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Solving Differential Equations with Long Division

4.3 More manipulations

Just as the fraction can be decomposed into , so can the fractions from differential equations:

This is just stating that solutions to

can be obtained by taking solutions from the equations

and adding them: f(x) = g(x) + h(x).

And this is just a generalization of the integration step,

which says merely that a solution of

can be obtained by solving (in order)

However, we won't have much need for this until we look at more advanced stuff.

5 Putting it all together

We now have three rules, integration, exponential shift, and long division, which combined give a powerful
means to tackle approximately half of the differential equations one faces in a first class in differential

To summarize, when faced with a differential equation   and where is some
constant coefficient polynomial in and h(x) is some sum of polynomials, exponentials, and polynomials
times exponentials, you can always find a sequence of steps and manipulations which will arrive at one

Let's demonstrate with one complicated example. The differential equation is,

which means our fraction is

Doing long divisions,

We arrive at a solution

6 General solutions

The techniques shown so far will produce a solution to an important class differential equations, but just as
in calculus, having one answer is not always enough. While   is a solution of   there
are other solutions which all have the form where C is some constant, i.e. "any two integrals differ
by no more than a constant".

When we solve differential equations, one or several integrations occur, and getting the most general
solution is a matter of keeping track of the Cs that come from them.

With the differential equations we are studying, this notion of a constant that solutions can differ by is
replaced with something more general. Say f1(x) and f2(x) are both solutions of , i.e.

then, by subtracting the two equations, we have

showing that two solutions differ by no more than a   solution. When trying to find just one solution,
not caring which, we have taken Finding all possible solutions of amounts to finding one
solution of   and then adding to it all solutions of .

Consider the example from integration,

since C is the most general solution of This is what we already know about indefinite integrals.
For a more complicated   we have to be more clever.

and we can check that   Thus

is the general solution. It is no coincidence that we took 0 to be in the above. If we had picked a
different exponential, , we would have gotten nowhere

Finding the right exponential shifts to make the s apparent requires a bit of trickery.

Lemma 6.1 (Roots of the denominator). If k is some number such that , then

Conversely, the only solutions of

are of the form   where k D(k) = 0 and p(x) is a polynomial with degree at most m where m
is the multiplicity of k in the factorization of D.

Proof.,so adding things up we have

The converse is beyond the scope of this discussion.
In other words, to find all the solutions of you need only consider those exponential shifts with
where D(k) = 0.

For example, k2 + k - 2 = 0 when k is either 1 or -2, so

where and   are constants which "suck up" the 3s and negative sign.
So what we have learned here is that to get general solutions, one has to find all the solutions of
which come from exponential shifts that cause a (a.k.a. an unknown constant) to appear.
So the general solution to the large example problem is

There are further simplifying tricks to finding all solutions of but you will have to work those out
on your own.

7 Problems

Here are some differential equations to try things out on.
These only require long division.

This requires long division and exponential shift.

This requires all the tricks.

After you find one solution, you might want to try finding the general solution (keeping track of the Cs).
When the highest derivative is there should be 2 of them and when the highest is there should be 1.

7.1 Advanced
Show that if   is a polynomial in , and D(k) ≠ 0, then

has the solution. An easy way is to plug this into the differential equation. A more interesting way
is to use long division. What solutions can you find if D(k) = 0?

The polynomial k2 + 3k + 2 factors into (k + 1)(k + 2). Show that

that is, first solve and then take the result and "divide" it by and check to see that it is a
solution to