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Thread: 10.09.15t solve the intial value problem

  1. #1
    Super Member bigwave's Avatar
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    10.09.15t solve the intial value problem

    $\tiny{10.09.15t}$
    $\textrm{ solve the intial value problem}$
    \begin{align*}\displaystyle
    \theta \frac{dy}{dx}+y&=\cos \theta,>0; \, y(\pi)=1 \\
    &=
    \end{align*}

    ok I'm not in a class right now so trying to self teach but ????
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  2. #2
    Junior Member Ahri's Avatar
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    Re: 10.09.15t solve the intial value problem

    Are you sure it's $$\theta \frac{dy}{dx} + y=\cos(\theta)$$ and not $$\theta \frac{dy}{d\theta} + y = \cos (\theta)$$
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  3. #3
    Super Member bigwave's Avatar
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    Re: 10.09.15t solve the intial value problem

    yeah you are correct ..... typo
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  4. #4
    Junior Member Ahri's Avatar
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    Re: 10.09.15t solve the intial value problem

    No problem

    So for $$\theta\frac{dy}{d \theta} + y = \cos (\theta)$$
    Write as
    $$\frac{d}{d\theta} \left( \theta y\right) = \cos (\theta)$$
    (Do you know where this comes from?...)

    Now, integrate both sides with respect to $\theta$. What do you get?
    Last edited by Ahri; Jun 15th 2017 at 03:53 PM.
    Thanks from bigwave and HallsofIvy
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  5. #5
    Super Member bigwave's Avatar
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    Re: 10.09.15t solve the intial value problem

    $\frac{d}{d\theta} \left( \theta y\right) = \cos (\theta)$

    is it

    $y=-sin(\theta)$
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  6. #6
    Junior Member Ahri's Avatar
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    Re: 10.09.15t solve the intial value problem

    If you were to apply the product rule to $\frac{d}{d\theta}(\theta y)$ you would get $y+ \theta \frac{dy}{d\theta}$, that is how I got that.

    Next step is to integrate both sides, so

    $$\int \frac{d}{d\theta} \left(\theta y\right) d\theta = \int \cos(\theta ) d\theta$$
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  7. #7
    Junior Member Ahri's Avatar
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    Re: 10.09.15t solve the intial value problem

    I'm logging off now, so here's a solution, I've assumed you meant $\theta >0$ in your original question.

    Try to work your way through it, and if you're unsure of any of the steps just leave a reply.

    $$\theta\frac{dy}{d\theta}+y=\cos(\theta),\ \theta >0,\ y(\pi)=1$$
    Write as
    $$\frac{d}{d\theta}(\theta y) = \cos(\theta)$$
    Integrate both sides with respect to $\theta$
    $$\int \frac{d}{d\theta}(\theta y) d\theta = \int \cos (\theta)d\theta$$
    Which gives
    $$\theta y = \sin(\theta) + C$$
    C is a constant. Now divide by $\theta$ to isolate $y$,
    $$y=\frac{\sin(\theta)+C}{\theta}$$
    Applying the initial condition:
    $$y(\pi) = \frac{\sin(\pi) + C}{\pi}=1$$
    $$\frac{C}{\pi}=1$$
    $$C=\pi$$
    So the solution is
    $$y(\theta)=\frac{\sin(\theta)+\pi}{\theta}$$
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  8. #8
    Super Member bigwave's Avatar
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    Re: 10.09.15t solve the intial value problem

    much mahalo
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