# y'''+y'=tan(x)

• Apr 29th 2013, 03:46 AM
loai
y'''+y'=tan(x)
I need a helpe to fine the Particular solution of the DE
Y'''+Y'=tan(x)
• Apr 29th 2013, 04:02 AM
JJacquelin
Re: y'''+y'=tan(x)
y''+y = ?
• Apr 29th 2013, 04:05 AM
loai
Re: y'''+y'=tan(x)
realy it's
y'''+y'=tanx
• Apr 29th 2013, 04:09 AM
Prove It
Re: y'''+y'=tan(x)
I would start by writing \displaystyle \displaystyle \begin{align*} u = y' \end{align*}, then you get \displaystyle \displaystyle \begin{align*} u'' + u = \tan{(x)} \end{align*}. You should be able to solve this second order constant coefficient DE using whatever method you're comfortable with. Then once you have u you can find y.
• Apr 29th 2013, 04:11 AM
loai
Re: y'''+y'=tan(x)
thanx
• Apr 29th 2013, 10:07 AM
JJacquelin
Re: y'''+y'=tan(x)
Quote:

Originally Posted by loai
realy it's
y'''+y'=tanx

Sure, the equation is y'''+y'=tanx
But, that was not what I was asking for :
When you integrate it, what do you obtain ?
y'' + y' = ?
• Apr 29th 2013, 05:47 PM
Prove It
Re: y'''+y'=tan(x)
Quote:

Originally Posted by JJacquelin
Sure, the equation is y'''+y'=tanx
But, that was not what I was asking for :
When you integrate it, what do you obtain ?
y'' + y' = ?

If you went for direct integration, you should really get \displaystyle \displaystyle \begin{align*} y'' + y = \int{\tan{(x)}\,dx} \end{align*}, NOT \displaystyle \displaystyle \begin{align*} y'' + y' = \textrm{ anything} \end{align*}. Having said that, I think my method would be easier, as I think it would be easier to find a particular solution to a second order DE with a RHS of \displaystyle \displaystyle \begin{align*} \tan{(x)} \end{align*} than \displaystyle \displaystyle \begin{align*} \ln{ \left| \sec{(x)} \right| } \end{align*}.
• Apr 29th 2013, 09:31 PM
JJacquelin
Re: y'''+y'=tan(x)
Quote:

Originally Posted by Prove It
If you went for direct integration, you should really get \displaystyle \displaystyle \begin{align*} y'' + y = \int{\tan{(x)}\,dx} \end{align*}, NOT \displaystyle \displaystyle \begin{align*} y'' + y' = \textrm{ anything} \end{align*}. Having said that, I think my method would be easier, as I think it would be easier to find a particular solution to a second order DE with a RHS of \displaystyle \displaystyle \begin{align*} \tan{(x)} \end{align*} than \displaystyle \displaystyle \begin{align*} \ln{ \left| \sec{(x)} \right| } \end{align*}.

Hi ProveIt !

You gave to Ioai the answer that he should have found by himself. (Smile)
• Apr 29th 2013, 11:26 PM
Prove It
Re: y'''+y'=tan(x)
Quote:

Originally Posted by JJacquelin
Hi ProveIt !

You gave to Ioai the answer that he should have found by himself. (Smile)

You seem to have missed my point. My point is that the method you suggested, while valid, will give a more difficult DE to solve than the method that I suggested. It's difficult to point this out without showing what I did :)
• Apr 30th 2013, 12:38 AM
JJacquelin
Re: y'''+y'=tan(x)
Quote:

Originally Posted by Prove It
You seem to have missed my point. My point is that the method you suggested, while valid, will give a more difficult DE to solve than the method that I suggested. It's difficult to point this out without showing what I did :)

I don't missed your point at all. In fact, I didn't want to get into an argument of few importance.
Of course, your method is easier at the first step leading to u(x). But after that, one have to integrate u(x) in order to obtain y(x). This step involves integrals which are as complicated as the integrals encountered in my method. So, I think that there is few difference between both methods.