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Math Help - first-order differential eq.

  1. #1
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    first-order differential eq.

    Solve the inital value problem for y(x):

     xy' + 8y = 3x^4

    with the initial condition:  y(1) = 4
    integrating factor would be e^{8x}

    \frac{d}{dx} (y* e^{8x}) = e^{8x}(3x^4)

    e^{8x}y = \int e(3x^4)^{8x}

    not sure if what im doing so far is right, having a hard time finishing it.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by viet View Post
    integrating factor would be e^{8x}

    \frac{d}{dx} (y* e^{8x}) = e^{8x}(3x^4)

    e^{8x}y = \int e(3x^4)^{8x}

    not sure if what im doing so far is right, having a hard time finishing it.
    The integrating factor is x^7

    multiply through by x^7

    \Rightarrow x^8 y' + 8x^7 y = 3x^{11}

    \Rightarrow \left( x^8 y \right)^{ \prime} = 3x^{11}

    Now continue
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by viet View Post
    integrating factor would be e^{8x}

    \frac{d}{dx} (y* e^{8x}) = e^{8x}(3x^4)

    e^{8x}y = \int e(3x^4)^{8x}

    not sure if what im doing so far is right, having a hard time finishing it.
    i will recheck this. i vaguely remember some method where you always use e^(some power) for the integrating factor, but i'm not sure if you did it correctly (i rarely remember formulas or routines verbatim, i just know how to work stuff out). i know the last line is wrong though, since 8x is the power of e, not 3x^4.

    if you want to know how i got my integrating factor, ask
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  4. #4
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    xy'+8y=3x^4 \mbox{ on }(0,\infty)

    Then,

    y' + \frac{8}{x} \cdot y = 3x^3

    The solution is given by,

    y = \frac{1}{I(x)}\cdot \left( \int I(x)Q(x) dx +C \right)

    Where I(x) = \exp \left( \int \frac{8}{x} dx \right) and Q(x) = 3x^3.
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  5. #5
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    ok i corrected this.

    the integrating factor i got is x^8 i got it by:
    xy' +8y = 3x^4

    y' + \frac{8}{x}y = 3x^3

    e^{{\int}\frac{8}{x}dx} = x^8

    multiply both sides by x^8
    x^9y' + 8x^8y = 3x^12

    \frac{d}{dx}(x^9y) = 3x^12

    x^9y = \int{3x^{12}}dx

    x^9y = \frac{3x^{13}}{13}+C

    y = (\frac{3x^{13}}{13}+C)x^{-9}

    y = \frac{3x^{4}}{13}+Cx^{-9}

    i guess now i just plug in y=1, x=4. to solve for C
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by viet View Post
    ok i corrected this.

    the integrating factor i got is x^8 i got it by:
    xy' +8y = 3x^4

    y' + \frac{8}{x}y = 3x^3

    e^{{\int}\frac{8}{x}dx} = x^8

    multiply both sides by x^8
    x^9y' + 8x^8y = 3x^12

    \frac{d}{dx}(x^9y) = 3x^12

    x^9y = \int{3x^{12}}dx

    x^9y = \frac{3x^{13}}{13}+C

    y = (\frac{3x^{13}}{13}+C)x^{-9}

    y = \frac{3x^{4}}{13}+Cx^{-9}

    i guess now i just plug in y=1, x=4. to solve for C
    when you multiply through by x^8 you should get what i got. don't multiply the original equation. multiply the one you simplified to when you dived through by x
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    when you multiply through by x^8 you should get what i got. don't multiply the original equation. multiply the one you simplified to when you dived through by x
    i continue on from what you got, and got:

     y = \frac{1}{4}x^4 + C{x^{-8}}
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by viet View Post
    i continue on from what you got, and got:

     y = \frac{1}{4}x^4 + C{x^{-8}}
    ok, so far. but don't forget the initial condition, they expect you to find C
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  9. #9
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    1 = 64 + C(4^{-8})

    -63 = C(\frac{1}{65536})

    -4128768 = C

    i think i did it wrong
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  10. #10
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    Quote Originally Posted by viet View Post
    1 = 64 + C(4^{-8})

    -63 = C(\frac{1}{65536})

    -4128768 = C

    i think i did it wrong
    yes you did

    it says y(1) = 4

    so y(1) = 4 = \frac {1}{4} (1)^4 + C(1)^{-8}

    \Rightarrow 4 = \frac {1}{4} + C

    Now continue
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  11. #11
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    thanks i feel pretty dumb.
    final answer is  y = \frac{1}{4}x^4 + \frac{15}{4}{x^{-8}}
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