Tag Archives: advice for the future

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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Study Guide to Bernoulli Equations

Posted on April 30, 2015 by | 1 comment

Overview

What are Bernoulli Equations?

Well to put it simply Bernoulli Equations are first order differential equations. What sets Bernoulli Equations apart from other first order differential equations is that they are nonlinear first order differential equation. When we was first introduced to first order differential equations we learned that the standard form  was :

y’ +p(t)y = g(t) ,  y(to) = yo

What separates Bernoulli Equations from other first order equations is that in standard form, it is not equal to some function that is linear but one that has an exact solution. What this means is that their is some power that is raised to the right side of the equation which we shall call n and n cannot be equal to 0 or 1. this is what makes Bernoulli Equations nonlinear. Now we can change the form of a standard first order linear equation into a nonlinear first order equation:

y’ +p(t)y = q(t)y^(n)  y(to)=yo

The above equation is now the standard form for a Bernoulli equation. What we can now understand from this equation is that  p(t) and q(t) are functions that are continuous. Since n is a real number that has to be n> 0 and n>1 for this  to be non linear.

Now the question is how do we solve Bernoulli Equations?

Sample Problem:

take an equation such as t^(2)y’ + 2ty – y^(3) = 0

and change it by adding y^(3) on both sides and we get:

t^(2)y’ + 2ty = y^(3)

Next we divide by t^(2) to on both sides of the equation so we can get it into standard form.

y’ + 2/yt = y^(3)/t^(2)       <————-STANDARD FORM

Now we can go into the steps to solve this equation;

1. Divide by y^(n)

y’/y^(3) + 2/yt/y^(3) =  (y^(3)/t^(2))/y^(3)

we get: (y’ + 2/yt)/y^(3) = 1/t^(2)

2. We substitute v = y^(1-n)

Since y^(3)

we get:v = y^(1-3) = y^(-2)

now we can rearrange the formula to so we can substitute v into it by moving the y^(2) to the numerator and we get:

y’/y^(3) + 2y^(-2)/t = 1/t^(2)

y’/y^(3) + 2v/t = 1/t^(2) <——————-Substitute v for y^(2)

Before we move to the solve Step we need to take the Derivative of the substitution v = y^(-2)

d/dy y^(-2)

dv/dt = -2y^(-3) dy/dt

(-1/2)v’ = -2y’/y^(3)(-1/2)

Now we can substitute for y’/y^(3) with the above equations and we get:

(-1/2)v’ + (2/t)v = 1/t^(2) <—————– First order linear equation

next we use the method of intergrating factor which states

mu = e^ intergral of (-4/t) which is equal to 1/t^(4)

we then multiply both sides b\by Mu and get the equation:

(dv(t)/dt)/t^(4) + d/dt(1/t^(4))v = (-2/t^(6))

Next we intergrate both sides and we come to the final answer of

y(t) = +- Sqrt(5t)/sqrt(ct^(5) + 2)

Videos:

https://www.youtube.com/watch?v=q4eUQsURq4k

 

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Solving Repeated Roots

Posted on April 29, 2015 by | Leave a comment

Once you know how to solve second order linear homogeneous differential equations with constant coefficients, real or complex, the next step is to solve with those that have repeated roots. When solving for repeated roots, you could either factor the polynomial or use the quadratic equation, if the solution has a repeated root it means that the two solutions for “x” or whatever variable are the same.

Theorem for Solving Repeated Roots

Let:   ay” + by’ + cy = 0

Be a differential equation such that the characterstic equation has the repeated root “r” That is :

(b^2)-4ac = 0

Then the general solution to the differential equation is given by

y  =  c1ert + c2tert 

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OpenLab #1: Advice from the Past

Posted on February 10, 2015 by | 67 comments

Last Spring I taught this same course for the first time.   At the end of the semester, I gave my students the following assignment:

Imagine that you are invited to speak on the first day of MAT 2680, to give advice to entering students. Write at least three sentences … describing what you would tell them.

To see the assignment and the students’ responses, follow this link.

Your assignment, due on Thursday 2/19 at the start of class, is to:

  1. Read through ALL the responses (there are 53 of them – many are quite short).
  2. Write a reply to this post (150 words minimum, not including quotations) responding to all of the following:
    1. What advice seemed most relevant to you personally? Why? (you can copy/paste a short statement, or put it in your own words)
    2. Based on this advice, what changes can you make right now to help you with this course?

Extra Credit. For extra credit, write a response to one of your classmates’ comments.  Do you have any advice?  Be kind.

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