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Math Help - Question regards integral calculus

  1. #1
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    Question regards integral calculus

    Hi its me again and I have return with more question that I currently got stuck on
    ************************************************** *********

    For the following differential equation

    \frac{dy}{dx}=\frac{2cos^2(x)-sin^2(x)+y^2}{2cos(x)} , \frac{-\pi}{2}<x<\frac{\pi}{2}

    show that the substitution y(x)=sin(x)+\frac{1}{u(x)} , yields the differential equation for u(x)

    \frac{du}{dx}=-u*tan(x)-\frac{1}{2}sec(x)

    Hence find the solution y(x) to the original differential equation that satisfies the condition y(0)=2. Find the interval on which the solution to the initial value problem is defined.

    ************************************************** **********
    So I first set out to integrate \frac{dy}{dx} first, however I have a problem that I do not know how to move y^2 to the otherside of the equation

    e.g. \frac{1}{y^2}dy=\frac{2cos^2(x)-sin^2(x)}{2cos(x)}dx


    what I did is

    \int 1 dy = \int cos(x)dx -\frac{1}{2}\int sin(x)tan(x)dx + \frac{y^2}{2}\int\frac{1}{cos(x)}dx


    I integrated \int sin(x)tan(x)dx by using integration by parts with u=tan(x) and \frac{dv}{dx}=sin(x)

    so the equation become

     y=sin(x)+\frac{cos(x)tan(x)}{2}-\frac{sin(x)}{2}+\frac{y^2}{2sin(x)} + C

    I'm certain that my integration is wrong, since y(0)=2 will result in \frac{y^2}{0}


    The same problem also applies to \frac{du}{dx}=-u*tan(x)-\frac{1}{2}sec(x)

    since I do not know how to move, the y and u to the other side of the equation. So can any body tell me how to integrate this question?

    Thanks
    Junks

    P.S. I finally posted this in latex


    edit: I'll try to use First Order, O.D.E but since y^2will first order O,D,E work there?

    By first order O.D. E

    \frac{du}{dx}=-u*tan(x)-\frac{1}{2}sec(x)

    I found my u to be  u=\frac{-1}{2}cos(x)tan(x)

    is this correct?
    Last edited by junkwisch; September 21st 2013 at 07:39 PM.
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  2. #2
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    Re: Question regards integral calculus

    Well when you make the substitution \displaystyle \begin{align*} y = \sin{(x)} + \frac{1}{u} \end{align*} then that means \displaystyle \begin{align*} \frac{dy}{dx} = \cos{(x)} - \frac{1}{u^2}\,\frac{du}{dx} \end{align*}, so substituting into your DE you get

    \displaystyle \begin{align*} \frac{dy}{dx} &= \frac{2\cos^2{(x)} - \sin^2{(x)} + y^2}{2\cos{(x)}} \\ \cos{(x)} - \frac{1}{u^2}\,\frac{du}{dx} &= \frac{2\cos^2{(x)} - \sin^2{(x)} + \left[ \sin{(x)} + \frac{1}{u} \right] ^2 }{2\cos{(x)}} \\ \cos{(x)} - \frac{1}{u^2}\,\frac{du}{dx} &= \frac{2\cos^2{(x)} - \sin^2{(x)} + \sin^2{(x)} + \frac{2}{u}\sin{(x)} + \frac{1}{u^2}}{2\cos{(x)}} \\ \cos{(x)} - \frac{1}{u^2}\,\frac{du}{dx} &= \frac{2\cos^2{(x)} + \frac{2}{u}\sin{(x)} + \frac{1}{u^2}}{2\cos{(x)}} \\ \cos{(x)} - \frac{1}{u^2}\,\frac{du}{dx} &= \cos{(x)} + \frac{1}{u}\,\frac{\sin{(x)}}{\cos{(x)}} + \frac{1}{u^2}\,\frac{1}{2\cos{(x)}} \\ -\frac{1}{u^2}\,\frac{du}{dx} &= \frac{1}{u}\,\frac{\sin{(x)}}{\cos{(x)}} + \frac{1}{u^2}\,\frac{1}{2\cos{(x)}} \\ \frac{du}{dx} &= -u\,\frac{\sin{(x)}}{\cos{(x)}} - \frac{2}{\cos{(x)}} \\ \frac{du}{dx} + \frac{\sin{(x)}}{\cos{(x)}}\,u &= -\frac{2}{\cos{(x)}} \end{align*}

    This is now first order linear and so can be solved using an integrating factor.
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  3. #3
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    Re: Question regards integral calculus

    Thank you, "Prove it" I got the idea of solving this problem now
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  4. #4
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    Re: Question regards integral calculus

    I have more question to ask in regards to this same question.

    so I got u=\frac{-1}{2}cos(x)tan(x)

    since y(x)=sin(x)+\frac{1}{u} then y(x)=sin(x)-\frac{-2}{cos(x)tan(x)}

    I subbed this value of y into  \frac{dy}{dx}=\frac{2cos^2(x)-sin^2(x)+y^2}{2cos(x)}

    the equation become \frac{dy}{dx}=cos(x)-\frac{sin^2(x)}{2cos(x)}+\frac{sin^2(x)+\frac{4}{s  in^2(x)} -4}{2cos(x)}

    \int cos(x) + \frac{4}{sin^2(x)2cos(x)}-\frac{2}{cos(x)} dx

    If I did not made any mistake in the process, how do you integrate this equation  \int\frac{4}{sin^2(x)2cos(x)} dx?

    ************************************************** *******************************************
    I think I need to integrate \frac{dy}{dx} so i can solve for this

    "Hence find the solution to the original differential equation that satisfies the condition . Find the interval on which the solution to the initial value problem is defined."
    Last edited by junkwisch; September 21st 2013 at 10:08 PM.
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  5. #5
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    Re: Question regards integral calculus

    Quote Originally Posted by junkwisch View Post
    If I did not made any mistake in the process, how do you integrate this equation  \int\frac{4}{sin^2(x)2cos(x)} dx?

    ************************************************** *******************************************
    I think I need to integrate \frac{dy}{dx} so i can solve for this

    "Hence find the solution to the original differential equation that satisfies the condition . Find the interval on which the solution to the initial value problem is defined."
    I haven't checked whether your working is correct, but use integration by parts to compute the above integral.
    \int u' v \ dx = uv - \int uv' \ dx

    2 \int \csc^2 x \sec x \ dx, \ u' = \csc^2 x, \ v = \sec x.

    Recall \int \csc^2 x \ dx = - \cot x, \ \frac{d}{dx} \sec x = \sec x \tan x.
    Last edited by FelixFelicis28; September 22nd 2013 at 06:05 AM.
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  6. #6
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    Re: Question regards integral calculus

    It should be du/dx = - tan x - sec x/2

    not the du/dx = -u tanx - 2sec x given above.

    you then take the onto the LHS and you now have a linear first order ODE.

    integrating factor is sec x.

    get down to u = -sinx/2 + C cos x

    from y= sinx +1/u, u = 1/(y-sinx)

    Im not sure where to go from here, can anyone help solving for y?
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  7. #7
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    Re: Question regards integral calculus

    I guess we didnt have exactly the same question

    Im not sure if what i did is right, but try sub u into y then y into dy/dx



    P.s. writing on my phone right now so I couldnt write a latex
    Last edited by junkwisch; September 22nd 2013 at 03:52 PM.
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