I need a helpe to fine the Particular solution of the DE

Y'''+Y'=tan(x)

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- Apr 29th 2013, 03:46 AMloaiy'''+y'=tan(x)
I need a helpe to fine the Particular solution of the DE

Y'''+Y'=tan(x) - Apr 29th 2013, 04:02 AMJJacquelinRe: y'''+y'=tan(x)
y''+y = ?

- Apr 29th 2013, 04:05 AMloaiRe: y'''+y'=tan(x)
realy it's

y'''+y'=tanx - Apr 29th 2013, 04:09 AMProve ItRe: 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 AMloaiRe: y'''+y'=tan(x)
thanx

- Apr 29th 2013, 10:07 AMJJacquelinRe: y'''+y'=tan(x)
- Apr 29th 2013, 05:47 PMProve ItRe: y'''+y'=tan(x)
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 PMJJacquelinRe: y'''+y'=tan(x)
- Apr 29th 2013, 11:26 PMProve ItRe: y'''+y'=tan(x)
- Apr 30th 2013, 12:38 AMJJacquelinRe: y'''+y'=tan(x)
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.