Differential equation

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• Feb 8th 2009, 12:10 PM
Konidias
Differential equation
For what \$\displaystyle nonzero\$ values of \$\displaystyle k\$ does the function \$\displaystyle y=sinhkt\$ satisfy the DE \$\displaystyle y''-25y=0\$?

I got \$\displaystyle k=5\$ and \$\displaystyle k=-5\$ which I think is correct, but I'm struggling with the other two parts of the question:

For those values of \$\displaystyle k\$, show that every member of the family of functions \$\displaystyle Asinhkt+Bcoshkt\$ (where\$\displaystyle A\$ and \$\displaystyle B\$ are constants) is also a solution of the DE.

Come up with a second order DE for which \$\displaystyle y=sinkt\$ is a solution for the same values of \$\displaystyle k\$ obtained before.
• Feb 8th 2009, 03:33 PM
HallsofIvy
Quote:

Originally Posted by Konidias
For what \$\displaystyle nonzero\$ values of \$\displaystyle k\$ does the function \$\displaystyle y=sinhkt\$ satisfy the DE \$\displaystyle y''-25y=0\$?

I got \$\displaystyle k=5\$ and \$\displaystyle k=-5\$ which I think is correct

Yes, it is. If y= sinh(kt) then y'= k cosh(kt) and y"= k^2 sinh(kt). Putting that into the differential equation, y"- 25y= k^2 sinh(kt)- 25 sinh(kt)= (k^2- 25)sinh(kt)= 0 for all t. Since sinh(kt) is not alway7s 0 itself, we must have k^2- 25= 0 so k= 5 or k= -5.

Quote:

but I'm struggling with the other two parts of the question:

For those values of \$\displaystyle k\$, show that every member of the family of functions \$\displaystyle Asinhkt+Bcoshkt\$ (where\$\displaystyle A\$ and \$\displaystyle B\$ are constants) is also a solution of the DE.
Just DO it! With k= 5, those functions are \$\displaystyle A sinh(5t)+ B cosh(5t)\$. Find the second derivative, plug it into the equation and see what happens! With k= -5, you haved \$\displaystyle A sinh(-5t)+ B cosh(-5t)\$. Since sinh is an odd function and cosh is an even function, that is just \$\displaystyle -A sinh(5t)+ B cosh(5t)\$ which is really the same thing since A itself can be positive or negative.

[/quote]Come up with a second order DE for which \$\displaystyle y=sinkt\$ is a solution for the same values of \$\displaystyle k\$ obtained before.[/QUOTE]
So now youy are working with \$\displaystyle y= sin(5t)\$. Take the second derivative, stare at it and think!(Clapping)