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## Differential equations

### Course: Differential equations > Unit 2

Lesson 1: Linear homogeneous equations# 2nd order linear homogeneous differential equations 4

Another example with initial conditions! Created by Sal Khan.

## Want to join the conversation?

- Now that I've been through this playlist, I have to ask, will the general solutions for all linear homogeneous differential equations include e^rx? I mean, are there usually no other types of solutions that appear in these problems (like maybe some arbitrary polynomial or perhaps even natural log), just the exponential function? Thanks in advance.(20 votes)
- In short, no. A polynomial kind of degrades as you take successive derivatives, so the derivatives cannot sum to 0. Differentiating natural log puts it in the form 1/x, so you definitely could not get things to cancel out.

However, there are*non*-homogeneous ODEs. These are of the form

Ay''+By'+Cy=f(x). These equations end up having a solution that can be a combination of the exponential function and polynomials, trigonometric, etc. functions. Check out the videos on the method of undetermined coefficients. They're really interesting!(27 votes)

- I tried doing some research online, and I kept on reading about something called the Wronskian. Could you please explain that?(18 votes)
- Right, so if we have n solutions to a differential equation that are each differentiable n-1 times, the Wronskian is the determinant of the matrix formed by the equations and each of their k derivatives. So, if we have two solutions, y_1 and y_2 (n =2), and they are differentiable so that we can get y_1' and y_2' then we can form the Wronskian as follows:

| y_1 y_2 |

W = | y_1' y_2' | = y_1*y_2' - y_1'*y_2

This is the simple case of n=2, and you can show that it expands to be true for all n if you assume the equations that form the solution to be 'nice enough'.(15 votes)

- What if a Constant (C1 or C2) ends up equaling 0... which e (and it's power) am suppose to eliminate?(4 votes)
- You look at your equation, C1 is the first C and C2 is the second. It doesn't actually doesn't matter if you labeled the first one as C2 or the second one as C8 as long as you continue using that placement of first and second once you start solving for C1,2,3....n(4 votes)

- Do you add og subtract when you do the thing with:

c1+c2=2

3/2c1+1/2c2=1/2

On this vid you subtract, and on the video before you add them together?(4 votes)- Adding or subtracting those equations depends on the sign of c1 or c2 . Our aim is to cancel out one of the constants (either c1 or c2) and thereby find the other. Here we have the sign of c1 positive in both the equations.c1 cancels out only when we subtract. hope you understand(1 vote)

- Using the same technique, can you solve first order differential equations in the same fashion, 0y'+3y'+6y=0? Can you solve a first order equation that way?(2 votes)
- Short answer, yes. In fact any constant coefficient equation like the one you gave, the one in this video, or one of any order is really only as hard as finding the roots of its characteristic equation. These get harder as their degree goes up, but you could cook up some easy third-order examples. Their characteristic equations would have a degree matching the order, so would be cubics. Hope this helps!(3 votes)

- 1:58You can solve the quadratic equation without using the quadratic formula like this:

4r^2 - 8t + 3 = 0 |Divide by four:

r^2 - (8/4)r + 3/4 = 0 |Multiply both numerator and denominator of 3/4 by four:

r^2 - (8/4)r + 12/16 = 0 |Figure out two numbers a,b that satisfy a + b = -8 and ab = 12:

a = -2, b = -6 |Substitute a,b into (r + a/4) (r + b/4) = 0:

(r - 2/4) (r - 6/4) = 0

(r - 1/2) (r - 3/2) = 0

r = 1/2 or r = 3/2(2 votes) - Are there any videos on this site about systems of differential equations?(2 votes)
- What if the values for r are the same? How do you solve for C1 and C2 then?

Example:

y(0) = C1 + C2 = 5

y'(0) = (-13/5)C1 + (-13/5)C2 = 5

How would you go about solving this one?(2 votes)- Interesting question!

The system of equations that you wrote has no solutions, because C1 + C2 = 5 would imply that (-13/5)C1 + (-13/5)C2 equals -13 instead of 5. So there's a problem.

I suspect from your system of equations that you got a double root of r = -13/5 when solving a second order differential linear homogeneous equation, and the initial conditions were y(0) = 5, y'(0) = 5.

In this situation, the general solution is y = C1 e^(-13x/5) + C2 x e^(-13x/5), because one of the two terms of the general solution needs a factor of x when there's a double root.

Then the initial conditions y(0) = 5 and y'(0) = 5 would lead to the system

C1 = 5

(-13/5)C1 + C2 = 5.

This system of equations is quite easy to solve.

Have a blessed, wonderful day!(1 vote)

- Hi, understand this. Thank you very much for the great explanation. I have a question thought:

can someone tell me what I would do with an equation in this form:

5y'' + 3y' +4y - 1 = 0 ?

What to do with the "-1" ? Can I neglect it? Am i missing an obvious algebraic transformation?(1 vote)- Just add 1 to both sides so it becomes 5y''+3y'+4y=1 then apply undetermined coefficients or laplace transforms to compute the answer.(3 votes)

- At6:04is multiplying only top equation right? It seems intuitive, but doesn't that affect the solution?(1 vote)
- No, that one of the most common ways to save a system of equations. Since you are multiplying the whole equation (both sides of the equal sign), you are not modifying the equation at all.(2 votes)

