IVP applications of second-order ODE

[zebrastripes]

Homework Statement

Given the equation mx''+cx=cAsin(Ωt) with the initial conditions x(0)=0 and x'(0)=0.
Solve the initial value problem for the case when Ω < ω and show that |x(t)| < H provided
A < H(1-(Ω/ω)).

Homework Equations

The Attempt at a Solution

For my solution to the equation I got x(t)=(Aω/(ω^2-Ω^2))[ωsin(Ωt)-Ωsin(ωt)]

So, I'm hoping that this is right which I found using x(t)=x_h+x_p.
But I'm completely confused about the second part to show that |x(t)| < H provided
A < H(1-(Ω/ω)).

This is my first post here, so apologies if I'm not doing it right. :shy:
Thanks in advance.

 

[LCKurtz]

Homework Statement

Given the equation mx''+cx=cAsin(Ωt) with the initial conditions x(0)=0 and x'(0)=0.
Solve the initial value problem for the case when Ω < ω and show that |x(t)| < H provided
A < H(1-(Ω/ω)).

The Attempt at a Solution

For my solution to the equation I got x(t)=(Aω/(ω^2-Ω^2))[ωsin(Ωt)-Ωsin(ωt)]

So, I'm hoping that this is right which I found using x(t)=x_h+x_p.
But I'm completely confused about the second part to show that |x(t)| < H provided
A < H(1-(Ω/ω)).

Good work so far; you are almost there. Take the absolute value of both sides of your $x(t)$ equatiion and use this:$$
|\omega \sin(\Omega t) - \Omega \sin(\omega t)|\le
|\omega| | \sin(\Omega t)| + |\Omega| | \sin(\omega t)|
\le \omega + \Omega$$and put in the overestimate given for A. See what drops out.

 

[zebrastripes]

Ahh Ok, so it's my modulus work that's really poor!

I'm thinking you're equation changes mine to: Aω/(ω^2-Ω^2)<H(ω+Ω)

So with the substitution for A I get (ω-Ω)/(ω^2-Ω^2)<(ω+Ω)

So from this, could I finish by showing this is true? Or have I gone wrong again? I'm thinking that my Aω/(ω^2-Ω^2) should have changed when I was taking the modulus?

 

[LCKurtz]

Ahh Ok, so it's my modulus work that's really poor!

I'm thinking you're equation changes mine to: Aω/(ω^2-Ω^2)<H(ω+Ω)

The contraction "you're" is a short form of "you are". The possessive case is "your". Also, while I'm commenting on your post, I should point out that you never told us what $\omega$ stands for. Best not to leave definitions out.

You will have less trouble if you will write out your full string of inequalities. You have$$
x(t) = \frac{A\omega}{\omega^2-\Omega^2}(\omega \sin(\Omega t) - \Omega \sin(\omega t))$$Start by taking the absolute value of both sides:$$
|x(t)|=\left| \frac{A\omega}{\omega^2-\Omega^2} \right| |\omega \sin(\Omega t) - \Omega \sin(\omega t)| \le ...$$Now continue the string on the right using what you know so far.