Show that ____ is a solution to the differential equation model?

[thename1000]

Homework Statement

Trying to go over this part of my test review. I'm not understanding how to do this. The specific problem which I actually have worked out in my notes:

A diff equation model of the motion of a spring where x is displacement from the spring's natural length, k is the spring constant, and m is the mass as follows:

(d^2*x)/(d*t^2) = k/m * x


Show that x(t)=sin(sqrt(k/m)*t) + 2cos(sqrt(k/m)*t) is a solution to the model.

The Attempt at a Solution

My notes appear to take the first, then the second dir. of the second function there. But that's all I can make sense of. Maybe I'm not understanding how to take the dir of these functions I'm not sure.

For example on the second to the last line, how am I supposed to simplify -k/m (sin(sqrt(k/m)t) + 2cos (sqrt(k/m)t)

to

-k/mx

thanks for any help!

 

[Office_Shredder]

Because you just said that x=(sin(sqrt(k/m)t) + 2cos (sqrt(k/m)t) ?

 

[thename1000]

Because you just said that x=(sin(sqrt(k/m)t) + 2cos (sqrt(k/m)t) ?

uhhh...thanks? Since somebody answered I won't get a real answer now. :(

 

[Office_Shredder]

Uhh... that was a real answer?

Show that x(t)=sin(sqrt(k/m)*t) + 2cos(sqrt(k/m)*t) is a solution to the model.

Notice how you assume that x is sin(sqrt(k/m)*t) + 2cos(sqrt(k/m)*t) to start your answer.

For example on the second to the last line, how am I supposed to simplify -k/m (sin(sqrt(k/m)t) + 2cos (sqrt(k/m)t)

to

-k/mx

Notice how you want to replace the bolded portion with x. Notice how x is assumed to be exactly the bolded portion.

 

[thename1000]

Yes, I see what you're saying, sorry.

I gave the second to the last line as an example of how I don't fully understand the process.

To take the derivative or second derivative of the function do I sub something for sqrt(k/m)?

sin(u*t) + 2cos(u)*t

Like that?

 

[HallsofIvy]

You can but you should be able to use the chain rule without having to substitute like that. It is, after all, just a constant: the derivative of f(ax) is af '(ax).