In this course, we'll be focusing on linear equations because those are

the equations that have analytical solutions.

That's what we'll be doing in this course,

is finding analytical solutions.

But the non-linear equations are also very important for

modeling problems in engineering and physics and other fields.

So how about these three equations?

The first equation is a linear equation.

By linear, we mean linear in q so,

there's wherever q enters,

it enters by itself.

So, there is a d squared q, dt squared,

a dq, dt and a q.

There is never a q squared,

there is never a function of q that has terms in the Taylor series that are above q.

In that case, then it's called linear.

The coefficients can be functions of time in a linear equation,

but you must only have q by itself.

The third equation is also a linear equation.

Here it's linear in u,

so there is no u squared term in this equation either.

However, the second equation is a non-linear equation.

There is a d squared Theta, dt squared.

There's a d Theta, dt.

Those two terms are linear but then there's this term here,

which is sine Theta.

So sine Theta has a Taylor series of Theta minus Theta cubed over three factorial.

So, it has terms of higher powers of Theta

in it and this is called a non-linear term in this equation.

So, this term is considered a non-linear equation.

Okay, so in this course,

we'll do first-order differential equations, second-order differential equations.

Mainly linear equations, we'll focus on linear equations.

We'll do mainly ordinary differential equations except for the last topic of this course,

where we'll tackle the solution of a partial differential equation.

So, I think in this first lecture,

I should at least solve a differential equation.

There is a very simple differential equation you can solve already just using calculus.

That's the equation for a mass falling under gravity without any air resistance,

that equation would take the form of d squared x,

dt squared equals minus g. This equation

says that the acceleration of that mass is a constant and in the downward direction,

g is the well-known 9.8 meters per second squared.

So, this type of equation because there is no other x dependence

besides the derivative term and there is no t dependence in this equation,

means you can solve this equation just by integrating.

So, if you integrate this equation twice,

you would end up with a solution that I can write down may be familiar to many of you.

X of t is equal to the initial position of the mass plus

the initial velocity of the mass times t

minus 1.5 gt squared.

Okay, that's just by integrating this equation twice.

You can see that the first derivative

of putting t equals 0 and this equation gives you x naught.

That's called an initial condition.

So, the initial conditions here,