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Thread: 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.
    $\displaystyle y'+y\sqrt t \cdot \sin t =0$.
    Using the integrating factor method, I have to calculate $\displaystyle \int \sqrt t \sin t dt$, which is a problem to me.
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  2. #2
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    It's separable: $\displaystyle \frac{dy}{y}=-\sqrt{t}\sin(t) dt$ so I'd just let $\displaystyle u=\sqrt{t}$ and end up with $\displaystyle -2\int u^2 \sin(u^2) du$ 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: $\displaystyle \frac{dy}{y}=-\sqrt{t}\sin(t) dt$ so I'd just let $\displaystyle u=\sqrt{t}$ and end up with $\displaystyle -2\int u^2 \sin(u^2) du$ which doesn't look bad to integrate via parts.
    Thanks I could follow you, I noticed that $\displaystyle dt=2udu$.
    I'm having problems with the IBP. What do you choose as f and g'?
    If I left $\displaystyle f=u^2$ and $\displaystyle g'=\sin u^2$, then I'll have to calculate $\displaystyle \int \sin u^2 du$ for which I've no clue about how to calculate it. I tried another IBP but with no success.
    If I let $\displaystyle f=\sin u^2$ and $\displaystyle g'=u^2$, I'll have to calculate $\displaystyle \int u^4 \cos (u^2)du$, which is worse than the original integral.

    So I'd opt for the first option, but I've some problems solving $\displaystyle \int \sin (u^2) du$. 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:

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

    $\displaystyle \int- \sqrt{t}\sin{t} dt=\sqrt{t} \text{Cos}[t]-\sqrt{\frac{\pi }{2}} \text{FresnelC}\left[\sqrt{\frac{2}{\pi }} \sqrt{t}\right]$
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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:

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

    $\displaystyle \int- \sqrt{t}\sin{t} dt=\sqrt{t} \text{Cos}[t]-\sqrt{\frac{\pi }{2}} \text{FresnelC}\left[\sqrt{\frac{2}{\pi }} \sqrt{t}\right]$
    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 $\displaystyle \int_0^x \cos(\pi x^2/2) dx$ and that's why the particular example above has the $\displaystyle \sqrt{2/\pi}$ in it. Really, we started with:

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

    which I get by parts:

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

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

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