# Differential equations

Content of the topic

Differential equations

Differential equations are equations where the unknowns are functions.

The equations contain various derivatives and exponentials of these functions.

If the differential equation contains a function of one variable, then it is called an ordinary differential equation (ODE); if it contains a function of two or more variables, then it is called a partial differential equation.

Here is the cast list:

The variable of the function

The function

shortly

And here is an equation

Order

It indicates the highest derivative of the function in the equation.

Linearity

If the equation includes the unknown function and its derivatives only in first-degree form, then the equation is linear.

Here, for instance, the order is 2.

This equation is not linear.

Now let’s get to the funniest matter: finding the solution.

We will group the differential equations based on their types, and then discuss the solution methods for each group.

But there is a snag.

Namely, some badass physicists don’t use x for the variable, instead, they use t, and then the function is not called y, but x.

The reason is that differential equations are often used to describe processes where the variable is time, commonly denoted by t.

If we denote the variable by t and the function by x, then the equation will be:

In this case the symbol for differentiation is a dot.

And now, let’s see how to solve these equations.

Separable differential equations

The reason is that differential equations are often used to describe processes where the variable is time, commonly denoted by t.

If we denote the variable by t and the function by x, then the equation will be:

In this case the symbol for differentiation is a dot.

And now, let’s see how to solve these equations.

is replaced by

Multiply by dx.

Now comes the separation: move all y terms to the dy side, and move all x terms to the dx side.

Integrate both sides to get the solution.

In this case it is enough to have +C only on one side.

GENERAL SOLUTION:

If constant y were zero, then here we couldn’t have done the division.

Let's see whether y=0 is a solution

It seems like it is.

PARTICULAR SOLUTION:

We get the particular solution if we give C a fix value.

It would be particularly satisfying to get a solution where y(0)=666

Here is another equation, let’s try this one, too.

Now, we get rid of the logarithms.

There is such a thing as

This way, bye-bye logarithm.

Here, C is some constant value, therefore ec is another constant, let’s call it D.

Now we have to check whether y=0 is a solution.

It seems like it is.

Here, too, the particular solution means that we fix the value of D to some number.

Let’s assume we want  to hold.

Here is something more entertaining.

There is such a thing as , this way, bye-bye tangent.

Well, this seems to be done.

And now let’s see another type of differential equation.

Homogenous type of first-order differential equations

Homogenous type of first-order differential equations

Let’ start by clarifying what being homogenous type means.

Here is this thing

well, this is a polynomial, but that is not the point.

If we substitute  into this,

then voila,  will appear in all terms.

This wonderful characteristic is called homogeneity.

This polynomial, for instance, is not homogenous.

This is because if , then x will have a different exponential in each term.

So, that’s it for homogeneity. Now let’s move on to applying this knowledge for solving differential equations.

Let’ solve this one.

The equation is not separable, because if we divided by ,...

then some  term will definitely remain on the  side.

And that is detrimental to obtaining the solution.

But if we don’t divide, then some y will remain on the  side.

Luckily, the degrees are homogenous.

The  part has a degree of two...

and so does the  part.

substitution, shortly

This equation is now separable, so here comes the separation.

We solve the separable equation, where instead of y, we are going after u.

And when we find u, we turn it back into y.

Let's see another one.

There is such a thing as , this way, bye-bye tangent.

Why don’t we check out another homogeneous equation?

The equation is not separable, but the degrees are homogenous.

It seems the degree is 4.

This is a good sign; we can use the usual substitution.

Now, we get rid of the logarithms.

Exact differential equations

Exact differential equations

This equation

is exact, if there exists a function  such that

Then, the solution for the equation is that function:

So, solving an exact differential equation means finding this  function.

However, first we should test the equation to see whether it is exact.

We can do it in two ways.

One is differentiation, the other is integration.

Well, this will be a lot more understandable once we look at the family tree of the participants.

Functions  and  in the equation should have a common ancestor, namely .

We can check that by integration.

At the same time, they also should have a common descendant:

(What a horrible incest!!!)

We can check that by differentiation.

Well, differentiation is easier.

