Tag Archives: characteristic equation

Repeated Roots Second Order Linear Homogeneous Equation

Posted on May 1, 2015 by | 4 comments

Overview

Second order linear equations become homogeneous when the linear function of y and y’ (which can be written in the form y” + p(t)y’ + q(t)y = g(t)) is equal to zero. In other words when g(t)=0.

This guide will be discussing how to solve homogeneous linear second order differential equation with constant coefficient, which is written in the following form:

y”+by’+cy = 0

The first step is to use the equation above to turn the differential equation into a characteristic    equation. The characteristic equation is written in the following form:

r2 +br+c = 0

Second to find the roots, or r1 and r2 you can either factor or use the quadratic formula:

 r2 = ± -b √b²-4ac

 2a

It is important to remember when to the particular equation above. There will problems where the variable a is not needed in the quadratic formula because there will be no a in the differential equation. In the cases where there is no a variable limit the a variable from the quadratic equation. The quadratic equation will the look like the following:

r2 = ± -b √b²-4c

2

Once you get repeated roots , or r1 and r2 from the characteristic equation then     y = ert is considered a solution of the differential equation.

The next step would be to plug r1 and r2 into the general equation:

y = C1 er1t +C2 er2t

Another important thing to realize and remember is that when solving a homogeneous equation for a repeated root the solution will end up cancelling out. In order to avoid this a “t” needs to be place in the general solution. The general solution for the repeated root will then be in the following form:

y = C1 er1t +C2t er2t

Sample Problem

Problem 1:

y”+12y’+36 = 0

Step 1: Turn the differential equation into a characteristic equation

r2 + br + c = 0

r2 + 12r + 36 = 0

Step 2: Factor the characteristic equation

r2 + 12r + 36 = 0

(r + 6) (r + 6) = 0

r1 = -6 r2 = -6

Step 3: Use y = ert as a solution for r1 and r2

y = ert

y = e-6t

Step 4: Plug r1 and r2 into the general solution for the repeated roots

y = C1 er1t +C2t er2t

y = ert

y = e-4t

Step 4: Plug r1 and r2 into the general solution for the repeated roots

y = C1 er1t +C2t er2t

y = C1 e-4t +C2t e-4t

y = C1 e-6t +C2t e-6t

 

Problem 2:

y”+8y’+16 = 0 , y(0) = 2   y’(0)= 6

Step 1: Turn the differential equation into a characteristic equation

r2 + br + c = 0

r2 + 8r + 16 = 0

Step 2: Factor the characteristic equation

r2 + 8r + 16 = 0

(r + 4) (r + 4) = 0

r1 = -4 r2 = -4

Step 3: Use y = ert as a solution for r1 and r2

y = ert

y = e-4t

Step 4: Plug r1 and r2 into the general solution for the repeated roots

y = C1 er1t +C2t er2t

y = C1 e-4t +C2t e-4t

The following three videos are form Khan Academy. I find these videos ever useful . I would recommend you watch them if you are still confused  about Repeated Roots of Second order linear homogeneous equation.

 

 

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Non-homogeneous Second Order Differential Equations Using Methods Of Undetermined Coefficients

Posted on May 1, 2015 by | 1 comment

Non-homogeneous differential equations are the same as homogeneous differential equations, However they can have terms involving only x, (and constants) on the right side. The interesting part of solving non homogeneous equations is having to guess your way through some parts of the solution process.
You also can write non-homogeneous differential equations in this format: y” + p(x)y’ + q(x)y = g(x).

The Reason I’ve chosen this problem is because it basically touches every aspect of a Non-homogeneous second order differential Equation using methods of undetermined coefficients.

y” + 6y’ + 9y = -578 sin 5t

The first step when dealing with undetermined or constant coefficients is getting the Characteristic equation. Which will later become the first half of our solution.

y” + 6y’ + 9y = 0
r^2 + 6r + 9 = 0
( r+3 ) ( r+3) = 0
R1 = -3 R2 = -3

When solving the characteristic equation sometimes we run into repeated roots. In this case we must use the following formula y = C1e^r1(t)+C2 (t)e^r2(t)

y (t) = C1e^-3t + C2te^-3t

After finding the characteristic equation our next step in this linear equation is to guess the y. In this case our y is going to include both sin and cosine. We make this guess because the cosine is going to come up eventually when finding the derivative and by adding it into the equation, it will allow us to cancel it out later on throughout the steps in the solution process. We also will need to find the y’ and y”.

y = C sin (5t) + D cos (5t)
y’ = 5 C cos (5t) – 5 D sin (5t)
y” = -25 C sin (5t) + 9 D cos (5t)

Now we plug it back into its original equation.

-25 C sin (5t) – 25 D cos (5t)
+30 C cos (5t) – 30 D sin (5t)
+9 C sin (5t) + 9 D sin (5t)
= -578 sin 5(t)

At this point we want to cancel and group any like terms.

(-16C -30D) sin (5t) + (-16D + 30C) cos (5t) = -578 sin (5t)

Resulting two equations.

(-16) (C) – (30) (D) = -578
(30) (C) – (16) (D) = 0
Now we must find Find C & D
(30) (C) – (16) (D) = 0
+16 (D) +16 (D)
C = (16/30) (D)

Now we plug in C back into the next equation.

