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MathSofiya

  • 2 years ago

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

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  1. MathSofiya
    • 2 years ago
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    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

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

  3. amistre64
    • 2 years ago
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    assume e^rx is a solution; there is an even baser steup that is gone thru that gets you to this part

  4. MathSofiya
    • 2 years ago
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    but they started with \[y'-9y=0\] where did the e come from?

  5. MathSofiya
    • 2 years ago
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    In Paul's example I mean

  6. amistre64
    • 2 years ago
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    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

  7. akitav
    • 2 years ago
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    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 \[y = e^{3x} \]

  8. MathSofiya
    • 2 years ago
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    @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 \[y'-9y=0\]

  9. MathSofiya
    • 2 years ago
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    It's a new idea I guess and my brain is being difficult

  10. amistre64
    • 2 years ago
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    there is a real analysis step that can get us to using e^rx for solutions

  11. MathSofiya
    • 2 years ago
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    let's see what it is

  12. amistre64
    • 2 years ago
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    its been awhile since i had to convince myself of it, so lets see what the google has in store :)

  13. MathSofiya
    • 2 years ago
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    sure :)

  14. amistre64
    • 2 years ago
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    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

  15. MathSofiya
    • 2 years ago
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    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... \[P(x)\frac{d^2y}{dx^2}+Q(x)\frac{dy}{dx}+R(x)y=c_1y_1(x)+c_2y_2(x)\]

  16. amistre64
    • 2 years ago
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    Thats just generalizing the setup to include functions as well as constants

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

  18. amistre64
    • 2 years ago
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    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

  19. MathSofiya
    • 2 years ago
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    Oh that's what it means

  20. amistre64
    • 2 years ago
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    im gonna run upstairs to check out some books for the real analysis part. internet explorer is being bad

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

  22. amistre64
    • 2 years ago
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    in a sense, yes, but x^m can also be used; and there are means of solutions by using power series as well

  23. MathSofiya
    • 2 years ago
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    ok so \[y=e^{xr}\] is a solution for \[y''+4y'+4y=0\] I just need to manipulate it?

  24. MathSofiya
    • 2 years ago
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    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 ... \[y=c_1e^{-2x}+c_2e^{-2x}\]

  25. MathSofiya
    • 2 years ago
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    I think I got it!!!!!

  26. v4xN0s
    • 2 years ago
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    off topic: WTF u guys r on second order diff EQ alredy?? its only been a week

  27. MathSofiya
    • 2 years ago
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    No it's my online class that I'm finishing this summer. School starts next week for me @v4xN0s

  28. MathSofiya
    • 2 years ago
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    So my final answer is correct right? Or do I need to do more?

  29. amistre64
    • 2 years ago
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    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\[y(x)=c_1e^{2x}+c_2~xe^{2x}\]

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

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