# Differential Equations Help

• Feb 5th 2009, 11:16 AM
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 $\displaystyle y=sinh (kt)$ satisfy the differential equation

$\displaystyle y'' - 25y = 0$

1b.

for those values of k, show that every member of the family of functions $\displaystyle Asinh(kt) + Bcosh(kt)$ (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 $\displaystyle y=sinh (kt)$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

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

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, 01:43 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 $\displaystyle y=sinh (kt)$ satisfy the differential equation

$\displaystyle y'' - 25y = 0$

1b.

for those values of k, show that every member of the family of functions $\displaystyle Asinh(kt) + Bcosh(kt)$ (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 $\displaystyle y=sinh (kt)$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

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

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

In general we have shown that $\displaystyle \sinh(kx)$ solves the differencial equation $\displaystyle y'' -k^2y=0$.
• Feb 5th 2009, 03:09 PM
HallsofIvy
Crucial point: you must have $\displaystyle sinh(kt)(25- k^2)= 0$ 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.