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Math Help - differential equation

  1. #1
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    differential equation

    solve  y'' - 3y' - 10 y = 0; \quad y = 1, y'=10 \;\textnormal{at}\; x = 0

     y(x) =


    heres what i did:
     r^2-3r-10 = 0

     (r-2) (r-5) = 0
    roots are : -2, 5

    General solution:  y = C_1 e^{-2x}+C_2 e^{5x}

     y' = -2C_1 e^{-2x}+5C_2 e^{5x}

    i need help applying the condition
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by viet View Post
    solve  y'' - 3y' - 10 y = 0; \quad y = 1, y'=10 \;\textnormal{at}\; x = 0

     y(x) =


    heres what i did:
     r^2-3r-10 = 0

     (r-2) (r-5) = 0
    roots are : -2, 5

    General solution:  y = C_1 e^{-2x}+C_2 e^{5x}

     y' = -2C_1 e^{-2x}+5C_2 e^{5x}

    i need help applying the condition
    The initial condition y(0)=1 gives:

    <br />
C_1+C_2=1<br />

    and the condition y'(0)=10 gives:

    <br />
-2~C_1+5~C_2=10<br />

    So now you have a pair of simultaneous equations for C_1 and C_2 to solve.

    RonL
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by viet View Post
    solve  y'' - 3y' - 10 y = 0; \quad y = 1, y'=10 \;\textnormal{at}\; x = 0

     y(x) =


    heres what i did:
     r^2-3r-10 = 0

     (r-2) (r-5) = 0
    roots are : -2, 5

    General solution:  y = C_1 e^{-2x}+C_2 e^{5x}

     y' = -2C_1 e^{-2x}+5C_2 e^{5x}

    i need help applying the condition
    For the record, the characteristic equation factors as
    (r + 2)(r - 5) = 0

    Given that you have the solution to the differential equation correct I would guess this is merely a typo?

    -Dan
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  4. #4
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    yeah it was a typo, thanks for pointing that out topsquark
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    The initial condition y(0)=1 gives:

    <br />
C_1+C_2=1<br />

    and the condition y'(0)=10 gives:

    <br />
-2~C_1+5~C_2=10<br />

    So now you have a pair of simultaneous equations for C_1 and C_2 to solve.

    RonL
    can you explain how to solve that? im pretty much lost.
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  6. #6
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    Quote Originally Posted by viet View Post
    can you explain how to solve that? im pretty much lost.
    This is just a linear system.

    Say,
    C_1+C_2=5
    C_1-C_2=1
    Add,
    2C_1=6\implies C_1=3
    And so on ....

    The same idea here.
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack View Post
    The initial condition y(0)=1 gives:

    <br />
C_1+C_2=1<br />

    and the condition y'(0)=10 gives:

    <br />
-2~C_1+5~C_2=10<br />

    So now you have a pair of simultaneous equations for C_1 and C_2 to solve.

    RonL
    To solve these multiply the firs equation by 2 to get:

    <br />
2~C_1+2~C_2=2<br />

    Now add the third equation to this (that is add the left hand sides, which
    sum is equal to the sum of the right hand sides):

    <br />
2~C_1+2~C_2-2~C_1+5~C_2=2+10<br />

    or:

    <br />
7~C_2=12<br />

    so:

    <br />
C_2=12/7<br />

    Now substitute this back into the first equation to get:

    <br />
C_1=-5/7<br />

    RonL
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