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Dif. eq. problem - dont know why it is wrong

  1. Feb 11, 2006 #1
    dif. eq. problem -- dont know why it is wrong

    I did this problem three times with the same wrong answer...

    solve for y:
    [tex]y'''-9y''+18y'=0[/tex]
    y(0)=5
    y'(0)=7
    y''(0)=6

    so I change it into the auxiliary equation then solve for the variables, i get the roots: 3 and 6. so the general solution would be:

    [tex]y'=c_1e^{3x}+c_2e^{6x}[/tex]
    solving for c1 and c2 i get:

    [tex]7=c_1+c_2[/tex]
    [tex]2=c_1+2c_2[/tex]
    [tex]c_1=12[/tex]
    [tex]c_2=-5[/tex]

    [tex]y=\int{12e^{3x}-5e^{6x}}[/tex]
    [tex]y=36e^{3x}-30e^{6x}+c[/tex]

    solving for c when x=0 and y=5, i get

    [tex]y=36e^{3x}-30e^{6x}-1[/tex]

    what am i donig wrong?
     
  2. jcsd
  3. Feb 11, 2006 #2

    HallsofIvy

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    Hey! y'''- 9y"+ 18y'= 0 is a third order differential equation!
    Its auxiliary equation, r3- 9r2+ 18r= r(r2- 9r+ 18)= r(r- 3)(r- 6) has three roots! They are 0, 3, and 6.

    The general solution for y is y(x)= C1e3x+ C2e6x+ C3 (since e0x= e0= 1). Use that to satisfy the initial conditions:
    y(0)= C1+ C2+ C3= 5. y'(0)= 3C1+ 6C2= 7, and y"(0)= 9C1+ 36C2= 6.

    That's how I would have done it. What you did was treat it as a second order d.e. for y' and then integrate. You would have gotten exactly right answer except the anti-derivative of ea is (1/a)eax, not aeax!
     
  4. Feb 11, 2006 #3

    jamesrc

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    Double check your integration:

    [tex]\int 12 e^{3x}=12\frac{e^{3x}}{3}+c [/tex]

    and so on...
     
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