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Math Help - Linear first order homogenous DE

  1. #1
    MHF Contributor arbolis's Avatar
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    Linear first order homogenous DE

    I think the following DE is of 1st order, linear and homogeneous, correct me if I'm wrong.
    I don't know which method to use to solve it.
    y'+y\sqrt t \cdot \sin t =0.
    Using the integrating factor method, I have to calculate \int \sqrt t \sin t dt, which is a problem to me.
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  2. #2
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    It's separable: \frac{dy}{y}=-\sqrt{t}\sin(t) dt so I'd just let u=\sqrt{t} and end up with -2\int u^2 \sin(u^2) du which doesn't look bad to integrate via parts.
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  3. #3
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by shawsend View Post
    It's separable: \frac{dy}{y}=-\sqrt{t}\sin(t) dt so I'd just let u=\sqrt{t} and end up with -2\int u^2 \sin(u^2) du which doesn't look bad to integrate via parts.
    Thanks I could follow you, I noticed that dt=2udu.
    I'm having problems with the IBP. What do you choose as f and g'?
    If I left f=u^2 and g'=\sin u^2, then I'll have to calculate \int \sin u^2 du for which I've no clue about how to calculate it. I tried another IBP but with no success.
    If I let f=\sin u^2 and g'=u^2, I'll have to calculate \int u^4 \cos (u^2)du, which is worse than the original integral.

    So I'd opt for the first option, but I've some problems solving \int \sin (u^2) du. Am I in the right direction? Am I missing something?
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    Ok, I made a mistake. Sorry. Not easy to integrate that. The reason I say that is I just used DSolve in Mathematica and it returns FrenselC functions which is just another name for the integral encountered in this problem but I personally think is ok in terms of an "exact" solution. So then I'd leave it as:

    y=ke^{-\int \sqrt{t}\sin(t) dt} or if you wish:

    \int- \sqrt{t}\sin{t} dt=\sqrt{t} \text{Cos}[t]-\sqrt{\frac{\pi }{2}} \text{FresnelC}\left[\sqrt{\frac{2}{\pi }} \sqrt{t}\right]
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  5. #5
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by shawsend View Post
    Ok, I made a mistake. Sorry. Not easy to integrate that. The reason I say that is I just used DSolve in Mathematica and it returns FrenselC functions which is just another name for the integral encountered in this problem but I personally think is ok in terms of an "exact" solution. So then I'd leave it as:

    y=ke^{-\int \sqrt{t}\sin(t) dt} or if you wish:

    \int- \sqrt{t}\sin{t} dt=\sqrt{t} \text{Cos}[t]-\sqrt{\frac{\pi }{2}} \text{FresnelC}\left[\sqrt{\frac{2}{\pi }} \sqrt{t}\right]
    Is there a way to reach this result with calculus level math? I'm surprised I have this exercise to solve if it seems so complicated. I never dealt with the Fresnel thing before.
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  6. #6
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    Ok, let me try to explain without getting into more trouble: That FresnelC thing is a Mathematica construct for the Fresnel integral \int_0^x \cos(\pi x^2/2) dx and that's why the particular example above has the \sqrt{2/\pi} in it. Really, we started with:

    \int_{y_0}^y \frac{dy}{y}=-\int_{t_0}^t \sqrt{t}\sin(t)dt

    which I get by parts:

    \ln|y|\biggr|_{y_0}^y=\sqrt{t}\cos(t)\biggr|_{t_0}  ^t-\int_{\sqrt{t_0}}^{\sqrt{t}} \cos(u^2)du

    and if t_0=0 and we use the standard definition of a Fresnel integral: C(x)=\int_0^{x}\cos(u^2)du then I could write the solution as:

    y(t)=k\exp\left\{\sqrt{t}\cos(t)-C(\sqrt{t})\right\}
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