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Math Help - Inital value problem

  1. #1
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    Inital value problem

    I need a little help as I am not sure of my footing
    I have to Solve the initial value problem

    dy/dx = cos(5x)/7+sin(5x), y=6 when x=0
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Inital value problem

    Note that the ODE is separable.
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  3. #3
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    Re: Inital value problem

    I do not understand what you mean
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Inital value problem

    I mean:

    \frac{dy}{dx} = \frac{\cos(5x)}{7}+\sin(5x)\ \mbox{with} \ y=6 \ \mbox{when} \ x=0
    \Leftrightarrow dy = \left[\frac{\cos(5x)}{7}+\sin(5x)\right]dx
    \Leftrightarrow \int dy = \int \left[\frac{\cos(5x)}{7}+\sin(5x)\right]dx

    Can you continue?
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  5. #5
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    Re: Inital value problem

    I can follow that but I think I wrote the initial problem badly it should read (cos⁡(5x))/(7+sin⁡(5x))
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  6. #6
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    Re: Inital value problem

    Hello, landmark!

    \frac{dy}{dx} \,=\, \frac{\cos(5x)}{7+\sin(5x)} \quad y=6 \text{ when }x=0.

    We have: . dy \;=\;\frac{\cos(5x)}{7 + \sin(5x)}\,dx

    Integrate: . \int dy \;=\;\int\frac{\cos(5x)\,dx}{7 + \sin(5x)}

    Let u \:=\: 7 + \sin(5x) \quad\Rightarrow\quad du \:=\:5\cos(5x)\,dx \quad\Rightarrow\quad \cos(5x)\,dx \:=\:\tfrac{1}{5}du

    Substitute: . y \:=\:\int\frac{\frac{1}{5}\,du}{u} \:=\:\tfrac{1}{5}\int\frac{du}{u} \:=\;\tfrac{1}{5}\ln |u|+C

    Back-substitute: . y \:=\:\tfrac{1}{5}\ln|7 + \sin(5x)|+ C


    Initial values: y = 6,\,x = 0

    . . 6 \;=\;\tfrac{1}{5}\ln|7 + \sin(0)| + C \quad\Rightarrow\quad C \:=\:6 - \tfrac{1}{5}\ln7


    Therefore: . y \;=\;\tfrac{1}{5}\ln(7 + \sin(5x)| + 6 - \tfrac{1}{5}\ln 7

    . . . . . . . . . y \;=\;\tfrac{1}{5}\left[\ln|7 + \sin(5x)| - \ln 7\right] + 6

    . . . . . . . . . y \;=\;\tfrac{1}{5}\ln \left|\frac{7+\sin(5x)}{7}\right| + 6

    . . . . . . . . . y \;=\;\tfrac{1}{5}\ln\left|1 + \tfrac{1}{7}\sin(5x)\right| + 6
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  7. #7
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    Smile Re: Inital value problem

    I was half way there, but was struggling.
    Thanks a million
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