- [Instructor] In the previous video,
you learned what linear homogenous,
higher order differential equations are.
In this video, you will focus on solving those types
of equations with constant coefficients using
the characteristic equation.
Let's review what this entails.
Let's say you have a constant coefficient,
higher order differential equation with the form of a n
of minus one of t multiplied by y to the n minus one
of t plus all the way to plus a1 of t multiplied
by y prime of t plus a0 of t multiplied by y
of t equals zero.
Note that this is for real constant values for a of zero
to a of n, where a n of t does not equal zero.
You then have the corresponding characteristic equation,
which is equal to a n of t multiplied by r
to the n plus a n of minus one of t multiplied by r
to the n minus one plus all the way to a1 of t multiplied
by r plus a0 of t equals zero.
In this video, I will have you focus on factoring
the characteristic equation since other methods tend
to be more complex to work through.
Since you can't use the quadratic formula
for more than two roots, unless if they can be factored
accordingly to use it, you'll then need
to identify the roots
as they appear in your characteristic equation in order
to properly factor them.
Remember, you can have multiple types
of roots show up in the result of your equation,
because it is higher than a second order.
Let's review those three root types.
For type one, you have real distinct roots
where the roots are real numbers
and r of one does not equal r of two.
So this means that
however many distinct roots you have,
they must not equal each other.
Root type number two, you have identical real roots,
which are real numbers as well,
but this time they equal each other such
as r1 equals r2.
So again, note, you might have more
than two identical real roots,
so be aware that they may show up.
And finally, root type number three is complex roots
and note that these can be distinct or identical,
having them equal each other or not,
and that they involve imaginary numbers such as i with roots
that contain real constant a and b where a can be zero
and they are formatted with r1 equals a plus i multiplied
by b and r2 equals a minus one multiplied by b.
Note that again, sometimes you could factor these,
or sometimes when you're factoring you can then factor
your characteristic equation accordingly
in order to use the quadratic formula to solve
for these types of roots.
Let's review the types
of solutions you would have for the three different roots.
So for real and distinct roots,
your solution will be in the format
of y equals C1 multiplied by e to the r1 multiplied
by x plus C2 multiplied by e to the r2 multiplied
by x plus as many of those values as you need
for the real and distinct roots.
So note that sometimes you might just have one,
sometimes you might have two, sometimes you might have five.
Next is identical real roots.
These have the solution format of y equals C1 multiplied
by e to the r multiplied by x plus C2 multiplied
by x multiplied by e to the r multiplied by x plus
as many as you need.
Note that these have the form,
or you'll be multiplying them by x to the n
to where you have essentially C1 doesn't
have an x with it.
C2 has x, C3 would have x squared,
C4 would have x cubed and so on.
Finally, with complex roots, they have the form
of y equals C1 multiplied by e to the a multiplied by x,
multiply that by cosine of b multiplied by x,
and then you'll add in C2 multiplied by e
to the a multiplied by x multiplied by sine
of b multiplied by x.
Remember, this is for your roots, r1 and r2
with the value of a plus or minus i multiplied
by b where you plug in those values
of a and b accordingly.
Note that with complex roots,
usually you have them paired up like this
where you have those values to get cosine and sine,
but sometimes you might have identical sets of these roots,
and so in that case you do what you would
with identical real roots where you just multiply them
by whatever value of x to the n.
So for example, if you had two of the same set,
you would just multiply one of them by x
for both your cosine and sine terms.
If you have initial conditions, then you can solve them
to get the constant values in order
to gather a particular solution
for your differential equation.
Note that if you have repetitiveness such
as multiple solutions with the same route,
then you can multiply them accordingly,
such as by x to the n
to ensure they are all distinct solutions,
making sure there is linear independence.
So note that this is what I was explaining
in the previous slide,
where again, whether it's with the complex or real roots,
whenever you have identical terms for them,
you want to multiply them by the x to the n.
Let's solve a few examples to solidify this knowledge of how
to find the various roots
with higher order differential equations.
So let's begin with the equation y triple prime minus six
multiplied by y double prime plus 11 multiplied
by y prime minus six multiplied by y equals zero.
This gets you the characteristic equation
of r cubed minus six multiplied
by r squared plus 11 multiplied by r minus six equals zero.
This equation can easily factor into r minus one,
r minus two, and r minus three equals zero.
This gets you three distinct real roots of r1 equals one,
r2 equals two and r3 equals three.
You can then plug this into your general solution
to get y equals C1 multiplied by e
to the r1 multiplied by x plus C2 multiplied by e
to the r2 multiplied by x plus C3 multiplied by e
to the r3 multiplied by x.
Once you plug in your three roots of one, two,
and three, you then get a final general solution
of y equals C1 multiplied by e to the one multiplied
by x plus C2 multiplied by e to the two multiplied
by x plus C3 multiplied by e to the three multiplied by x.
Note that typically you would then simplify that first term
to just be e to the x versus e to the one multiplied by x.
Let's solve another equation.
This time, let's solve the differential equation,
two multiplied by y4,
which is a fourth derivative, plus 11 multiplied
by y triple prime plus 18 multiplied
by y double prime plus four multiplied
by y prime minus eight multiplied by y equals zero.
