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Math Help - Finding the complementary equation of ...

  1. #1
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    Exclamation Finding the complementary equation of ...

    L[y]=cy''+(2-c)y'-y=e^c

    I'm not sure how to approach this equation. I've already found the particular solution for this, which is Y(t)=e^c

    I tried to find the complementary solution for it by using the quadratic equation and trying to find the required numbers that way, but I've been told that that method doesn't work. But here's the work i've done so far:

    Work: tu^2+(2-t)u-1=0
    tu^2-tu+2u-1=0

    (let t=c)

    when t=/0, u=+-(an ugly imaginary no.)

    when t=0, 2u-1=0
    u=1/2

    thus, com. function is y=c1e^1/2t+c2e^1/2t




    I've been told that this is wrong. I'm still really confused as to what to do...

    EDIT: e^c not e^t
    Last edited by TheFirstOrder; March 26th 2010 at 10:22 PM.
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  2. #2
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    Quote Originally Posted by TheFirstOrder View Post
    cy''+(2-c)y'-y=e^t

    I'm not sure how to approach this equation. I've already found the particular solution for this, which is Y(t)=e^t.

    I tried to find the complementary solution for it by using the quadratic equation and trying to find the required numbers that way, but I've been told that that method doesn't work. But here's the work i've done so far:

    Work: tu^2+(2-t)u-1=0
    tu^2-tu+2u-1=0

    when t=/0, u=+-(an ugly imaginary no.)

    when t=0, 2u-1=0
    u=1/2

    thus, com. function is y=c1e^1/2t+c2e^1/2t




    I've been told that this is wrong. I'm still really confused as to what to do...
    Dear TheFirstOrder,

    cy''+(2-c)y'-y=e^t

    As you have found out e^{t} is a particular integral.

    Also, u=\frac{\sqrt{(-c+2)^2+4 c}+c-2}{2 c}~or~u=\frac{-\sqrt{(-c+2)^2+4 c}+c-2}{2 c} ; where u is the complimentary function.

    Therefore, the general solution is,

    y=e^{y}+Ae^{\frac{\sqrt{(-c+2)^2+4 c}+c-2}{2 c}}+Be^{\frac{-\sqrt{(-c+2)^2+4 c}+c-2)}{2 c}} ; where A and B are arbitary constants.

    Hope this will help you.
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  3. #3
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    [quote=Sudharaka;481123]Dear TheFirstOrder,

    cy''+(2-c)y'-y=e^t

    As you have found out e^{t} is a particular integral.

    Also, u=\frac{\sqrt{(-c+2)^2+4 c}+c-2}{2 c}~or~u=\frac{-\sqrt{(-c+2)^2+4 c}+c-2}{2 c} ; where u is the complimentary function.

    Therefore, the general solution is,

    y=e^{y}+Ae^{\frac{\sqrt{(-c+2)^2+4 c}+c-2}{2 c}}+Be^{\frac{-\sqrt{(-c+2)^2+4 c}+c-2)}{2 c}} ; where A and B are arbitary constants.

    Hope this will help you.[/quote)

    But can we use this method? (the coefficients are not constant...)
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  4. #4
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    Quote Originally Posted by TheFirstOrder View Post
    cy''+(2-c)y'-y=e^t

    I'm not sure how to approach this equation. I've already found the particular solution for this, which is Y(t)=e^t.

    I tried to find the complementary solution for it by using the quadratic equation and trying to find the required numbers that way, but I've been told that that method doesn't work. But here's the work i've done so far:

    Work: tu^2+(2-t)u-1=0
    tu^2-tu+2u-1=0

    when t=/0, u=+-(an ugly imaginary no.)

    when t=0, 2u-1=0
    u=1/2

    thus, com. function is y=c1e^1/2t+c2e^1/2t




    I've been told that this is wrong. I'm still really confused as to what to do...
    Is c a constant, or a function of x or y?

    If it is a constant, you can use this method.

    If not, you will need another way.
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