(-16)(16/30D) – (30) (D) = -578
(-256/30)(D) – (30) (D) = -578
(-578/15)D = -578
D = 1514304785053481528808497

When calculating the (-256/30)(D) – (30) (D).
We use the calculator to find the fraction value by entering math > frac > enter.
Plug in D

C = (16/30) (D)
C = (16/30) (15)
C = 8

After finding all values we’re finally ready to plug in all variables and combine the homogeneous equation with its general solution.
Resulting in:
y (t) = C1e^-3(t) + C2(t)e^-3(t) + 8 sin 5(t) + 15 cos 5(t)
If we were given the case of finding a particular solution, we will have to take this a few steps further to and plug in the (t) value. But fortunately for us this general solution will suffice.

Below I have included videos that has helped me understand how to solve Non-homogeneous second order differential Equations using methods of undetermined coefficients. Khan academy has been an incredible help in the understanding of all things in relation to differential equations.

 

Problem listed above was taken from the midterm review, question 9 by Professor JReitz.

 

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Repeated Roots, Second Order Linear

Posted on May 1, 2015 by | Leave a comment

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”)

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Differential Equations – Complex Roots

Posted on April 27, 2015 by | 3 comments

In order to achieve complex roots, we have to look at the differential equation:
Ay” + By’ + Cy = 0

Then we look at the roots of the characteristic equation:
Ar² + Br + C = 0

After solving the characteristic equation the form of the complex roots of r1 and r2 should be:
λ ± μi

We refer back to the characteristic equation, we then assume that all the solution to the differential equation will be:
y(t) = e^(rt)

By plugging in our two roots into the general formula of the solution, we get:
y1(t) = e^(λ + μi)t
y2(t) = e^(λ – μi)t

Since these two functions are still in complex form, and we started the differential equation with real numbers. It would best if our solution is also real numbers. In order to transform the complex solution into a real solution, we need to use the Euler’s Formula.
e^(iƟ) = cosƟ + isinƟ
Another form of the Euler’s Formula is:
e^(-iƟ) = cos(-Ɵ) + isin(-Ɵ)
e^(-iƟ) = cosƟ – isin(Ɵ)

Now we split up both of the solutions into two parts one with real exponent and one with an imaginary exponent. Afterwards, we then applied Euler’s Formula.
y1(t) = e^(λt) e^(μit) = e^(λt) [cos(μt) + isin(μt)]
y2(t) = e^(λt) e^(-μit) = e^(λt) [cos(μt) – isin(μt)]

After doing all of that work, we are still left with a part of a complex solution. But we are able to get rid of the complex part. We can rewrite this solution as:
y(t) = c1y1(t) + c2y2(t)
y1(t) + y2(t) = 2e^(λt) cos(μt)
Simply it further:
u(t) = ½y1(t) + ½y2(t) = e^(λt) cos(μt)
Therefore:
c1 = c2 = ½

We can get the second part of the solution by subtracting the two original solutions:
y1(t) – y2(t) = 2ie^(λt) sin(μt)

At a glance, it still look like a complex solution, but looking at the two constants c1 and c2 we can result in a real solution by dividing it by 2i.
c1 = ½i
c2 = -½i

The second solution would be:
v(t) = ½i y1(t) – ½i y2(t) = e^(λt) sin(μt)

All of the work just to achieve the two real solutions:
u(t) = e^(λt) cos(μt)
v(t) = e^(λt) sin(μt)

These two real solutions is also a general solution. So if the roots of the characteristic equation result in r1,2 = λ ± μi, the general solution to this would be:
y(t) = c1 e^(λt) cos(μt) + c2 e^(λt) sin(μt)

EXAMPLE
y” – 10y’ + 29y = 0, y(0) = 1, y'(0) = 3

The characteristic equation would be:
r² – 10r + 29 = 0

The roots would be:
5 ± 2i

General Solution:
y(t) = e^(5t)[c1 cos(2t) + c2 sin(2t)]
y'(t) = 5e^(5t)[cos(2t) + c2 sin(2t)] + e^(5t)[-2 sin(2t) + 2c2 cos(2t)]

We use the initial values to find the constants.
y(0) = 1
We substitute all the t’s with 0’s and set it to equal 1.
y(0) = e^(5(0))[c1 cos(2(0)) + c2 sin(2(0))] = 1
1[c1(1) + c2(0)] = 1
“c1 = 1

y'(0) = 3
We substitute all the t’s with 0’s and set it to equal 3.
y'(0) = 5e^(5(0))[cos(2(0)) + c2 sin(2(0))] + e^(5(0))[-2 sin(2(0)) + 2c2 cos(2(0))] = 3
5[1 + 0] + 1[0 + 2c2] = 3
5 + 2c2 = 3
c2 = -1

Particular Solution:
y = e^(5t)[cos(2t) – sin(2t)]

If you still do not understand this study guide you may go visit these websites or you can personally ask me, they go into more detail solving the differential equation of complex root solutions with characteristic equation. Thank you.

Videos from Khan Academy
Part I

Part II

Part III

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Second order ordinary homogeneous differential equations with constant coefficients

Posted on March 17, 2015 by | Leave a comment

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.

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