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Math Help - Silly De questions to muck up my brain!

  1. #1
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    Silly De questions to muck up my brain!

    I have two questions, trying to learn to do DE's using substitution....my problems ussally come out looking like horrbile bog monsters of intergation (like a parts of x^9 (e^x)) i could intergrate it but id like to keep my work to under 3 pages .....anyway:

    i have a (dy/dx) = cos (x+y)

    and

    dy/dx - y/x = y^3

    also i have one wth the dentotation "any appropriate method" and i trie excat and came out with the above explained bog monster....

    it starts ...(x(dy/dx))-4y=x^6(e^x)


    Thanks agian all, any help would be appreciated....
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by neven87 View Post
    I have two questions, trying to learn to do DE's using substitution....my problems ussally come out looking like horrbile bog monsters of intergation (like a parts of x^9 (e^x)) i could intergrate it but id like to keep my work to under 3 pages .....anyway:

    i have a (dy/dx) = cos (x+y)
    Put u=x+y, then du/dx=1+dy/dx so the original equation becomes:

    du/dx -1 = cos(u),

    which is of variables seperable type:

    integral 1/(cos(u) +1) du = integral dx

    RonL
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  3. #3
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    Hello, neven87!

    The last one can be solved with an integrating factor.


    x\frac{dy}{dx} - 4y \:=\:x^6e^x

    Divide by x\!:\;\;\frac{dy}{dx} - \frac{4}{x}y \:=\:x^5e^x

    Then: . I \;=\;e^{-4\int\frac{dx}{x}} \;=\;e^{-4\ln x} \;=\;e^{\ln(x^{-4})} \;=\;x^{-4}\;=\;\frac{1}{x^4}

    Multiply by I\!:\;\;\frac{1}{x^4}\frac{dy}{dx} - \frac{4}{x^5} \;=\;xe^x

    And we have: . \frac{d}{dx}\left(\frac{1}{x^4}y\right) \;=\;xe^x

    Integrate: . \int \frac{d}{dx}\left(\frac{1}{x^4}y\right) \;=\;\underbrace{\int xe^x\,dx}_{\text{by parts}}

    And we have: . \frac{1}{x^4}y \;=\;e^x(x - 1) + C

    Therefore: . y \;=\;e^xx^4(x-1) + Cx^4

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