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Explinations needed, second degree linear equations with a constant

  1. Oct 5, 2011 #1

    C.S

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    Hello,

    I am in need of explinations on how to resolve second degree linear equations with a constant.
    I had a maths test last saturday and this was the last question :
    Resolve y"+3y'+2=cos(t)
    Having almost fallen into the trap of resolving y"+3y'+2y=cos(t), I passed the 2 over by cos(t) and tried to resolve y"+3y'=cos(t)-2, however I doubt that that's possible (maybe I'm wrong).
    Anyhow I am looking for some help in resolving this equation because I have the same equation given as a test tomorrow morning.

    Many thanks,

    Callum
     
  2. jcsd
  3. Oct 5, 2011 #2
    Do you know how to solve the homogeneous equation?
     
  4. Oct 5, 2011 #3

    C.S

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    Yes I do, y0(t)=C1.e-3t+C2 with (C1,C2) →ℝ
    I believe that's correct.
     
  5. Oct 5, 2011 #4
    Not quite. If y = ert, then working out y' and y'' gives you ert(r2+3r+2) = 0, so the roots -1 amd -2 give you two solutions. Remember, there are n solutions to an nth order ODE.

    Now you just need to find the particular solution. Have you learned that yet?
     
  6. Oct 5, 2011 #5

    C.S

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    I thought ay"+by'+cy=0 gave us ar^2+br+c=0, in my case c=0.
    I do know how to find particular solutions.
    I'm going back into class, I will check back later. Thank you for your help.
     
  7. Oct 5, 2011 #6
    Oops! You're right. I read it wrong as y'' + 3y' +2y. My bad.
     
  8. Oct 5, 2011 #7

    C.S

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    Most people in my class did the same, any ideas though?
     
  9. Oct 5, 2011 #8
    You have the right idea - just find the particular solution for the -2 (should be easy to see it's a polynomial in t) and the cosine term.
     
  10. Oct 5, 2011 #9

    lurflurf

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    Homework Helper

    y"+3y'=-2+cos(t)
    clearly
    (-2+cos(t))'''+(-2+cos(t))'=0
    so
    y(5)+y(3)+3y=0
    or in operator form
    (D^2+3)y=-2+cos(t)
    (D^5+4D^3+3D)y=D(D^2+3)(D^2+1)y=0
    so the inhomogeneous problem reduces to a homogeneous problem.
    A solution to the new problem will be a solution to the old one.
     
  11. Oct 5, 2011 #10
    Hi,

    So you want to solve the second-order differential equation [itex]y''+3y'=\cos(t)-2[/itex]. The corresponding homogeneous equation is [itex]y''+3y'=0[/itex], the solution of which is [itex]C_1+C_2\exp(-3t)[/itex], where [itex]C_1[/itex] and [itex]C_2[/itex] are undetermined coefficients. The forcing function [itex]\cos(t)-2[/itex] can be expressed as the sum of two terms, [itex]f_1(t)=\cos(t)[/itex] and [itex]f_2(t)=2[/itex]. By the superposition principle, the particular solution [itex]y_p(t)[/itex] can be expressed as the sum of the particular solutions [itex]y_{p1}(t)[/itex] and [itex]y_{p2}(t)[/itex] corresponding to [itex]f_1(t)[/itex] and [itex]f_2(t)[/itex], respectively. The individual particular solutions can be found using the method of undetermined coefficients.

    Consider first [itex]y_{p1}(t)[/itex]. We choose a trial solution [itex]y_{t1}(t)=A_0\cos(t)+A_1\sin(t)[/itex], where [itex]A_0[/itex] and [itex]A_1[/itex] are undetermined coefficients. Filling in [itex]y_{t1}''+3y_{t1}'=\cos(t)[/itex] and comparing terms leads to the pair of simultaneous equations [itex]-A_0+3A_1=1[/itex] and [itex]-3A_0-A_1=0[/itex], the solution of which is [itex]A_0=-\frac{1}{10}[/itex] and [itex]A_1=\frac{3}{10}[/itex]. So [itex]y_{p1}(t)=-\frac{1}{10}\cos(t)+\frac{3}{10}\sin(t)[/itex].

    Next, consdier [itex]y_{p2}(t)[/itex]. The forcing function is a zeroth order polynomial. As explained by http://books.google.com/books?id=Pj...method of undetermined coefficients:&f=false", since the homogeneous equation lacks a damping term (i.e., one proportional to [itex]y[/itex]), we should choose as our trial solution a polynomial with degree one order higher than the forcing function, so we choose [itex]y_{t2}(t)=B_0+B_1t[/itex], with [itex]B_0[/itex] and [itex]B_1[/itex] undetermined coefficients. Filling in [itex]y_{t2}''+3y_{t2}'=-2[/itex] then leads to [itex]B_0=0[/itex] and [itex]B_1=-\frac{2}{3}[/itex]. So [itex]y_{p2}(t)=-\frac{2}{3}t[/itex].

    Combining the above results, I obtain [tex]y(t)=y_h(t)+y_{p1}(t)+y_{p2}(t)= C_1+C_2\exp(-3t)-\frac{1}{10}\cos(t)+\frac{3}{10}\sin(t)-\frac{2}{3}t[/tex] as the final general solution of the differential equation. I verified this solution with Maple.

    Hope this helps!

    Cheers, Kurt
     
    Last edited by a moderator: Apr 26, 2017
  12. Oct 5, 2011 #11

    C.S

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    Thanks for all your help, the maths teacher said it was a last add in the test so that no one could get it all right, no wonder especially in the time given.
    I will try to meditate all of that and see if I can do it again.
     
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