1. Differential Equation help

Hi,

For what $\displaystyle nonzero$ values of $\displaystyle k$ does the function $\displaystyle y = sinh (kt)$ satisfy the diff erential equation $\displaystyle y'' - 25y = 0$?

I don't really understand what the general method of solving this differential equation is. Do I differentiate w.r.t to $\displaystyle k$??
The only thing I understand regarding DE's is separating the variables but I dont think it's used here..is it? :S.

What I've tried:

$\displaystyle y' = k cosh (kt)$

$\displaystyle y'' = k^2 sinh (kt) + cosh (kt)$
Since
$\displaystyle y'' = 25y$

$\displaystyle k^2 sinh (kt) + cosh (kt) = 25 sinh (kt)$

$\displaystyle (k^2 - 25) sinh (kt) + cosh (kt) = 0$

--
But I don't know where to go from there...or if I've gone in a completely wrong direction. Explanation and/or answer would be greatly appreciated ASAP. Thanks!

2. Originally Posted by Solo
Hi,

For what $\displaystyle nonzero$ values of $\displaystyle k$ does the function $\displaystyle y = sinh (kt)$ satisfy the differential equation $\displaystyle y'' - 25y = 0$?

I don't really understand what the general method of solving this differential equation is. Do I differentiate w.r.t to $\displaystyle k$??
The only thing I understand regarding DE's is separating the variables but I dont think it's used here..is it? :S.

What I've tried:

$\displaystyle y' = k cosh (kt)$

$\displaystyle \color{red}y'' = k^2 sinh (kt) + cosh (kt)$
where'd you get that from?

your approach is good, but you messed up where i indicated

3. Ah I think I've differentiated first WRT t, and then WRT k, which is worng

Should it just be:

$\displaystyle y'' = k^2 sinh (kt)$

then:

$\displaystyle k^2 sinh (kt) = 25 sinh (kt)$
$\displaystyle k^2 = 25$

k = ± 5

Is that right? Surely it can't be that easy?

4. Originally Posted by Solo
Ah I think I've differentiated first WRT t, and then WRT k, which is worng

Should it just be:

$\displaystyle y'' = k^2 sinh (kt)$

then:

$\displaystyle k^2 sinh (kt) = 25 sinh (kt)$
$\displaystyle k^2 = 25$

k = ± 5

Is that right? Surely it can't be that easy?
yup, that's it

(note that zero also works, but they asked for nonzero solutions, so we deliberately forget about it)

5. Cheers

Now I'm stuck on the next part:

Come up with a second order di fferential equation for which $\displaystyle y = sin (kt)$ is a solution for the same values of k obtained in part (a).

For $\displaystyle k = 5$

$\displaystyle y = sin (5t)$
$\displaystyle y' = 5 cos (5t)$
$\displaystyle y'' = -25 sin (5t)$
Equation: $\displaystyle y'' + 25y = 0$

For $\displaystyle k = -5$

$\displaystyle y = sin (-5t)$
$\displaystyle y' = -5 cos (-5t)$
$\displaystyle y'' = 25 sin (-5t)$
Equation: $\displaystyle y'' - 25y = 0$

The question only asks for one differential equation...What have I done wrong?

6. $\displaystyle y=sin(-5t)=-sin(5t)$ since $\displaystyle sin(x)$ is an odd function i.e. $\displaystyle sin(-x)=-sin(x)$

$\displaystyle y=-sin(5t)$
$\displaystyle y'=-5cos(5t)$
$\displaystyle y''=25sin(5t)$

$\displaystyle 25sin(5t)-25sin(5t)=0$

7. Originally Posted by Solo
Come up with a second order di fferential equation for which $\displaystyle y = sin (kt)$ is a solution for the same values of k obtained in part (a).
.
.
.
Equation: $\displaystyle y'' + 25y = 0$
this equation works for both the values of k.