Linear first order homogenous DE

**arbolis** · February 23rd 2010, 10:40 AM

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.

**shawsend** · February 23rd 2010, 12:40 PM

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.

**arbolis** · February 23rd 2010, 01:20 PM

Originally Posted by shawsend

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?

**shawsend** · February 23rd 2010, 01:56 PM

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:

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

**arbolis** · February 23rd 2010, 02:41 PM

Originally Posted by shawsend

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.

**shawsend** · February 23rd 2010, 03:36 PM

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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