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Differential equation problem

  1. Jun 7, 2005 #1
    Solve the following differential equation:

    [tex](D + 2)(D + 3)y = 4t + 5e^t; y(0)=4, y'(0)=5[/tex]

    I have the following as the answer is it correct?

    [tex]y=17e^{-2t} -13e^{-3t} + \frac{4}{11}t + \frac{5}{12}e^t[/tex]
  2. jcsd
  3. Jun 7, 2005 #2


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    You try plugging your answer back into the equation, to see if it works out right?
  4. Jun 7, 2005 #3


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    Nope.Here's what Maple gives as the general solution

    [tex]\frac{d^2 y}{dx^2}+5\frac{dy}{dx}+6y=4x+5e^x [/tex] Exact solution is :

    [tex] y\left( x\right) =\frac{2}{3}x-\frac{5}{9}+\frac{5}{12}e^x+C_1 e^{-3x}+C_2 e^{-2x} [/tex]

    I get some nasty looking coeff.

  5. Jun 7, 2005 #4
    i am way off then. by the way could matlab solve this problem?
  6. Jun 7, 2005 #5
    I'm not understanding what my professor calls an annihilator. what i have is

    this D.E.

    [tex](D + 2)(D + 3)y = 4t + 5e^t;y(0)=4,y'(0)=5[/tex]

    I need to find a differential opperator that will make the right side of this D.E.

    zero. So i think that this differential opperator (annihilator) should be this.


    So to solve this problem i need to solve this homogenous D.E. first:

    [tex](D + 2)(D + 3)y=0[/tex] and its solutions are this:

    [tex]y_c=C_1e^{-2t} + C_2e^{-3t}[/tex]

    now i need to find the other part which involves the differential opperator. so

    now i need to solve this part [tex]D^2(D-1)=0[/tex]

    which yields [tex] C_3 + tC_4 + C_5e^t[/tex]

    if all of the above is correct i need to find out what the arbitrary constants of

    [tex] C_3 + tC_4 + C_5e^t[/tex] are

    so what i do is say that this [tex] C_3 + tC_4 + C_5e^t[/tex] is equal to


    so now i have [tex]y_p= C_3 + tC_4 + C_5e^t[/tex]

    now i find [tex]y'_p[/tex] and [tex]y''_p[/tex]

    [tex]y'_p= C_4 +C_5e^t [/tex]


    now what i do is plug in all the y-sub-p's into this [tex]y'' + 5y' +6y= 4t +5e^t[/tex]

    and match up the the coef. to find out what the constants are

    what i get is this

    [tex]12C_5e^t + 5C_4 + 6tC_4 + 6C_3= 4t + 5e^t[/tex]

    now my problem is that the C-sub-4's are not the same. one of them is

    multiplied by t which is a problem because i can't get the coef. to match

    what have i done wrong?
  7. Jun 7, 2005 #6


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    You have this system

    [tex] \left\{\begin{array}{c} 5C_{4}+6C_{3}=0\\6C_{4}=4 \end{array} \right [/tex]

    Solve it.

  8. Jun 7, 2005 #7

    i don't see why could you elaborate?
  9. Jun 7, 2005 #8


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    That should read

    [tex] 12C_{5} e^{t}+6C_{4}t+\left(5C_{4}+6C_{3}\right)\equiv 5e^{t}+4t+0 [/tex]

    Do you see where the system i posted comes from...?

  10. Jun 7, 2005 #9
    ahh!! yes i see now. i made a simple mistake, but to be truthful i did not know that i could combine csub4 and csub3 but now i see why i can.
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