Stationary solution to differential equation

[Twinflower]

Homework Statement

What is the stationary amplitude to the sine wave?

4 \frac{dy}{dt} + 3y = sin(t) , y(0) = 0


2. The attempt at a solution

I figure that the stationary solution to this eq. is found when the derivate is 0.

0+3y = sin(x)
y = \frac{sin(x)}{3}

The amplitude to this function is \frac{1}{3}. The correct answer is supposed to be 0.2

So where did I take the wrong turn?

 

[I like Serena]

Hi Twinflower,

I'm afraid that a stationary amplitude to the sine wave does not imply that you have a stationary point in the solution for your DE.

Basically you need to solve the DE.
Do you know how to find the particular solution?

 

[Twinflower]

To be honest, I thougt that the stationary amplitude was given by the particular solution.

But I'll get back to this after dinner :) Hopefully with some more calculations

 

[I like Serena]

To be honest, I thougt that the stationary amplitude was given by the particular solution.

But I'll get back to this after dinner :) Hopefully with some more calculations

It is! :)
That is, you can deduce the stationary amplitude from the particular solution.
It's just that sin(t)/3 is not a particular solution.

 

[Twinflower]

Ok, I am a little confused now.

In my earlier exercises, (with DC voltage), the stationary result was always the particular one.
And the particular one was always the "original" equation where the derivative was 0.

Is something different here because of the sine function, or have I misunderstood something?

 

[I like Serena]

Yes, I guess something is different because of the sine function.

In other cases you probably had a nice constant result for y which would indeed be an amplitude.
In this case you find that the amplitude of y equals sin(t)/3, but this is not a constant expression, and you can't tell yet for which value of t you get the amplitude.
The only thing you can tell, is that the amplitude is at most 1/3.

 

[Twinflower]

Yes, I indeed have had nice constants :)
Made my life easy and uptil now I felt that I had somewhat in control of differential equations.

Suppose this sine function messed it up a bit for me.

What exactly is a constant expression? Do I have to derivate the particular one?

 

[I like Serena]

"1" is a constant expression.
Or "C", assuming that C is a constant (meaning it does not depend on t or y).

What you need to find is a solution to your DE.
Since the RHS is a sine, it seems likely that a combination of a cosine and a sine with unknown amplitudes would do the trick...

 

[Twinflower]

Ah, Ok. So I have to solve the homogenous (or something) part and use the "expected solution format", and then find the unknown new constant using the given start condition? (y(0) = 0).

So, the expected solution for the homogeous part is A sin(t) + B cos (t) or something?
(We never learn to find those expected solutions, unfortunately)

 

[I like Serena]

Yep!

(It's called an inhomogeneous DE, and a single solution for it is called a particular solution.
This method to solve it is called the method of undetermined coefficients.)

 

[ehild]

The solution of the homogeneous part of this ode tends to zero with time, leaving the partial solution only. The partial solution is a general sine function with the same angular frequency as the sine function on the right-hand side, but with unknown amplitude and phase Y=Asin(ωt+θ) (ω=1 in the example). This oscillatory solution is stationary in the sense that the amplitude and phase constant does not change with time. The problem asks the amplitude A.

ehild

 

[Twinflower]

OK, I feel I'm getting a few steps in the right direction, but I'm not completely sure where I'm headed.

This is what I think so far:

Particular solution:
y_p = \frac{sin(t)}{3}Homogeneous solution
y_h = 4 \frac{dy}{dx} + 3y = 0

Converting the derivate io \lambda and the variable (y) into 1:
y_h = 4 \lambda +3 = 0
\lambda = -\frac{3}{4}

OK now. The first thing I am uncertain of is where to use \lambda.
So let's just try something:

Since the inhomogeneous part is k sin (\alpha x) the guessed form of the homogeneous part would be K cos (\alpha x) + M sin (\alpha x)
(referring to http://en.wikipedia.org/wiki/Method...ents#Typical_forms_of_the_particular_integral )

Now, since the \alpha is used the same way I am used to use \lambda, I am going to try to put lambda where alpha is.K cos (-\frac{3}{4} t) + M sin (-\frac{3}{4} t)And I need both parts here, that is y(t) = y_p + y_h

Thus:
y(0) = 0 = \frac{sin(t)}{3} + K cos (-\frac{3}{4} t) + M sin (-\frac{3}{4} t)
Because t = 0, most parts say poof and disappear.
That is, leaving me with 0 = K because cos(0) = 1

Now it looks like I have to dig another step down to find the M coefficient.
That means that I have to find y'(0), and that's where I get really unsure of what to do..

