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Math Help - 2 first order d.e problems

  1. #1
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    2 first order d.e problems

    Urg, im really struggling with D.E...

    1.
    nvm i found what i was doing wrong in this one


    2.
    \frac{dy(sin x)}{dx} + ysecx = cos^2x
    I get an answer with a bunch of e's in it :S.
    Answer is :
    y = cosx + kcotx

    I have nooo idea about this ques :S

    Any help for either questions would be much appriciated
    Last edited by AshleyT; April 5th 2009 at 08:05 AM.
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  2. #2
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    Quote Originally Posted by AshleyT View Post
    Urg, im really struggling with D.E...

    1.
    \frac{dy}{dx} \frac{-3y}{x} = x
    I get:
    Where the integrating factor is -x^3
    Thus:
    \frac{d(-yx^3)}{dx} = -x^4
    And i get for an answer:
    y = \frac{x^2}{5} - \frac{c}{x^3}
    But the answer:
    y = kx^3 - x^2

    I don't get what im doing wrong :S

    And this one i get a completely wrong answer
    2.
    \frac{dy(sin x)}{dx} + ysecx = cos^2x
    I get an answer with a bunch of e's in it :S.
    Answer is :
    y = cosx + kcotx

    Any help for either questions would be much appriciated
    For the first, you are correct, teh integrating factor is x^{-3} but there's a mistake in what follows

    \frac{1}{x^3} \frac{dy}{dx} - 3 \frac{1}{x^4} = \frac{1}{x^2} so \frac{d}{dx}\left( \frac{y}{x^3}\right) = \frac{1}{x^2}.

    Integrating gives

    \frac{y}{x^3} = - \frac{1}{x} + k which leads to the answer.

    For the second problem, first put in standard form

    \frac{dy}{dx} + \frac{y}{\sin x \cos x} = \frac{\cos^2x}{\sin x}

    the integrating factor is \tan x so

    \frac{d}{dx} \left( \tan x \cdot y \right) = \cos x which you can integrate.
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  3. #3
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    Quote Originally Posted by danny arrigo View Post
    For the first, you are correct, teh integrating factor is x^{-3} but there's a mistake in what follows

    \frac{1}{x^3} \frac{dy}{dx} - 3 \frac{1}{x^4} = \frac{1}{x^2} so \frac{d}{dx}\left( \frac{y}{x^3}\right) = \frac{1}{x^2}.

    Integrating gives

    \frac{y}{x^3} = - \frac{1}{x} + k which leads to the answer.

    For the second problem, first put in standard form

    \frac{dy}{dx} + \frac{y}{\sin x \cos x} = \frac{\cos^2x}{\sin x}

    the integrating factor is \tan x so

    \frac{d}{dx} \left( \tan x \cdot y \right) = \cos x which you can integrate.
    Thankyou very much for your reply .
    However how do you achieve tan x?
    You integrate cosecxsecx right? I thought this was sec x?
    Then to e ^ sec x

    This must have been what i was doing wrong :S
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  4. #4
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    Quote Originally Posted by AshleyT View Post
    Thankyou very much for your reply .
    However how do you achieve tan x?
    You integrate cosecxsecx right? I thought this was sec x?
    Then to e ^ sec x

    This must have been what i was doing wrong :S
    For your integrating factor \int \sec x \csc x\, dx = \ln \tan x
    just to note \int \sec x \tan x\, dx = \sec x
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  5. #5
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    Quote Originally Posted by danny arrigo View Post
    For your integrating factor \int \sec x \csc x\, dx = \ln \tan x
    just to note \int \sec x \tan x\, dx = \sec x
    oic.
    Thank-you
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