Finding the general solution of this ODE

In summary, the conversation involves solving a differential equation for unforced under-damped oscillatory motion with given constants p and omega. The general solution can be expressed in the form x = {\mathop{\rm Re}\nolimits} ^{ - pt} \cos \left( {\sqrt {\omega ^2 - p^2 } t - \alpha } \right), and it can be simplified using the trig identity Cos(a+ b)= Cos(a)Cos(b)- Sin(a)Sin(b). The solution also involves finding the values of C1, C2, and R.
  • #1
Benny
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0
Hi can someone help me out with the following question?

Q. The differential equation for unforced under-damped oscillatory motion can be written [tex]\mathop x\limits^{..} + 2p\mathop x\limits^. + \omega ^2 x = 0[/tex] where the constants p and omega satisfy [tex]0 < p < \omega [/tex].

Find the general solution of this differential equation, and show that this solution can be expressed in the form [tex]x = {\mathop{\rm Re}\nolimits} ^{ - pt} \cos \left( {\sqrt {\omega ^2 - p^2 } t - \alpha } \right)[/tex] where R and alpha are constants.

[tex]
x\left( t \right) = e^{ - pt} \left( {c_1 \cos \left( {t\sqrt {p^2 - \omega ^2 } } \right) + c_2 \sin \left( {t\sqrt {p^2 - \omega ^2 } } \right)} \right)
[/tex] by using 0 < p < omega.

As can be seen I haven't been able to get too far. I can't think of a way to get to the answer. Can someone please help me out?
 
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  • #2
Yes, the characteristic equation is r2+ 2pr+ &omega;= 0 (You have &omega;2 in the differential equation but your solution seems to be for &omega; only) and the solutions to that are [itex]-p\pm\sqrt{p^2-\omega^2}= -p\pm\sqrt{\omega^2-p^}i[/itex].
Yes, the general solution to the differential equation is
[tex]e^{-pt}\left(C_1 Cos(\sqrt{\omega^2-p^2}t)+C_2 Sin(\sqrt{\omega^2-p^2}t\right)[/tex].

Now, all you want to do is show that that can be expressed as a single cosine.

Use the trig identity: Cos(a+ b)= Cos(a)Cos(b)- Sin(a)Sin(b).

Obviously, you want [itex]b= \sqrt{\omega^2-p^2}[/itex]. You will need to find a such that Cos(a)= C1, Sin(a)= C2. Obviously that can't happen unless the sum of their squares is equal to 1: that's where R comes in: What is the sum of the squares now?
 
  • #3


Sure, I would be happy to help you with this question! First, let's start by looking at the general form of the given differential equation:

\mathop x\limits^{..} + 2p\mathop x\limits^. + \omega ^2 x = 0

We can see that this is a second-order linear homogeneous differential equation, which means that the general solution will have the form:

x(t) = c1e^rt + c2e^st

where r and s are the roots of the characteristic equation:

r^2 + 2pr + \omega^2 = 0

Solving this quadratic equation, we get:

r = -p ± \sqrt{p^2 - \omega^2}

Since we are given that 0 < p < \omega, we know that p^2 - \omega^2 < 0, which means that the roots will be complex numbers. This indicates that the solution will involve trigonometric functions.

Now, let's use the method of undetermined coefficients to find the particular solution. We will assume that the particular solution has the form:

x(t) = A\cos(\alpha t) + B\sin(\alpha t)

where A and B are constants to be determined and \alpha is a constant to be determined as well.

Plugging this into the differential equation, we get:

-A\alpha^2\cos(\alpha t) - B\alpha^2\sin(\alpha t) + 2pA\alpha\sin(\alpha t) - 2pB\alpha\cos(\alpha t) + \omega^2(A\cos(\alpha t) + B\sin(\alpha t)) = 0

Equating the coefficients of \cos(\alpha t) and \sin(\alpha t) on both sides, we get the following system of equations:

-A\alpha^2 + \omega^2A + 2pB\alpha = 0
-B\alpha^2 + \omega^2B - 2pA\alpha = 0

Solving this system of equations, we get:

A = \frac{\omega^2}{\alpha^2 - \omega^2}B
and
\alpha = \pm\sqrt{p^2 - \omega^2}

Plugging these values back into the particular solution, we get:

x(t)
 

1. What is the general solution of an ODE?

The general solution of an ODE (ordinary differential equation) is a formula that gives all possible solutions to the equation. It is a family of solutions that includes all possible initial conditions.

2. How do you find the general solution of an ODE?

To find the general solution of an ODE, you need to solve the equation using various methods such as separation of variables, integrating factors, or substitution. Once you have found the general solution, you can then plug in specific initial conditions to find a particular solution.

3. What is the difference between the general solution and a particular solution of an ODE?

The general solution is a formula that includes all possible solutions to an ODE, while a particular solution is a specific solution that satisfies the given initial conditions. The general solution represents a family of solutions, while a particular solution represents a single solution.

4. Can an ODE have more than one general solution?

No, an ODE can only have one general solution. However, the general solution may include constants that can take on different values, resulting in different particular solutions.

5. Why is finding the general solution important in solving ODEs?

Finding the general solution is important because it allows us to find all possible solutions to an ODE, instead of just a single solution. This is especially useful in real-world applications where there may be multiple solutions that satisfy the given initial conditions.

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