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Thread: First order ode

  1. #1
    Sep 2013

    First order ode

    For the following diferential equation

    dy/dx = (2 co2x - sin2x + y2)/ 2 cos x , -pi/2<x<pi/2

    show that the substitution y(x) = sin x + (1/u(x)) yields the deferential equation for u(x),

    du/dx = -u tanx - (1/2) sec x

    Hence fi nd the solution y(x) to the original deferential equation that satisfies the condition
    y(0) = 2. Find the interval on which the solution to the initial value problem is defi ned.

    Im having trouble doing this question. I have had about 4 attempts and can not seem to get to the du/dx they have. I'm solving the y= sinx +1/u to give dy/du= cosx -u-2
    and then letting that = the original dy/dx. Is this the way to go about it or am i missing something? i also rearrange the substitution so u= 1/(y-sin x).
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  2. #2
    Junior Member
    May 2013

    Re: First order ode

    I just realised that we have the exactly same question lol

    Question regards integral calculus

    but I think our method is different, I'm working backward by integrate $\displaystyle \frac{dy}{dx}$ to give me the equation for y, then I also integrate $\displaystyle \frac{du}{dx}$ to get u
    so i can sub them into $\displaystyle y(x)=sin(x)+\frac{-1}{u(x)}$

    look at my link "Prove it" made a reply there, which made sense
    Last edited by junkwisch; Sep 21st 2013 at 07:46 PM.
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