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Math Help - cos^2(t)y'+4y=5

  1. #1
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    cos^2(t)y'+4y=5

    I'm kind of stuck on this problem. The problem given is cos2(t)y' + 4y = 5 , where y(0) = 6

    So what I got are no where near the choices give.

    y' + \frac{4y}{cos^2(t)} = 5cos^2(t)
    \frac{dy}{dt} + \frac{4y}{cos^2(t)} = 5cos^2(t)
    e^{\int\frac{4}{cos^2(t)}dt} = e^{4tan(t)}
    e^{4tan(t)} \frac{dy}{dt} + e^{4tan(t)}\frac{4}{cos^2(t)} y = 5cos^2(t)
    \int \frac{d}{dt} e^{4tan(t)}y = \int 5cos^2(t)
    e^{4tan(t)} y = \frac{\frac{5}{2}(t+sin(t)cos(t))+c}{e^{4tan(t)}}

    When I Plug in y(0)=6, I get C = 6....

    The choices for the answers are
    y(t) = 4+2etan(t)
    y(t) = \frac{5}{4}+\frac{19}{4}e^{-4tan(t)}
    y(t) = \frac{5}{4}+\frac{19}{4}e^{-tan(t)}
    y(t) = 4+2e-tan(t)
    Last edited by JC05; February 14th 2014 at 10:05 PM.
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  2. #2
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    Re: cos^2(t)y'+4y=5

    which is it?

    4t or 4y?
    Last edited by romsek; February 14th 2014 at 09:55 PM.
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  3. #3
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    Re: cos^2(t)y'+4y=5

    ok nm it's clearly 4y

    \cos^2(t)y'+4y=5

    I'm seeing the answer is y(t)=\frac{1}{4} e^{-4 \tan (t)} \left(5 e^{4 \tan (t)}+19\right)

    but I have no idea yet how that is obtained. Maybe using an integrating factor. One of the serious math guys here can probably shed some light on it.
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  4. #4
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    Re: cos^2(t)y'+4y=5

    Yes, it's an integrating factor type, and that attempt in #1 is along the right lines except that it should be 5sec^2 (t) on the rhs.
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  5. #5
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    Re: cos^2(t)y'+4y=5

    Quote Originally Posted by JC05 View Post
    I'm kind of stuck on this problem. The problem given is cos2(t)y' + 4y = 5 , where y(0) = 6

    So what I got are no where near the choices give.
    \cos^2(t) y'+4y=5

    y'+\frac{4y}{\cos^2(t)}=\frac{5}{\cos^2(t)}

    use integrating factor

    \mu(t)=e^{\int \frac{4}{\cos^2(t)}dt}=e^{4 \tan(t)}

    \mu(t) y' + \mu(t)\frac{4y}{\cos^2(t)}=\mu(t)\frac{5}{\cos^2(t  )}

    \frac{d}{dt}\left(\mu(t)y\right)=\frac{5 e^{4 \tan(t)}}{\cos^2(t)}

    \mu(t) y=\int\frac{5 e^{4 \tan(t)}}{\cos^2(t)}dt=\frac{5}{4}e^{4\tan(t)}+C

    y=\frac{5}{4}+Ce^{-4\tan(t)}

    y(0)=\frac{5}{4}+C=6 \Rightarrow C=\frac{19}{4}

    y=\frac{1}{4}\left(5+19e^{-\tan(t)}\right)

    so your third choice is correct.
    Last edited by romsek; February 15th 2014 at 02:29 PM.
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