Differential Equation Q? Show d^2x/dt^2 + 9x = 0 satisfied by x = 6sin3t + 4 cos3t

[limaecho1988]

Show that the differential equation
d^2x/dt^2 + 9x = 0

is satisfied by, x = 6sin3t + 4 cos3t

I can get to the point of: x = Asin3t + Bcos3t

Cant work out how to do the satisfied bit? Any help appreciated.

 

[MarkFL]

Given that:

\(\displaystyle x(t)=6\sin(3t)+4\cos(3t)\)

What is \(\displaystyle \displaystyle \frac{d^2x}{dt^2}\) ?

What to you get when you substitute for \(\displaystyle x\) and \(\displaystyle \displaystyle \frac{d^2x}{dt^2}\) into the given ODE?

 

[HallsofIvy]

Show that the differential equation
d^2x/dt^2 + 9x = 0

is satisfied by, x = 6sin3t + 4 cos3t

I can get to the point of: x = Asin3t + Bcos3t

Why? Why not set A= 6 and B= 4 as this problem says?

Cant work out how to do the satisfied bit? Any help appreciated.

The problem does not ask you to solve the equation, just show that the given function satisfies the equation.

If you were given the polynomial equation, \(\displaystyle x^5- 3x^4+ x^3+ x+6= 0\) and asked to solve it, that would be a very difficult problem! Buy if you are asked to "show that x= 2 is a solution" is just basic integer arithmetic!

 

[limaecho1988]

Thanks for the clue.

d^2x//dt^2 = -54sin(3t) - 36cos(3t)

substituting x into the ODE satisfies it.

 

[MarkFL]

In general, we will find that for:

\(\displaystyle x(t)=A\cos(\omega t)+B\sin(\omega t)\)

then:

\(\displaystyle x''(t)=-\omega^2x(t)\implies x''(t)+\omega^2x(t)=0\)