There are quite a number of ways of seeing this.

It is probably worth pointing out a different problem. Suppose you have

,

and you want to send this back. You look in the table and see that

but isn’t a square… or is it? Any positive number can be written as a square via , and so,

.

This is how you could spot that as .

OK, I have four ways of getting .

1. First of all, just by noticing that it is a difference of squares. We have

,

and indeed any such factor can be factored like this using the above trick:

.

You asked about other ways.

2. The other way is via the Factor Theorem:

a root a factor,

therefore we look for the roots of i.e. we solve it equal to zero:

.

Therefore we have roots and and so factors and :

.

3. The same except that at we use the ‘-b’ formula with :

,

and the rest follows similarly.

4. Pragmatically, you have but the answer has . Multiply this out:

,

so they are equal.

Regards,

J.P.

Is there a long way of doing this out? I won’t get this in the exam if something like this comes up. Do you just have to know that changes to and ?

]]>I’m afraid your solution is incorrect in many ways.

Looking at the auxiliary equation

is correct. However the factors are

.

This gives a homogeneous solution:

.

Now we look for a particular solution. We trial . This has first and second derivatives and . Force this into the differential equation

.

To satisfy the differential equation this must equal :

, and so we have a particular solution

.

So that the general solution is

.

Now we need to find and such that the initial conditions are satisfied. I will leave this to you.

Regards,

J.P.