Particular integral problem for a Differential Equation.

[Cornishman]

Homework Statement

Hi this is my first post here... just started a new course, and in the prep material I've been given this question which is doing my head in!

The DE

$$\frac{d^2 x}{dy^2}+\frac{dx}{dt}+x=sin(wt)$$

describes the deflection x of some suspension as a function of t, in response to a bumby surface. Here w>0 is a parameter that describes the frequency of the bumps.
(hopefully the LaTeX equation looks ok, because for some reason this works computer doesn't want to show me it!)

Homework Equations

The first question asks me to solve the Associated Homogeneous equation,

for which I got $$-\frac{1}{2}+-\sqrt{\frac{3}{4}}i$$ as a conjugate pair of solutions of the aux. equation. This gives me the General solution,

$$x_{c}=e^{-1/2t}(C cos\sqrt{\frac{3}{4}}t+D sin\sqrt{\frac{3}{4}}t)$$

Hopefully I'm all good so far?

The Attempt at a Solution

So the next part asks me to find a particular integral which satisfies the DE and hence obtain the general solution...(I should add, I've only ever learn't the method of undertermined coefficients from previous courses, so the method can't be to exotic.)

So I guessed a solution of the form:
$$x=p cos(wt)+q sin(wt)$$

Differentiating (twice), we end up with
$$\frac{dx}{dt}=-wp sin(wt)+wq cos(wt)$$
and
$$\frac{d^2 x}{dt^2}=-w^2 p cos(wt)-w^2 q sin(wt)$$

I then substituted these back into the equation, which after some jiggery pokery I got

$$(-w^2 p+wq+p)cos(wt)+(-w^2 q-wp+q)sin(wt)=sin(wt)$$

So

$$(-w^2 p+wq+p)=0$$
and
$$(-w^2 q-wp+q)=1$$

But now I'm stuck, how do I find the values for p and q? Can anyone give me a clue as to how to find them?

 

[Mark44]

Everything looks fine in what you did, based on a quick scan. Your last two equations are really in two variables, p and q, and those are what you are solving for. w is merely a parameter.

Here's your system a little more explicitly. You can find p and q, right?

(1 - w²)p + wq = 0
-wp + (1 - w²)q = 1

 

[Cornishman]

Yeah thanks... rearrange the first equation to get a term for q, then substitute that into the second to get a equation for p interms of w. This can the be resubstituted into my first equation to solve q in terms of w.

For some reason, I did this way back... and thought that there was a different way that I could determine actual values for p and q, like 2 and 3 or something... not sure why I thought that now?

Thanks again... Might be back for the next question soon!

 

[Cornishman]

Right still got problems...Heres my general solution,

$$x_c+x_p=e^{-1/2 t} (C cos(\sqrt{\frac{3}{4}}t)+D sin(\sqrt{\frac{3}{4}}t))-\frac{w}{1-w^2+w^4} cos(wt)+\frac{1-w^2}{1-w^2+w^4}sin(wt)$$

Hopefully that's ok. Now the question,

as $$t\rightarrow \infty$$ the solution approaches $$x(t) = k sin(w(t-t_0))$$
where $$t_0$$ and $$k$$ depend on $$w$$

Show that, $$k=\frac{1}{\sqrt{w^4-w^2+1}}$$

I assume, seeing as the transient portion of my general equation disappears as t approaches infinity, they are asking me to equate

$$-\frac{w}{1-w^2+w^4} cos(wt)+\frac{1-w^2}{1-w^2+w^4}sin(wt) =k sin(w(t-t_0))$$

I tried using a trig identity, [tex] sin(x+y)=sin(x)cos(y)+cos(x)sin(y) [\tex]

but quickly get bogged down, and can't seem to work through the re-arrangement? Am I on the right lines? IS there an easier way?

 

[Cornishman]

Any help?

 

[vela]

Whenever you have something that looks like $A\cos\alpha+B\sin\alpha$, you can consider A and B to be the length of the legs of a right triangle so that

$$\frac{A}{\sqrt{A^2+B^2}} = \sin\beta$$

$$\frac{B}{\sqrt{A^2+B^2}} = \cos\beta$$

So you can write

$$A\cos\alpha+B\sin\alpha = \sqrt{A^2+B^2}(\sin\beta\cos\alpha+\cos\beta\sin\alpha) = \sqrt{A^2+B^2}\sin(\alpha+\beta)$$

where

$$\beta = \arctan\frac{A}{B}$$