1. ## 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

2. 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 $\displaystyle y'' -k^2y=0$.

3. 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.