 

[Dick]

The value of lambda=(-3/4) tells you the homogeneous solution is e^(-3t/4). And you aren't following your rule to guess a particular form. Since the inhomogeneous part is sin(t), I would guess the form y=K*cos(t)+M*sin(t). Now find K and M. Put your guessed form into 4*dy/dt+3y=sin(t).

 

[I like Serena]

OK, I feel I'm getting a few steps in the right direction, but I'm not completely sure where I'm headed.

I'm impressed that you're diving in!

This is what I think so far:

Particular solution:
y_p = \frac{sin(t)}{3}

A particular solution should be a solution to the (inhomogeneous) DE.
Can you substitute this solution in the DE? See if it fits?

Homogeneous solution
y_h: 4 \frac{dy}{dx} + 3y = 0

Converting the derivate io \lambda and the variable (y) into 1:
y_h: 4 \lambda +3 = 0
\lambda = -\frac{3}{4}

OK now. The first thing I am uncertain of is where to use \lambda.

This is the homogeneous part of the DE, with the RHS (that is independent of y' and y) replaced by zero.
You have the right method, solving it with λ, but you should know that the resulting homogeneous solution is:
y_h = C_1 e^{\lambda t}

So let's just try something:

Since the inhomogeneous part is k sin (\alpha x) the guessed form of the homogeneous part would be K cos (\alpha x) + M sin (\alpha x)
(referring to http://en.wikipedia.org/wiki/Method...ents#Typical_forms_of_the_particular_integral )

In your case the inhomogeneous part is \sin t.
You should match this with k sin (\alpha x), deducing k and alpha.
The particular solution would then be of the form:
y_p = K \cos (\alpha t) + M \sin (\alpha t)

 

[Twinflower]

Aha, so I need to "transform" the particular part as well?
I've thought that the method of undetermined coefficients was only used in the homogeneous part.

But am I right in asuming that I can disregard the homogeneous part as the problem ask about the stationary condition and the homogeneous part "disappears" after about 5 time constants?

 

[Twinflower]

Meaning that
y(t) = y_p + y_h
y(t) = K cos (\alpha t) + M sin (\alpha t) + D e^{\lambda t}

But first I have to find the undetermined coefficients K and M.
Det undet. constant D in the homogeneous part is later found when K and M is found, and when inserting the starting condition y(0) = 0.

So basicly I am stuck at the particular part.
Following is a few suggestions/questions:y_p = K cos (\alpha t) + M sin (\alpha t) = ?
What should I put on the RHS to deduce K and M?

Am I right in asuming that I cannot put the derivate to zero because the resulting funtion is a sine where the derivative is constantly changing?

 

[ehild]

y_p = K cos (\alpha t) + M sin (\alpha t) = ?
What should I put on the RHS to deduce K and M?

You should put the function with the actual alpha value - what is it?

Then substitute yp into the original de and find the constants K and M.

Am I right in asuming that I cannot put the derivate to zero because the resulting funtion is a sine where the derivative is constantly changing?

Yes, that is right.

ehild

 

[Twinflower]

Is the function with the actual alpha value just sin(t) ?
And in that case, is alpha 1 ?

 

[ehild]

Yes, alpha=1.

ehild

 

[Twinflower]

Ok, thanks ehild.

Still leaves me with 2 unknowns in one equation.
So, if t = 0, I get that K = 0.

Should I derive the Kcos(αt)+Msin(αt) and insert K = 0 to find M ?

I am completely out in the bushes on this one :/

 

[ehild]

No, Yp is not equal to sin(t) and K is not zero.

