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Math Help - Solving the initial value first order differential equation

  1. #1
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    Post Solving the initial value first order differential equation

    \cos (x)y' + \sin (x)y = 2\cos (x)\sin (x) - 1<br />
    y(\frac{\pi }<br />
{4}) = 3\sqrt 2 ,0 \leqslant x \leqslant \frac{\pi }<br />
{4}<br />
    I've tried to get y' and y by itself and got this mess:
    \int {\frac{{dy}}<br />
{y}}  = \int {\frac{{2{{\cos }^3}(x)\sin (x) - 1 - \sin (x)dx}}<br />
{{\cos (x)}}} <br />
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  2. #2
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    Quote Originally Posted by genlovesmusic09 View Post
    \cos (x)y' + \sin (x)y = 2\cos (x)\sin (x) - 1<br />
    y(\frac{\pi }<br />
{4}) = 3\sqrt 2 ,0 \leqslant x \leqslant \frac{\pi }<br />
{4}<br />
    I've tried to get y' and y by itself and got this mess:
    \int {\frac{{dy}}<br />
{y}} = \int {\frac{{2{{\cos }^3}(x)\sin (x) - 1 - \sin (x)dx}}<br />
{{\cos (x)}}} <br />
    The DE can be re-written as \frac{dy}{dx} + (\tan x) y = 2 \sin x - \frac{1}{\cos x}.

    Use the integrating factor technique to solve it.
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  3. #3
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    where did cos^3(x) go?
    I understand how everything else changed
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  4. #4
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    Quote Originally Posted by genlovesmusic09 View Post
    where did cos^3(x) go?
    I understand how everything else changed
    This is the equation you posted (check the OP yourself):
    \cos (x)y' + \sin (x)y = 2\cos (x)\sin (x) - 1
    so I have no idea what you mean by
    where did cos^3(x) go?
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  5. #5
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    yes I see the confusion. I mistyped the problem it is:
    \cos (x)y' + \sin (x)y = 2{\cos ^3}(x)\sin (x) - 1<br />
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  6. #6
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    Quote Originally Posted by genlovesmusic09 View Post
    yes I see the confusion. I mistyped the problem it is:
    \cos (x)y' + \sin (x)y = 2{\cos ^3}(x)\sin (x) - 1<br />
    Well then, all you need to do is make the appropriate small change in my original post and then continue from there.
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