# Differential Equations Help

• Feb 5th 2009, 12:16 PM
jordanrs
Differential Equations Help
Stuck on the following problem, unsure of how to go around it

1a.

For what none values of k does the function $y=sinh (kt)$ satisfy the differential equation

$y'' - 25y = 0$

1b.

for those values of k, show that every member of the family of functions $Asinh(kt) + Bcosh(kt)$ (where A and B are constants) is also a solution to the differential equation.

1c.

come up with a second order differntial equation for which $y=sinh (kt)$is a solution for the same values of k obtained in part (a)

im usually ok with differentail equations but unsure when it comes to the worded versions

for 1a i tryed the following but werent sure where else to go

$y=sinh (kt)$
$y'' = k^2sinh(kt)$
$25y = k^2sinh(kt)$
$25sinh(kt) = k^2sinh(kt)$
$sinh(kt)(25-k^2) = 0$

so one possible solution is k = +/- 5 but im not sure if that is the right thing to do

Any help is most appreciated
• Feb 5th 2009, 02:43 PM
ThePerfectHacker
Quote:

Originally Posted by jordanrs
Stuck on the following problem, unsure of how to go around it

1a.

For what none values of k does the function $y=sinh (kt)$ satisfy the differential equation

$y'' - 25y = 0$

1b.

for those values of k, show that every member of the family of functions $Asinh(kt) + Bcosh(kt)$ (where A and B are constants) is also a solution to the differential equation.

1c.

come up with a second order differntial equation for which $y=sinh (kt)$is a solution for the same values of k obtained in part (a)

im usually ok with differentail equations but unsure when it comes to the worded versions

for 1a i tryed the following but werent sure where else to go

$y=sinh (kt)$
$y'' = k^2sinh(kt)$
$25y = k^2sinh(kt)$
$25sinh(kt) = k^2sinh(kt)$
$sinh(kt)(25-k^2) = 0$

so one possible solution is k = +/- 5 but im not sure if that is the right thing to do

Any help is most appreciated

We are told that $y=\sinh(kx)$ solves $y'' - 25y=0$. Now, $y'' = k^2\sinh(kx)$, therefore we want $k^2\sinh(x) - 25\sinh(x) = 0 \implies \sinh(x) (k^2 - 25) = 0$. Thus, $k=\pm 5$ and so $\sinh(5x)$ and $-\sinh(5x)$ solve this differencial equation.

In general we have shown that $\sinh(kx)$ solves the differencial equation $y'' -k^2y=0$.
• Feb 5th 2009, 04:09 PM
HallsofIvy
Crucial point: you must have $sinh(kt)(25- k^2)= 0$ for all t. Since there exist values of t such that sinh(kt) is not 0, we must have 25- k^2= 0 or k^2= 25.