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Initial value problem D.E. (confused of a weird sol.)

  1. Dec 13, 2008 #1
    1. The problem statement, all variables and given/known data
    solve:

    y" + 4y' + 4y = (3 + x)e-2x ... (1); y(0) = 2 & y'(0) = 5

    2. Relevant equations
    using Undetermined Coefficients Method


    3. The attempt at a solution

    solving the associated homog. eq of (1) : y" + 4y' + 4y = 0 ... (2)

    we get: yc = c1 e-2x as m=-2

    let f(x) = (3 + x)e-2x

    the UC set of :

    a) 3e-2x = {e-2x} = S1

    b) xe-2x = {xe-2x , e-2x} = S2

    as S1 is included in S2 then we omit S1

    but e-2x is a solution of (2) as its in yc then we multiply the 2ns set by x to get:

    S'2 = {x2 e-2x, xe-2x } where both of the elements are NOT sol.'s of (2)

    hence we get yp = Axe-2x + Bx2 e-2x

    y'p = Ae-2x -2Axe-2x + 2Bxe-2x - 2Bx2 e-2x

    & y"p = -4Ae-2x + 4Axe-2x + 2Be-2x - 8Bxe-2x +4Bx2 e-2x

    substitution in (1) one gets:

    2Be-2x = (3+x)e-2x so B = 3+x/2 ? ??

    where is A? why is B interms of x?

    plz help
    thx
     
  2. jcsd
  3. Dec 13, 2008 #2

    gabbagabbahey

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    Homework Helper
    Gold Member

    Sure, m=-2 but with what multiplicity?:wink:...Remember, your homog DE is second order, so you expect two linearly independent solutions. You seem to only be using one of them.

    This will also affect y_p.
     
  4. Dec 13, 2008 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    This is a second order equation. You need TWO independent solutions to form the general solution. Since m= -2 is a double root of the characteristic equation, xe-x is the other, independent, solution.

    A has disappeared and you can't solve for the constant B because both Ae-2x and Bxe-2x are already solutions to the homogeneous equation. In order not to have either e-2x or xe-2x you will need to multiply by x2.

    Try y= (Ax2+ Bx3) e-2x and solve for A and B.
     
  5. Dec 13, 2008 #4
    thx. i had in mind that i might have done a mistake in writing y_c but didn't bother to correct it. now i see how it caused the 2nd problem.
    now after resolving it i got A =3/2 & B = 1/6

    thus we now have:

    y = c1 e-2x + c2 xe-2x + (3/2)x2 e-2x + (1/6)x3 e-2x

    solving for the conditions i got:

    c1 = 2 & c2 = 9

    thus y becomes:

    y = 2e-2x + 9xe-2x + (3/2)x2 e-2x + (1/6)x3 e-2x


    i think this is the general solution. thx 4 the help
     
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