So, first differentiate  and , to see if the equation is exact indeed,

and then integrate them to get the solution.

That’s a brilliant plan, so let’s solve a problem.

Here is an equation:

Let's see whether it is exact.

This equation is exact, if

Well, it seems yes.

The solution of exact equations is  where

We obtain the solution by integration:

Integrate with respect to x,

in this case y behaves like a constant.

But because y behaves just like a constant, it is possible that this  is not simply , but it contains y as well.

We can verify this by differentiating this with respect to y, and see what we get as a result.

Well, theoretically we should get .

Let’s compare to the original.

It seems that

Well, this seems to be done.

We could try writing the solution in explicit form, too,

namely, in a form where y is expressed.

Sadly, this cannot always be done.

But now it does work.

Let's see another one, too.

Let’s check whether it is exact.

It seems to be exact, so let’s get down to solving it.

Integrate with respect to x,

in this case y behaves like a constant.

And here comes this , where it is possible that this is not simply a , but it contains y as well.

Let's see what it is this time.

Here, y cannot be expressed, so we cannot write the solution in explicit form.

Finally, let's see one more equation.

First check whether the equation is exact.

Sadly, these are not equal, so the equation is not exact.

Let’ see what can be done in such cases.

The next slides will discuss that.

The integrating factor

Here is this equation

which is sadly, not exact, because

Our job is to use some magic and make it exact.

Let’s try multiplying the equation by x.

Let’s see whether this multiplication by x was good for the equation.

It seems that it was.

This is now an exact equation, and its solution is:

At the end we find out what  is.

Well, it seems it was effective to multiply by x.

This is lovely, but begs the question: why we chose x for the multiplication.

If the equation is not exact, then we try to make it exact by using an integrating factor.

To find the integrating factor, first we compute these:

There is nothing to worry about, everything will be all right soon.

If the first one of these contains only y,

or the second one contains only x,

then there is hope for finding the integrating factor.

Now the first one contains both x and y, so that is not useful for us.

But the second one is good.

And the integrating factor...

Finding the integrating factor

Here comes another equation.

Let’s check whether it is exact.

Well, not really.

So, here comes the integrating factor.

The first one should only contain x...

so, sadly, this is not good.

The second one is promising...

Now, this equation is exact.

So, let’s get down to solving it:

Bad news. This is partial integration.

And there is this  as well.

Well, it seems , therefore  is just some constant.

Here is an equation.

Let’s check whether it is exact.

It seems it is not exact.

No problem, here comes the integrating factor.

After some transformation...

Now this equation is exact.

So, let’s get down to solving it:

Finally, let’s differentiate this with respect to y, to see what is going on with .

And then we have this equation:

Let's see whether it is exact.

Well, no.

No problem, here comes the integrating factor.

Here, it doesn’t matter which one we use.

But this one looks easier.

Now we are ready for the solution.

The integrating factor 2.0

Here is this equation

which is sadly, not exact, because

Our job is to use some magic and make it exact.

Let’s try multiplying the equation by x.

Let’s see whether this multiplication by x was good for the equation.

It seems that it was.

This is now an exact equation, and its solution is:

At the end we find out what  is.

Well, it seems it was effective to multiply by x.

This is lovely, but begs the question: why we chose x for the multiplication.

If the equation is not exact, then we try to make it exact by using an integrating factor.

To find the integrating factor, first we compute these:

There is nothing to worry about, everything will be all right soon.

If the first one of these contains only y,

or the second one contains only x,

then there is hope for finding the integrating factor.

Now the first one contains both x and y, so that is not useful for us.

But the second one is good.

And the integrating factor...

Finding the integrating factor

Here comes another equation.

Let’s check whether it is exact.

Well, not really.

So, here comes the integrating factor.

The first one should only contain x...

so, sadly, this is not good.

The second one is promising...

Now, this equation is exact.

So, let’s get down to solving it:

Bad news. This is partial integration.

And there is this  as well.

Well, it seems , therefore  is just some constant.

Here is an equation.

Let’s check whether it is exact.

It seems it is not exact.

