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Math Help - Two questions

  1. #1
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    Two questions

    I have some more questions.
    dy/dx = 1 + x - y - xy, solve the differential equation.

    I've been thinking how I can separate the variables before doing the integration, but I could not figure out how because of the xy. Please give me some hint to start! TQ =)

    Also, if you are kind enough, please help with this question
    (cosx)^2* dy/dx = y
    i'm stucked at integral of du / 1+ cos u = integral of dy / y. with 2x substituted as u.
    I do not know how to proceed from there.
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  2. #2
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    Quote Originally Posted by custer View Post
    I have some more questions.
    dy/dx = 1 + x - y - xy, solve the differential equation.

    I've been thinking how I can separate the variables before doing the integration, but I could not figure out how because of the xy. Please give me some hint to start! TQ =)

    [snip]
    The differential equation re-arranges into \frac{dy}{dx} + (x + 1) y = x + 1.

    This is solved using the integrating factor method.

    Quote Originally Posted by custer View Post
    [snip]
    Also, if you are kind enough, please help with this question
    (cosx)^2* dy/dx = y
    i'm stucked at integral of du / 1+ cos u = integral of dy / y. with 2x substituted as u.
    I do not know how to proceed from there.
    The differential equation re-arranges into \int \! \frac{1}{y} \, dy = \int \! \sec^2 x \, dx.

    Both integrals are trivial.
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  3. #3
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    sorry but I do not see how you get (sec x)^2
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  4. #4
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    Quote Originally Posted by custer View Post
    sorry but I do not see how you get (sec x)^2
    By definition, \frac{1}{\cos^2 x} = \sec^2 x. I wrote \sec^2 x rather than \frac{1}{\cos^2 x} because the former is how the derivative of \tan x is usually stated ....
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