Yp=Kcos(t)+Msin(t) is a solution of the original ode:

4 \frac{dy}{dt} + 3y = sin(t)

Find the derivative of Yp and substitute dYp/dt for dy/dt,
and Yp for y.

ehild

 

[I like Serena]

Start with:
4 \frac{dy}{dt} + 3y = sin(t)
Substitute:
y=K\cos t + M\sin t
What is the derivative of this expression? For you need to substitute that as well.

 

[Twinflower]

Start with:
4 \frac{dy}{dt} + 3y = sin(t)
Substitute:
y=K\cos t + M\sin t
What is the derivative of this expression? For you need to substitute that as well.

Ok..

the derivate of y = K cos(t) + M sin(t) equals M cos(t) - K sin(t)
This is now dYp/dt which I am now supposed to insert as dy/dt in the original equation:
(also y is swapped with yp)

4\times M cos(t) - K sin(t) + 3( K cos(t) + M sin(t)) = sin(t)

(still in the bushes picking cherries)

 

[ehild]

Ok..

the derivate of y = K cos(t) + M sin(t) equals M cos(t) - K sin(t)


4\times M cos(t) - K sin(t) + 3( K cos(t) + M sin(t)) = sin(t)

(still in the bushes picking cherries)

You miss some parentheses.

ehild

 

[ehild]

After you have put in the missing parentheses, collect the cos(t) terms and also the sin(t) terms. The result is something like a cos(t)+b sin(t)=0.
What did yo get?
(If you find any cherries send over some...)

ehild

 

[Twinflower]

After you have put in the missing parentheses, collect the cos(t) terms and also the sin(t) terms. The result is something like a cos(t)+b sin(t)=0.
What did yo get?
(If you find any cherries send over some...)

ehild

Ok, new parantheses is ordered and installed as requested:


4\times (M cos(t) - K sin(t)) + 3( K cos(t) + M sin(t)) = sin(t)

Moving the RHS to the coolhandside:

4\times (M cos(t) - K sin(t)) + 3\times (K cos(t) + M sin(t)) - sin(t) = 0



(And by the way:

 

[ehild]

Ok, new parantheses is ordered and installed as requested:4\times (M cos(t) - K sin(t)) + 3( K cos(t) + M sin(t)) = sin(t)

Moving the RHS to the coolhandside:

4\times (M cos(t) - K sin(t)) + 3\times (K cos(t) + M sin(t)) - sin(t) = 0(And by the way:

Thank you for the cherries! Sweet and juicy and red and big...
But what about collecting the terms with sin(t) and cos(t)?

I meant cos(t)(4M+3K) +sin(t)(3M-4K-1)= 0

This equation must be valid for every t. What do you get for the unknow K and M in case t=?
And what comes out if t=pi/2, that is cos(t)=0?ehild

 

[Twinflower]

Ok, so that's going to be something like this:

7\times M cos(t) -(K+1) sin (t) = 0...right?

 

[ehild]

Ok, so that's going to be something like this:

7\times M cos(t) -(K+1) sin (t) = 0...right?

No... I did it in the previous post already. Read it. I have to leave for a while.

cos(t)(4M+3K) +sin(t)(3M-4K-1)= 0Solve for K and M.

 

[I like Serena]

 

[Twinflower]

I am more on this level of cherry picking:

http://tothevillagesquare.org/images/cherry-picking-numbers.jpg

I also regret to inform you that I have officially given up. This kind of problem is unlikely to show up on my exam tomorrow, and *if* one pops up, I can live with not completing it.
It won't ruin my A (or B or C or D for that matter).

I'm sorry to dissappoint you guys, after all your effort. Hopefully my teacher will cover this topic more thouroughly during the next semester. (which I've heard is almost a dedicated differential equation semester)

 

[I like Serena]

Aaaah, you like 'em big!

Will you let us know how your exam went?
(And if something like this popped up?)

 

[Twinflower]

Of course :)

(and btw, sorry about the huge dimmensions )

 

[Twinflower]

It went very well for my exam, and luckily no differential equations containing those spooky sine functions :)

After studying the solution which was published earlier this evening, I feel like I'm in the right spot for an A :)

 

[I like Serena]

Good!

And I guess we'll get back to DE's later. ;)