No problem, here comes the integrating factor.

After some transformation...

Now this equation is exact.

So, let’s get down to solving it:

Finally, let’s differentiate this with respect to y, to see what is going on with .

And then we have this equation:

Let's see whether it is exact.

Well, no.

No problem, here comes the integrating factor.

Here, it doesn’t matter which one we use.

But this one looks easier.

Now we are ready for the solution.

First-order linear differential equations

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.

First-order constant coefficient linear differential equations

A First-order constant coefficient linear differential equations

This type is a special case of the first-order linear differential equations.

They are called “constant coefficient” because in these equations the function  is some constant.

Let’s see a completely new solution method for this special type.

We could solve it the same way as we did in the previous slideshow, but this solution is much more amusing.

The first step is to solve this so called homogeneous equation:

This is a very simple equation.

The homogeneous equation is:

The homogeneous solution is:

We can get the general solution of the equation by adding the particular solution to the homogeneous solution.

We can obtain the particular solution based on the function on the right side, using a very funny procedure called “Method of Undetermined Coefficients”, or “Trial Functions Method”.

We find the particular solution using the Undetermined Coefficients Method:

exponential expression:

sine or cosine:

Here is this equation:

Now we start looking for the particular solution.

The exact nature of this particular solution always depends on the function on the right side.

It seems that now there will be sine and cosine in the particular solution.

We substitute this into the original equation.

And then we figure out what A and B is.

Here, the particular solution will be polynomial.

We plug this back into the original equation, and figure out the values of A, B and C.

And then we figure out what A and B is.

In cases where the particular solution includes exponential expressions, well, we could face some problems.

The next slideshow will discuss that.

If the particular solution includes an  term, then we may face some problems when trying to solve it.

The first step is to solve this so called homogeneous equation:

Then we proceed to the particular solution.

We substitute this into the original equation:

Next, let's see the meaning of resonance.

First-order constant coefficient linear differential equations - The resonance

In cases where the particular solution includes exponential expressions, well, we could face some problems.

The next slideshow will discuss that.

If the particular solution includes an  term, then we may face some problems when trying to solve it.

The first step is to solve this so called homogeneous equation:

Then we proceed to the particular solution.

We substitute this into the original equation:

Next, let's see the meaning of resonance.

This occurs when there is  in the particular solution, and its exponent is exactly the same as the exponent of the homogeneous solution.

This time the exponents are not the same, so there is no resonance.

But now, there is.

Let’ see what happens now.

So, it is equal to the homogeneous solution.

This is what we call “resonance”.

And in this case an x comes here.

Let's see another one.

The homogeneous solution is the usual one:

The particular solution will include a linear expression,

an ,

and another  where there is resonance.

Second-order homogeneous linear differential equations with constant coefficients

Second-order homogeneous linear differential equations with constant coefficients

Here we have this equation.

Using the previous methods we can’t expect much success in solving this equation, since it is of second order.

Well, this is not a very encouraging outlook on the solution.

Equations like this are usually darn hard to solve.

But luckily, this type is an exception.

Let’s see what we should do with it.

This is the general form of the equation. The solution of such equations is always some

Let’s substitute this into the equation, and see what happens.

This equation is called the characteristic equation.

In order to solve the differential equation, we have to solve this quadratic equation.

The solution of the differential equation is:

If the characteristic equation has two different real solutions:  and , then

If the characteristic equation has one real solution, then

If the characteristic equation has two different complex solutions:

And now, let's see the solution.

The characteristic equation is:

It seems we solved this one. Let's see another one.

Here comes the characteristic equation:

This wasn’t too hard either.

Finally let’ see the third type.

Well, there is a little trouble here.

There is a negative number under the root symbol, which means there is no real solution for the characteristic equation.

However, there is a complex solution. All we need to know is this:

And now let's see the solution.

It gets really exciting if the equation is nonhomogeneous.

Let’ see what happens in that case.