This time you have your characteristic equation
of two multiplied by r to the fourth plus 11 multiplied
by r cubed plus 18 multiplied
by r squared plus four multiplied
by r minus eight equals zero.
You can then factor this equation to get two multiplied
by r minus one and then r plus two cubed.
So that means that r plus two would show up three times,
and again that's equal to zero.
This gets you one distinct real root
of r1 equals one half
and three identical real roots of r2, r3
and r4 all equal to negative two.
This will then get you a general solution in the format
of y equals C1 multiplied by e to the r1 multiplied
by x plus C2 multiplied by e to the r2 multiplied
by x plus C3 multiplied by x multiplied by e
to the r3 multiplied by x plus C4 multiplied
by x squared multiplied by e to the r4 multiplied by x.
So note that you have those extra x
and x squared terms to account
for the identical real roots you are dealing with.
If you plug in your roots,
you then get a final general solution
of y equals C1 multiplied by e to the one half multiplied
by x plus C2 multiplied by e
to the negative two multiplied by x plus C3 multiplied
by x multiplied by e to the negative two multiplied
by x plus C4 multiplied by x squared multiplied
by e to the negative two multiplied by x.
Let's look at one more example,
but this time let's level it up in the complexity.
This time you have y to the five, which is going
to be your fifth derivative plus 12 multiplied by y
to the four, which is a fourth derivative plus 104
multiplied by y triple prime plus 408, multiplied
by y double prime plus 1,156 multiplied
by y prime equals zero.
You can then plug this
and get your characteristic equation of y
to the fifth plus 12 multiplied by r
to the fourth plus 104 multiplied by r cubed
plus 408 multiplied
by r squared plus 1,156 multiplied by r equal to zero.
You can then factor this characteristic equation
to get r multiplied by r squared plus six multiplied
by r plus 34,
and that part is going to be squared,
meaning it shows up twice and again that's equal to zero.
You can actually use the quadratic formula to solve
for the r squared plus six multiplied
by r plus 34 terms in your characteristic equation.
So in this case, your a is equal to one, b is equal to six,
and c is equal to 34,
and then you have your quadratic formula
of r equals negative b plus,
or minus the square root of b squared minus four multiplied
by a, multiplied by c divided by two, multiplied by a,
Plugging in those values for a, b and c,
you then get r equals negative six plus or minus
the square root of six squared minus four multiplied
by one multiplied by 34,
divide all that by two multiplied by one.
This then solves to be r equals negative three plus,
or minus five multiplied by i.
Now, when you solve your characteristic equation
of r multiplied by r squared plus six multiplied
by r plus 34,
and that portion is squared equals zero,
you then get one distinct real root of r1 equal to zero.
So that's the r term that's on the outside,
and then you get two sets of the same complex root pair.
So you'll have r2
and r4 equal to negative three plus five multiplied by i,
and then you'll have your roots r3 and r5 equal
to negative three minus five multiplied by i.
So your general solution will be in the following format
and note that it is on two lines,
because it's fairly lengthy.
So you'll have y equals C1 multiplied by e
to the r1 multiplied by x,
and that's for your distinct real root.
And then the rest of this is for those two sets
of complex roots.
So first you have plus C2 multiplied by e
to the a multiplied by x, multiplied by cosine of b
multiplied by x plus C3 multiplied by e to the a multiplied
by x multiplied by sine of b multiplied by x.
So that'll be for your first set of complex roots.
Then for your second set of complex roots,
you have plus C4 multiplied by x multiplied by e
to the a multiplied by x, multiply that by cosine of b
multiplied by x plus C5 multiplied
by x multiplied by e to the a multiplied
by x multiplied by sine of b multiplied by x.
So this is a great scenario where you have two sets
of the same complex roots,
and so again, you have that first set
and then the second set just adds that extra x term
that's multiplied in to make it linearly independent.
So note that for your complex roots, your a value is equal
to negative three and your b value is equal to five in order
to plug into those natural exponents and your sine
and cosine values.
Something else to note before I show you the final solution
is let's solve the very first term of that solution
where you have C1 multiplied by e
to the r1 multiplied by x.
If you plug in your root of r1 equals zero,
you then get C1 multiplied by e to the zero multiplied
by x, which equals just C1 multiplied by one,
which is just C1.
This will then simplify your final general solution
to be y equals C1 plus C2, multiplied by e
to the negative three multiplied by x, multiplied by cosine
of five multiplied by x plus C3 multiplied by e
to the negative three multiplied by x multiplied by sine
of five, multiplied by x plus C4 multiplied
by x multiplied by e to the negative three multiplied by x
multiplied by cosine of five multiplied
by x plus C5 multiplied by x multiplied by e
to the negative three multiplied by x, multiply that by sine
of five multiplied by x.
So again, pretty lengthy solution for this one,
but it's great seeing how all these different terms
come together.
Like I mentioned before, it is usually easiest
to use this technique
for higher order differential equations when you can factor
the characteristic equation.
Otherwise, you will need to use more advanced methods
such as numerical methods and Cardano's formula
to find your roots.
Note that you can also gather the particular solution
for these examples,
or others like them by using the same process as before.
Up next, let's explore reduction of order
with higher order differential equations.