## Video transcript

Let's solve another 2nd order
linear homogeneous differential equation. And this one-- well, I won't
give you the details before I actually write it down. So the differential equation is
4 times the 2nd derivative of y with respect to x, minus
8 times the 1st derivative, plus 3 times the function
times y, is equal to 0. And we have our initial
conditions y of 0 is equal to 2. And we have y prime of
0 is equal to 1/2. Now I could go into the whole
thing y is equal to e to the rx is a solution, substitute
it in, then factor out e to the rx, and have the
characteristic equation. And if you want to see all of
that over again, you might want to watch the previous
video, just to see where that characteristic equation
comes from. But in this video, I'm just
going to show you, literally, how quickly you can do these
type of problems mechanically. So if this is our original
differential equation, the characteristic equation is going
to be-- and I'll do this in a different color-- 4r
squared minus 8r plus 3r is equal to 0. And watch the previous video if
you don't know where this characteristic equation
comes from. But if you want to do these
problems really quick, you just substitute the 2nd
derivatives with an r squared, the first derivatives with an
r, and then the function with-- oh sorry, no. This is supposed to
be a constant-- And then the coefficient on the
original function is just a constant, right? I think you see what I did. 2nd derivative r squared. 1st derivative r. No derivative-- you could say
that's r to the 0, or just 1. But this is our characteristic
equation. And now we can just figure
out its roots. This is not a trivial one for
me to factor so, if it's not trivial, you can just use
the quadratic equation. So we could say the solution
of this is r is equal to negative b-- b is negative 8, so
it's positive 8-- 8 plus or minus the square root
of b squared. So that's 64, minus 4
times a which is 4, times c which is 3. All of that over 2a. 2 times 4 is 8. That equals 8 plus or minus
square root of 64 minus-- what's 16 times 3-- minus 48. All of that over 8. What's 64 minus 48? Let's see, it's 16, right? Right. 10 plus 48 is 58, then
another-- so it's 16. So we have r is equal to 8 plus
or minus the square root of 16, over 8, is equal to
8 plus or minus 4 over 8. That equals 1 plus
or minus 1/2. So the two solutions of this
characteristic equation-- ignore that, let me scratch that
out in black so you know that's not like a 30 or
something-- the two solutions of this characteristic equation
are r is equal to-- well 1 plus 1/2 is equal to
3/2-- and r is equal to 1 minus 1/2, is equal to 1/2. So we know our two r's, and we
know that, from previous experience in the last video,
that y is equal to c times e to the rx is a solution. So the general solution of this
differential equation is y is equal to c1 times e-- let's
use our first r-- e to the 3/2 x, plus c2 times
e to the 1/2 x. This differential equations
problem was literally just a problem in using the
quadratic equation. And once you figure out
the r's you have your general solution. And now we just have to use
our initial conditions. So to know the initial
conditions, we need to know y of x, and we need to
know y prime of x. Let's just do that right now. So what's y prime? y prime of our general solution
is equal to 3/2 times c1 e to the 3/2 x, plus--
derivative of the inside-- 1/2 times c2 e to the 1/2 x. And now let's use our actual
initial conditions. I don't want to lose them-- let
me rewrite them down here so I can scroll down. So we know that y of 0 is equal
to 2, and y prime of 0 is equal to 1/2. Those are our initial
conditions. So let's use that information. So y of 0-- what happens
when you substitute x is equal to 0 here.? You get c1 times e to the 0,
essentially, so that's just 1, plus c2-- well that's just e to
the 0 again, because x is 0-- is equal to-- so this is,
when x is equal to 0, what is y? y is equal to 2. Y of 0 is equal to 2. And then let's use the
second equation. So when we substitute x is equal
to 0 in the derivative-- so when x is 0 we get 3/2 c1--
this goes to 1 again-- plus 1/2 c2-- this is 1 again, e to
the 1/2 half times 0 is e to the 0, which is 1-- is equal
to-- so when x is 0 for the derivative, y is equal to 1/2,
or the derivative is 1/2 at that point, or the slope
is 1/2 at that point. And now we have two equations
and two unknowns, and we could solve it a ton of ways. I think you know how
to solve them. Let's multiply the top
equation-- I don't know-- let's multiply it by 3/2,
and what do we get? We get-- I'll do it in a
different color-- we get 3/2 c1 plus 3/2 c2 is equal to--
what's 3/2 times 2? It's equal to 3. And now, let's subtract-- well,
I don't want to confuse you, so let's just subtract the
bottom from the top, so this cancels out. What's 1/2 minus 3/2? 1/2 minus 1 and 1/2. Well, that's just
minus 1, right? So minus c2 is equal to--
what's 1/2 minus 3? It's minus 2 and 1/2,
or minus 5/2. And so we get c2 is
equal to 5/2. And we can substitute back
in this top equation. c1 plus 5/2 is equal to 2, or c1
is equal to 2, which is the same thing as 4/2, minus 5/2,
which is equal to minus 1/2. And now we can just substitute
c1 and c2 back into our general solution and we have
found the particular solution of this differential equation,
which is y is equal to c1-- c1 is minus 1/2-- minus 1/2 e to
the 3/2 x plus c2-- c2 is 5/2-- plus c2, which is 5/2, e
to the 1/2 x, and we are done. And it might seem
really fancy. We're solving a differential
equation. Our solution has e in it. We're taking derivatives
and we're doing all sorts of things. But really the meat of this
problem was solving a quadratic, which was our
characteristic equation. And watch the previous video
just to see why this characteristic equation works. But it's very easy to come up
with the characteristic equation, right? I think you obviously see that
y prime turns into r squared, y prime turns into r, and then
y just turns into 1, essentially. So you solve a quadratic. And then after doing that,
you just have to take one derivative-- because after
solving the quadratic, you immediately have the general
solution-- then you take its derivative, use your
initial conditions. You have a system of linear
equations which is Algebra I. And then you solve them for the
two constants, c1 and c2, and you end up with your
particular solution. And that's all there is to it. I will see you in
the next video.