Tag Archives: constant coefficients

Repeated Roots, Second Order Linear

Overview:

In a regular second order linear homogeneous equation the solution is easy to solve for after learning it in class. The equation should usually be in the following form when solving:(homogeneous=0). Where A, B, and C are constants and the highest degree should be the 2nd.

Ay”+By+Cy=0

The next step would be to get the characteristic equation of the differential equation from above or manage to put it into this form:

r²+Br+C

From here we need to be able to solve for the characteristic equation. This can be done in two ways. Either by factoring the equation or if it is not factor-able you may use the quadratic equation.

After getting the roots of the equation; we will call them r1 and r2, you can now proceed to solving the equation. The general solution of the characteristic equation is:

y=Ae^(r1*t)+Be^(r2*t)

This case is true for normal characteristic equations where the roots are real and different from each other , but in the case of this topic we would like to look at repeated roots; this would be when r1=r2.

If we try to solve the homogeneous equation using normal means with the repeated root the answer would come up to zero; that is not the correct answer because the values we are trying to solve for get cancelled out.

So in order to solve a equation with repeated roots we need to do one thing which is add a “t” in front of one of the original terms in the solution. Meaning the new equation we need to plug in values to would be:

 y=Ae^(r1*t)+tBe^(r2*t)

This would the final equation we use in order to solve an equation with repeated roots where A and B are the constants we want to solve for.

Example/Sample Problem:

Now lets try solving for a sample problem with repeated roots.

y”+18y’+81y=0 ; y(0)=-5, y'(0)=48

So the first step is to set up the characteristic equation:

r²+18r+81

Now we need to factor or solve for the values of r1 and r2.

(r+9)*(r+9)

r1=-9, r2=-9

Now as we can see the roots are exactly the same or repeated roots. So we have to plug it into the equation which would look like this:

 y=Ae^(-9t)+tBe^(-9t)

The next step would be to solve for the derivative of y.

y’=-9Ae^(-9t)-9tBe^(-9t)+Be^(-9t)

Now that we have both y and y’ we can now plug in the inital values of y(0)=-5 and y'(0)=48 to solve for the values of A and B.

For y(o)>>>  -5=Ae^(-9*0)+0*Be^(-9*0)

-5=A

For y'(0)>>> 48=-9Ae^(-9*0)-9*0Be^(-9*0)+Be^(-9*0)

48=-9A+B

48=-9(-5)+B

B=3

So now that we solved for both A and B we can now put it back into the original equation.

 y=-5e^(-9t)+3te^(-9t)

This is the final answer for the repeated roots problem.

3. In case I wasn’t clear enough with the explanation or steps needed to solve the equation, or even if you just didn’t understand because I skipped some steps while solving the equation you can refer to these two YouTube links for more help. The videos could help to provide an extra problem that you can work out while watching the video. The YouTuber provides excellent visuals and a great explanation of how the problems should be solved.

(Disclaimer: Not owned by me, all credits go to “Khan Academy”)

Second order linear homogeneous equations with constant coefficients

You have probably seen first order linear equations. Now we”ll move to second order linear equations. What makes it linear? A differential equation is linear if it is a linear function of the variables y, y’, y” and so on.

The standard form of the second order linear equation is

y” + p(t)y’ + q(t)y = g(t)

where p(t), q(t), and g(t) are constant coefficients.
There can also be a constant coefficient in front of the y”.
When g(t) = 0, we can call the differential equation homogeneous, otherwise it is a non-homogeneous.
We can write the second order linear homogeneous equation can be written as:

ay”+by’+cy=0

From that, we can write that characteristic equation:

ar^2+br+c=0

From the characteristic equation, we can find the roots. The roots can come out to be real roots, repeated roots, or complex roots. But we will only focus on real roots.
How do we know if the roots are real roots? We can find out by find out the discriminant. To find out the discriminant, we use b^2-4ac. For the roots to be real, it must follow this condition:

b^2-4ac > 0

After finding out the roots, we can plug it into the general solution which is:

y(t)=C1e^(r1*t)+C2e^(r2*t)

Now we can figure out the derivative of the general solution which we’ll use later.

y'(t)=(r1)C1e^(r1*t)+(r2)C2e^(r2*t))

With problems like these, we will have initial condition, y(0) and y'(0), which will be used to find C1 and C2 to determine the particular solution.
You plug in the initial to determine C1 and C2.
Once you determined C1 and C2, plug them back into the general solution to get the particular solution satisfying the initial condition.
The initial conditions gives us a unique solution.
Example 1:      y” + 5y’ + 6y = 0         y(0) = 2     y'(0) = 3
We start out with the characteristic equation which is:

r^2 + 5r + 5 = 0

Now find the roots anyway you feel comfortable with.
One way is by using the quadratic formula.
Now we got

r1 = (-2)
r2 = (-3)

We can plug it into the general solution

y(t)=C1e^(-2t)+C2e^(-3t)

Now that we have to figure out the derivative of general solution which will be:

y'(t)=(-2)C1e^(-2t)-3C2e^(-3t)

Now we can plug in the initial values which are y(0) = 2     y'(0) = 3
For the initial condition y(0) = 2:

2=C1e^(-2*0)+C2e^(-3*0)
2=C1+C2

For the initial condition y'(0) = 3:

3=(-2)C1e^(-2*0)-3C2e^(-3*0)
3=-2C1-3C2

Now that we have 2 equations, we can use them to solve for C1 and C2:

2=C1+C2
3=-2C1-3C2

Multiply the first equation by 2 so the C1 can cancel out.

4=2C1+2C2
3=-2C1-3C2
7=-1C2
C2=(-7)

Now that we know C2, we can plug it back into any of the two equations to find C1.

2=(-7)+C1
C1=9

Lastly we plug the C1 and C2 into the general solution to determine the particular solution.

y(t)=9e^(-2t)-7e^(-3t)

Videos to help with Second order linear homogeneous equations with constant coefficients:
This video is a transition from first order linear differential equations to second order linear differential equations.


This video goes into more detail when the roots are repeated or complex. It only goes up the general solution


In this video, he does a whole example using initial conditions.

Second order ordinary homogeneous differential equations with constant coefficients

Here are some video resources showing complete examples (from Khan Academy).  I hope they help!

-Prof. Reitz

Example 1 — in which the characteristic equation has two distinct real roots.

 

Example 2 — in which the characteristic equation has one repeated (real) root.

 

Example 3 — in which the characteristic equation has complex roots.