Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

MathSofiya Group Title

Big Round of Applause!!!! second-order linear equations \[y''+4y'+4y=0\] solve the differential equation

  • one year ago
  • one year ago

  • This Question is Closed
  1. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I'm looking at http://tutorial.math.lamar.edu/Classes/DE/SecondOrderConcepts.aspx I don't understand how they came to the e^3t as part of their solution

    • one year ago
  2. akitav Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    when \[y=e^{3x}\]\[\frac{dy}{dx} = 3e^{3x}\]\[\frac{d^2y}{dx^2}= 9e^{3x}\]This equation works similarly works for the function \[y=e^{-3x}\]

    • one year ago
  3. amistre64 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    assume e^rx is a solution; there is an even baser steup that is gone thru that gets you to this part

    • one year ago
  4. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    but they started with \[y'-9y=0\] where did the e come from?

    • one year ago
  5. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    In Paul's example I mean

    • one year ago
  6. amistre64 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    y = e^rx y' = r e^rx y'' = r^2 e^rx fill it in r^2 e^rx +4r e^rx +4 e^rx =0 factor out the e^rx e^rx (r^2 +4r +4) =0 e^rx is never zero, so that leaves us with the quadratic to solve for zeros, solve for r

    • one year ago
  7. akitav Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    From inspection, e^x is a basic function that yields itself as a result of differentiation. So we can directly use this property to assume that it would form a part of the solution to the equation that Paul is discussing. But since we need the function to give 9 times itself after two successive differentiation he used \[y = e^{3x} \]

    • one year ago
  8. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    @amistre64 Is what you have written what Paul did, or is it in response to my question? I'm sorry I'm not trying to be difficult. This is my first DE problem and this is what I'm seeing: \[y'-9y=0\rightarrow y=e^{3x}\] of course with the intermediate steps shown. Lets' just say that the differentiation for y makes sense, but I wouldn't come to think of e^{whatever} when looking at \[y'-9y=0\]

    • one year ago
  9. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    It's a new idea I guess and my brain is being difficult

    • one year ago
  10. amistre64 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    there is a real analysis step that can get us to using e^rx for solutions

    • one year ago
  11. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    let's see what it is

    • one year ago
  12. amistre64 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    its been awhile since i had to convince myself of it, so lets see what the google has in store :)

    • one year ago
  13. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    sure :)

    • one year ago
  14. amistre64 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    this gives the handwaving that i did, but it might be a better read than what i gave :) still looking for the other stuff http://www.stewartcalculus.com/data/CALCULUS%20Concepts%20and%20Contexts/upfiles/3c3-2ndOrderLinearEqns_Stu.pdf

    • one year ago
  15. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    uhm... Can you tell me what the relationship between \[P(x)\frac{d^2y}{dx^2}+Q(x)\frac{dy}{dx}+R(x)y=0\] and \[y(x)=c_1y_1(x)+c_2y_2(x)\] is? It says that they are the solution for the linear homogeneous equation, so... \[P(x)\frac{d^2y}{dx^2}+Q(x)\frac{dy}{dx}+R(x)y=c_1y_1(x)+c_2y_2(x)\]

    • one year ago
  16. amistre64 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Thats just generalizing the setup to include functions as well as constants

    • one year ago
  17. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    would this statement be correct? \[P(x)\frac{d^2y}{dx^2}+Q(x)\frac{dy}{dx}+R(x)y=c_1y_1(x)+c_2y_2(x)\]

    • one year ago
  18. amistre64 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    second order diffy qs have at most 2 solutions that need to be accounted for; and any linear combination of those solutions is also a solution

    • one year ago
  19. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh that's what it means

    • one year ago
  20. amistre64 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    im gonna run upstairs to check out some books for the real analysis part. internet explorer is being bad

    • one year ago
  21. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Sure, sounds good. Ok should I just accept that \[y=e^{rx}\] is always a solution?

    • one year ago
  22. amistre64 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    in a sense, yes, but x^m can also be used; and there are means of solutions by using power series as well

    • one year ago
  23. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    ok so \[y=e^{xr}\] is a solution for \[y''+4y'+4y=0\] I just need to manipulate it?

    • one year ago
  24. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    ok lets see here, the auxilary equation is \[r^2+4r+4=0\] that factors into \[(r+2)(r+2)=0\] so r=-2 and r=-2 and therefore ... \[y=c_1e^{-2x}+c_2e^{-2x}\]

    • one year ago
  25. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I think I got it!!!!!

    • one year ago
  26. v4xN0s Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    off topic: WTF u guys r on second order diff EQ alredy?? its only been a week

    • one year ago
  27. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    No it's my online class that I'm finishing this summer. School starts next week for me @v4xN0s

    • one year ago
  28. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    So my final answer is correct right? Or do I need to do more?

    • one year ago
  29. amistre64 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    your solution is "almost" correct notice that the "r" you found for the root of the characteristic equation (also called the auxillary equation) is just double. 2 and 2.\[y_1(x)=c_1e^{2x}\] and all of its constant c1s are a solution; the second y(x) we need to make sure we have ALL the solutions of a "second" degree diffyQ is to simply include an extra "x" in there.\[y_2(x)=c_2~xe^{2x}\] add the two of them together for a complete set of linear combinations of solutions\[y(x)=c_1e^{2x}+c_2~xe^{2x}\]

    • one year ago
  30. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @amistre64 is "almost" correct too... \[y'' + 4y' + 4y = 0\] \[\implies r^2 + 4r + 4 = 0\] \[\implies (r+2)(r+2) = 0\] \[\implies r = -2 \; (2\times)\] \[\large \implies y = y_c = c_1 e^{-2x} + c_2 xe^{-2x}\]

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.