Welcome to a lesson on constant coefficient
higher order linear homogeneous differential equations.
When we have a higher order constant coefficient
homogeneous linear equation, the procedure is the same
as it was for second order.
We just need to find more solutions
if the equation is and the order we need to find
in linearly independent solutions.
Let's look at an example.
Let's determine the general solution
to Y triple prime minus three y, double prime minus y
prime plus three Y equals zero.
Just like we did for the second order equation
we will assume the solution is Y equals E, the power of RX.
Next we determine the first
through third derivatives using the chain roll
and then perform substitution into the differential
equation which gives us the equation shown here.
And now we divide both sides by E, the power of Rx notice E
the power of Rx is never zero
and therefore we don't lose any solutions
and we could also factor out E
of the power of RX instead if we wanted to.
The result is the following characteristic equation
which is R cubed
minus three R squared minus R plus three equals zero.
And now we need to find the roots, which we can do
by factoring or solve graphically.
To solve by factoring
we use the technique of grouping where we cut the polynomial
in half on the left and then we factor
out the greatest common factor on the left and right
and we should have a common binomial factor
factoring R squared from R cubed minus three R squared
we have R squared times the quantity R minus three.
Next we factor out negative one from negative R plus three
which gives us negative one or minus one times the quantity.
R minus three equals zero.
Notice we do have a common binomial factor of R minus three.
We now factor R minus three from the left
which gives us the quantity R
minus three times the quantity.
R squared minus one equals zero.
R squared minus one is a difference of squares giving us
a factor of R minus one and a factor of R plus one.
The three roots are solutions are R equals three
R equals one and R equals negative one.
Notice we have three distinct real roots,
on the left if we were to graph the characteristic equation
notice how we do have roots of negative one, one
and three when we had a second order linear homogeneous
equation with constant coefficients.
Recall when we had two distinct real roots
the general solution was Y of X equals C sub one times E
the power of R sub one times x plus csub two times E
the power of R sub two times x.
And because we have three distinct real roots
the general solution is Y of X equals csub one times E.
The power of negative x using the root of negative one
plus C sub two times either of the X using the root
of positive one plus C sub three times E of power
three x using the root of positive three.
Let's look at a second example.
Let's find the general solution to the fourth order.
Constant coefficient, linear homogeneous defense equation.
Going straight to the characteristic equation.
We have R, the fourth minus three
R cubed plus three R squared minus R equals zero.
Notice how we have a common factor of R factoring out.
R gives us this equation.
And then from here we need to find one root
for R cubed minus three R squared plus three R minus one
which we can do by analyzing the graph on the left.
Notice how the graph on the left does show the root
of zero from the factor of R.
And notice one is also a root
which means R minus one must be a factor
of the given degree four polynomial.
From here we can perform synthetic division
which I've shown here below.
Where for R cubed minus three R squared
plus three R minus one, we have one negative three
three negative one because one is a root.
We have a one here on the left
and then we perform synthetic division, which for review
we bring down the one multiply by the root of one.
One times one is one, and then we add vertically.
Negative three plus one is negative two
multiply by one again, which gives us negative two.
Add get positive one, multiply by one.
Again, add we get zero.
The one negative two one represents the factor
of R squared minus two R plus one.
Which factors again?
Notice how we have R times three factors of our minus one.
This indicates the roots are zero
and one where one has a multiplicity of three.
So again, if we go back
to what we know about second order constant coefficient
linear homogeneous differential equations.
If we have two real equal roots
the general solution was YEX equals C sub one times
either power of RX plus csub
two times X times E power of RX.
So for our general solution we have Y of X equals
using the root of zero.
We have csub one times E of the zero and then plus
because the root of one has multiplicity of three
we have csub two times either the X
plus csub three times x e to the X
plus C sub four times x squared E to the X.
Notice for this third term here we have two factors
of x simplifying.
We have the general solution of Y of X equals C
sub one plus C sub two E to the X plus C sub three
x e to the X, plus C sub four x squared E to the X.
Let's look at one more example where we have complex roots.
Let's find the general solution
to Y triple prime minus two y double prime
plus nine Y prime minus 18 Y equals zero.
The characteristic equation is R cubed minus two R squared
plus nine R minus 18 equals zero.
And analyzing the graph on the left
we can see one root is R equals two
which gives us a factor of R minus two.
And then to determine the remaining factor we perform
to that division using two.
So notice how again the coefficients are one negative two
nine and negative 18, and the root of two is outside.
We bring down the one multiply by two, add multiply by two
add multiply by two add the result is 1090
which indicates the remaining factor is R squared plus nine.
So now in factored form we have the quantity R
minus two times the quantity R squared
plus nine equals zero.
This indicates R equals two, which we already knew
from the graph and then R squared
plus nine must equal zero subtracting nine
and square rooting both sides.
We get R equals plus or minus three I.
Recall when the equation was second order
and we had complex roots or complex solutions.
The general solution was in the form
of Y of X equals C sub one E, the power
of alpha X cosign and beta x plus csub two times E.
The power of alpha X times sine beta X
which means our general solution first using the root
of two, we have Y of X equals csub one times E
the power of two x, and then using the two complex solutions
or imaginary solutions where alpha zero and beta is three.
We have CSUB two, either the zero times cosign three x
plus csub three, either the zero sign, three x, simplifying.
We have the general solution of Y of X equals csub one
either the two x plus C sub two cosign three x
plus C sub three, sine three x.
We'll look at some more examples in the next several videos.
I hope you found this helpful.