second derivative of the position.

The Schrödinger equation may seem rather messy at first, so let's start with something more tangible. Let's say that we drop an elephant from the roof of a skyscraper. Its height off the ground will depend on the force of gravity, which is proportional to it's acceleration. That gives us a second order differential equation.

The solution is rather impressive: it's not just a number, like for the equations we are used to. It is a function, describing its position at all times!

## Math

This is a general second order differential equation, where and are some constants:

We solve this by solving a polynomial equation in , which looks similar to the differential equation above:

Using your favorite method to solve this, if the solutions and are real and , then the solution to the differential equation is:

This may look like black magic, but plugging this solution into the differential equation, you will see that it's true.

## Second order linear differential equations

### What is it?

Differential equations are equations which relate a quantity to its derivative(s). We shall often call the variable , as many of the applications lie within the realm of time-dependent systems.

A second order linear differential equation has the generic form:

Differential equations can for example be used to calculate when you need to fold out your parachute in order to avoid hitting the ground really hard. Given some data like what forces affect your fall, at what level off the ground you jumped and how long the parachute needs to unfold, solving the differential equation for the parachute can tell you at which height you will be at a certain time.

### Components and terminology

Let's dissect the equation and disentangle some terminology before we go solution hunting.

The coefficients and will always be constant in this course.

The term second order in the name refers to the highest order derivative in the equation, which is in our case. There exist differential equations of arbitrarily high order out there.

### Solutions

When the right hand side is zero, we say that the equation is homogeneous:

Its solution is referred to as .

In contrast, an inhomogeneous equation has some given function on one side, which slightly complicates things. In addition to the , we need to add a second expression , called the particular solution, to obtain the general form of the solution.

Notice that all of the terms , or in the equation stand by themselves. For example, there is no or . This is what we mean by linear in the name. The linearity is required for the solution methods we will use to work.

By the linearity property, we can find first by letting . Then we proceed, finding a solution so that the condition for the right hand side is satisfied. Finally, the complete solution is:

The linearity property also says that if and both solve the homogeneous equation, then is also a solution, given that and are two arbitrary constants.

## Homogeneous differential equations

Let's visit the real world for a while.

A toy wagon is attached to a wall with a spring and a damper. Get familiar with this kind of set-up. It will appear over and over again, but with slight variations. Sorry for the lack of imagination. The wagon's motion can be described by the equation:

where is the position as a function of time. and are just some values, like, for example and . Since there aren't any external forces, the right-hand side of the equation is .

Next, solve for the roots of . This is called the characteristic equation.

There are three cases for the solution:

• Both roots are distinct: then .

• If there's just one root: then .

• If there are two complex roots : then .

The Gods of math love the number !

To ensure that these are solutions to the differential equation, differentiate them and plug them into the original expression, trust in the math!

Alright, we'll do the first case for you. Here we go. Let's begin by differentiating :

If you'd insert everything into the original equation , you'd get:

Shuffling around the terms, we have:

Try memorizing each of these cases - it's a good investment.

## Inhomogeneous differential equations

Physicists love problems with wagons. It's so much fun watching the wagon move back and forth, watching the wagon spring expand and contract! Better yet, what if we have someone applying another force? In this example, there's a toddler applying a force to the wagon. The force varies, and it can be written as .

Let's call the position . After a bit of physics, we might end up with an equation like this:

This is an inhomogeneous differential equation, because the right-hand side isn't . Instead, we've got a term. Yuck.

The general solution can be written as:

The homogeneous solution is the solution to given by the characteristic polynomial:

So we get something like .

But what about the particular solution? Since shows up on the right-hand side, we might expect the particular solution to contain a term. There's nothing to lose in trying, right?

Let's tentatively use . Then , , and:

This means that:

Hence, the particular solution is:

Luck favors those who try. And this is often the case when it comes to differential equations. To find the particular solution, you must make a qualified guess.

In general, if the right-hand side is:

1. A polynomial: let be a polynomial of the same degree. If contains solutions to the homogeneous problem, multiply by .

2. A trigonometric function: let be , where is the same as in the trigonometric function to the right. If contains solutions to the homogeneous problem, multiply by .

3. An exponential function : let be .

If your right side is a product of either (1), (2) or (3), let be such a product, without specifying the coefficients.

After expanding whatever you've got on the left-hand side, you can determine the coefficients.

## Initial or boundary conditions

As we solve a second order linear differential equation, we end up with a solution which contains two constants. No matter the shape of the solution, they are always there. As long as the constants remain undecided, there is an infinite number of solutions to the equation.

The constants are decided by initial values: some information about the system at a specific point in time. We need as many initial values as the order of the equation.

For a differential equation of th order, we need initial values to determine all constants

This is no accident. It so happens to be, that the order determines the number of coefficients, and each coefficient requires an initial or boundary value for finding it.

A differential equation together with the required initial or boundary conditions is referred to as an initial value problem or boundary value problem. If the differential equation can be solved, this setup promises a completely determined solution.

### The wagon: the grand finale

So far, we have determined the solution to the wagon problem up to the constants. Its position is:

But that's not very informative unless we know what the constants are.

To determine the constants, we need two initial conditions. That means that we need to know two pieces of any of the following information:

1. the position ,

2. the velocity , or

3. the acceleration

at some given time.

Now, a little bird whispers in my ear the wagon's position at two different moments in time:

Plug these one at a time into the solution for and watch as all falls into place.

Using the second condition, we finish it off:

Plugging any value of into the equation now gives the exact position of the wagon for that time!   