Verifying That A Function Is A Solution To DE

[Bashyboy]

Homework Statement

Hello, I am suppose to verify that the indicated function
[itex]y = \phi (x)[/itex] is an explicit solution of the given first-order
differential equation. Then I am suppose to consider [itex]\phi[/itex] simply as a function, giving its domain; and then I am suppose to consider it as a solution, giving at least one interval of definition.

The differential equation: [itex]y' = 25 + y^2[/itex]

The possible solution: [itex]y = 5 \tan 5x[/itex]

Homework Equations

The Attempt at a Solution

I was able to determine the domain to be [itex]\displaystyle ... \cup (\frac{(2k -1) \pi)}{10},\frac{(2k +1) \pi)}{10}) \cup (\frac{(2k +1) \pi)}{10},\frac{(2k +3) \pi)}{10}) \cup (\frac{(2k -3) \pi)}{10},\frac{(2k +5) \pi)}{10}) \cup ... [/itex]

And I was able to show that the function satisfied the DE, thus being solution:

[itex]\displaystyle \frac{d}{dx} [5 \tan 5x] = 25 + (5 \tan 5x)^2[/itex]

[itex]25 \sec^2 5x = 25(1 + \tan^2 5x)[/itex]

[itex]25 \sec^2 5x =25 \sec^2 5x [/itex], which is a true statement.

What I am unsure of is, what should the interval of solution be? Does it have to in any way reflect the domain restrictions of the sec function, or only the tan?

 

[LCKurtz]

Since you need "at least one interval", I would suggest just giving the largest open interval about ##x=0## such that neither the tangent nor secant has a singularity.

 

[Bashyboy]

Okay, so I am suppose to find an interval for x such that differential equation and its solution are defined; that is, I have to take into account both tan and sec. I wasn't sure if this was the case or not. Thank you, LCkurtz.