Differential equations

**Apprentice123** · December 1st 2009, 08:14 AM

Originally Posted by lvleph

I guess I don't understand what you are asking. If we assume that $-A\sin wt + B \cos wt \ne 0$ this leads you to conclude the only time this is zero is when $mw^2 + cw + k = 0.$ Therefore, the we must solve for the roots of the characteristic equatioin $mw^2 + cw + k = 0.$ Again, I am not sure what you are having issues with, and so I am sorry I am not of much help.

After finding the roots
$w = \dfrac{-c \pm \sqrt{c^2 - 4mk}}{2m}$

How do I determine the equation:
$x(t) = C_1 e^{w_1t} + C_2e^{w_2t}$
???

**lvleph** · December 1st 2009, 08:17 AM

Refer to posts #9 and #11.

**Apprentice123** · December 1st 2009, 09:12 AM

Originally Posted by lvleph

Refer to posts #9 and #11.

Post #9 and #11 explain how to obtain C1 and C2. I wonder how after finding the roots get the equation

**lvleph** · December 1st 2009, 09:24 AM

Oh okay, sorry for the confusion. The characteristic polynomial comes from assuming a solution, i.e., we assume $x(t) = e^{wt}$ and plug this into the originial DEQ. Thus we obtain the following
$m \frac{d^2x}{dt^2} \left[e^{wt}\right] + c \frac{dx}{dt}\left[e^{wt}\right] + k e^{wt}$
which leads us to
$(mw^2 + cw + k)(e^{wt}) = 0.$
This tells us the characteristic equation is
$mw^2 + cw + k = 0.$
When we find the roots to this equation it tells us that we have one solution containing $w_1$ and another solution containing $w_2$ , i.e., we have one solution $x_1(t) = C_1 e^{w_1t}$ and a second solution $x_2 = C_2 e^{w_2 t}.$ However, since both these are solutions to the differential equation we know that a linear combination of the two is also a solution to the differential equation, and thus we obtain the general solution
$<br /> x(t) = x_1(t) + x_2(t) = C_1 e^{w_1t} + C_2e^{w_2t}.<br />$
I really hope I finally answered your question. I suppose after all this you will probably understand the problem more thoroughly.

**Apprentice123** · December 1st 2009, 09:54 AM

Originally Posted by lvleph

Oh okay, sorry for the confusion. The characteristic polynomial comes from assuming a solution, i.e., we assume $x(t) = e^{wt}$ and plug this into the originial DEQ. Thus we obtain the following
$m \frac{d^2x}{dt^2} \left[e^{wt}\right] + k \frac{dx}{dt}\left[e^{wt}\right] + c e^{wt}$
which leads us to
$(mw^2 + kw + c)(e^{wt}) = 0.$
This tells us the characteristic equation is
$mw^2 + kw + c = 0.$
When we find the roots to this equation it tells us that we have one solution containing $w_1$ and another solution containing $w_2$ , i.e., we have one solution $x_1(t) = C_1 e^{w_1t}$ and a second solution $x_2 = C_2 e^{w_2 t}.$ However, since both these are solutions to the differential equation we know that a linear combination of the two is also a solution to the differential equation, and thus we obtain the general solution
$<br /> x(t) = x_1(t) + x_2(t) = C_1 e^{w_1t} + C_2e^{w_2t}.<br />$
I really hope I finally answered your question. I suppose after all this you will probably understand the problem more thoroughly.

Thank you helping me a lot.Sorry, but why we assume $x(t) = e^{wt}$ ???

The equation would be:
$m \frac{d^2x}{dt^2} \left[e^{wt}\right] + c \frac{dx}{dt}\left[e^{wt}\right] + k e^{wt}$ ???

**lvleph** · December 1st 2009, 10:00 AM

Your absolutely correct, I made a mistake and have corrected the post. We assume this is the solution, because we want
$m x''(t) + c x'(t) + k x(t) = 0.$
This will only occur if $x'',x',x$ have the same form. Well, we know that the derivative of $e^t$ is itself, so this clues us into the type of solution we need, i.e., $e^{wt}$ .

**Apprentice123** · December 1st 2009, 10:14 AM

Originally Posted by lvleph

Your absolutely correct, I made a mistake and have corrected the post. We assume this is the solution, because we want
$m x''(t) + c x'(t) + k x(t) = 0.$
This will only occur if $x'',x',x$ have the same form. Well, we know that the derivative of $e^t$ is itself, so this clues us into the type of solution we need, i.e., $e^{wt}$ .

Thank you very much. To determine C1 and C2 need to know the initial conditions? I know you already explained, the more I do not get the initial solutions in terms of f (x) and g (x)

**lvleph** · December 1st 2009, 10:17 AM

The $f(x),g(x)$ are just general forms for the intial conditions. They can be constants or functions of $x$ .

**Apprentice123** · December 1st 2009, 10:27 AM

Originally Posted by lvleph

The $f(x),g(x)$ are just general forms for the intial conditions. They can be constants or functions of $x$ .

The first I understood
$x(0) = C_1e^{w_1 \cdot 0} + C_2e^{w_2\cdot 0} = C_1 + C_2 = L$ OK

But the second condition
$x_t(0) = w_1\cdot C_1 e^{w_1\cdot 0} + w_2 \cdot C_2 e^{w_2\cdot 0} = w_1\cdot C_1 + w_2 \cdot C_2 = 0$

How did you find this equation?

**lvleph** · December 1st 2009, 01:13 PM

Well, I was given an example where $g(x) = 0$ and so I took the derivative of $x(t) = C_1 e^{w_1t} + C_2 e^{w_2t}$
Then I plugged in $t= 0$ , but I knew from that $x(0) = 0$ . Putting this all together we could solve for $C_1,C_2$

**Apprentice123** · December 1st 2009, 03:15 PM

Originally Posted by lvleph

Well, I was given an example where $g(x) = 0$ and so I took the derivative of $x(t) = C_1 e^{w_1t} + C_2 e^{w_2t}$
Then I plugged in $t= 0$ , but I knew from that $x(0) = 0$ . Putting this all together we could solve for $C_1,C_2$

Thank you friend. You did x'(0), initial velocity, I correct ?

**lvleph** · December 1st 2009, 03:18 PM

Yes, but how to solve for $C_1,C_2$ will always depend on the problem and what is given for initial conditions. Typically you will be given $x(0),x'(0)$

**Apprentice123** · December 2nd 2009, 09:00 AM

Originally Posted by lvleph

Yes, but how to solve for $C_1,C_2$ will always depend on the problem and what is given for initial conditions. Typically you will be given $x(0),x'(0)$

In the underdamped system, we have $c^2-4mk<0$ , Complex Roots.

How can we determine the equation of the position in function of time. x(t) ?

**lvleph** · December 2nd 2009, 09:08 AM

The solution will be of the form
$x(t) = C_1 \cos wt + C_2 \sin wt$ ,
where $w = \dfrac{-c+\sqrt{c^2 - 4mk}}{2m}$ . This comes from eulers formula $e^{i\theta} = \cos \theta + i \sin \theta$ .

**Apprentice123** · December 2nd 2009, 09:21 AM

Originally Posted by lvleph

The solution will be of the form
$x(t) = C_1 \cos wt + C_2 \sin wt$ ,
where $w = \dfrac{-c+\sqrt{c^2 - 4mk}}{2m}$ . This comes from eulers formula $e^{i\theta} = \cos \theta + i \sin \theta$ .