Contents of this Calculus 2 episode:

Linear equations, Multiply by v(x), Integrating factor, General form of v(x), Solution, Integration.

First-order linear differential equations

In general, a first-order linear differential equation contains a and it also contains a first-order .

The solution for the equation will be funny; it will require a bit of hocus-pocus.

We multiply the equation by a function ,

and on the left side, envision the differentiation rule for products of functions.

But there is a snag. The first part matches,

but the second part...

well, for that we need

This is a simple separable equation that we can solve.

Let’s pick the positive one.

We start the solution by multiplying the equation by that , and this way a derivative of a product appears on the left side.

This is the first step.

Compute function :

Multiply the equation by , so that the left side is a derivative of a product.

And then integrate.

The last step is to integrate both sides.

Let's see an example of this.

Here comes this function:

Let's see how we could integrate

Well, we can do something like this:

Only there is a little trouble, because

But this can be helped.

Let’s pick the positive one.

Now, that we found function , we can do the multiplication.

And now let’s stop for a moment.

The left side of the equation is , that’s what we have been working on.

This is splendid, now all we have left is integration...

and it’s done.

Let's see another one.

Let's see :

It seems we have to multiply by x.

Well, here we returned to the original equation, but there is no need to worry, we are on the right path.

And now we can get to the integral.

Well, we solved this one, too, didn’t we?

Finally here comes one more equation.

And now we are ready for the multiplication.

Calculus 2 episode