Second-order nonhomogeneous linear differential equations with constant coefficients

The homogeneous equation and its solution:

If it has two real solutions:

If it has one real solution:

If it has two complex solutions:

Particular solution (Undetermined Coefficients Method)

Here is this equation that is nonhomogeneous:

In such cases we solve the homogeneous equation first,

and then find the particular solution using the Undetermined Coefficients Method.

To solve the homogeneous equation, we solve the usual

characteristic equation.

And now we are ready for the particular solution.

We can obtain the particular solution based on the function on the right side.

This time the function on the right side happens to be a polynomial, so we try to find the particular solution in that form, too.

But the right-side function could be exponential,

or even trigonometric.

The particular solution is:

Let's see what we get if we substitute this into the original equation.

And the general solution is:

Then, we have this other nonhomogeneous equation.

But there is a snag.

Just like with first-order equations, there could be resonance here, too.

Resonance occurs if a term of the homogeneous solution is equal to a term of the particular solution.

Now there is no resonance,

but there will be some in the next slideshow.

Compared to first-order equations, this resonance business gets a bit more complicated at second-order equations.

Here is this equation:

The solution of the homogeneous equation is:

And now we are ready for the particular solution.

We always figure out the particular solution based on the function on the right side.

One term of the homogeneous solution is equal to a term of the particular solution. That means that sadly, there is resonance.

In this case the constant multiplier doesn’t matter.

Due to the resonance, an x will come in here.

Now we compute the first and second derivative of the  particular solution.

Next, we substitute these into the original equation.

When the characteristic equation has only one real solution, there could be dual resonance.

The resonance appeared.

Therefore, a multiplier of x will be needed in the particular solution.

But then there will be resonance with the second term...

so we need another x multiplier.

This is what we call dual resonance.

From here, the solution proceeds as usual.

Boring, as usual.

So let’s not solve it now. Instead, let’s see what kind of resonance could appear when the characteristic equation has two complex roots.

Here are these two equations:

The characteristic equations are:

For the complex solution all we need to know is this:

In such cases resonance occurs if:

And then is the trial function.

Second-order nonhom. lin. diff. equations with constant coeff. - The resonance

The homogeneous equation and its solution:

If it has two real solutions:

If it has one real solution:

If it has two complex solutions:

Particular solution (Undetermined Coefficients Method)

Here is this equation that is nonhomogeneous:

In such cases we solve the homogeneous equation first,

and then find the particular solution using the Undetermined Coefficients Method.

To solve the homogeneous equation, we solve the usual

characteristic equation.

And now we are ready for the particular solution.

We can obtain the particular solution based on the function on the right side.

This time the function on the right side happens to be a polynomial, so we try to find the particular solution in that form, too.

But the right-side function could be exponential,

or even trigonometric.

The particular solution is:

Let's see what we get if we substitute this into the original equation.

And the general solution is:

Then, we have this other nonhomogeneous equation.

But there is a snag.

Just like with first-order equations, there could be resonance here, too.

Resonance occurs if a term of the homogeneous solution is equal to a term of the particular solution.

Now there is no resonance,

but there will be some in the next slideshow.

Compared to first-order equations, this resonance business gets a bit more complicated at second-order equations.

Here is this equation:

The solution of the homogeneous equation is:

And now we are ready for the particular solution.

We always figure out the particular solution based on the function on the right side.

One term of the homogeneous solution is equal to a term of the particular solution. That means that sadly, there is resonance.

In this case the constant multiplier doesn’t matter.

Due to the resonance, an x will come in here.

Now we compute the first and second derivative of the  particular solution.

Next, we substitute these into the original equation.

When the characteristic equation has only one real solution, there could be dual resonance.

The resonance appeared.

Therefore, a multiplier of x will be needed in the particular solution.

But then there will be resonance with the second term...

so we need another x multiplier.

This is what we call dual resonance.

From here, the solution proceeds as usual.

Boring, as usual.

So let’s not solve it now. Instead, let’s see what kind of resonance could appear when the characteristic equation has two complex roots.

Here are these two equations:

The characteristic equations are:

For the complex solution all we need to know is this:

In such cases resonance occurs if:

And then the trial function is done.

Differential equations | Problem 13

Differential equations | Problem 14