Question on showing general formula of solution

In summary, this person is struggling to solve a homework problem and is in need of a hint. They have found a quadratic formula for solving for d, but if alpha^2 - omega^2 < 0, then their d is not real. They have tried substituting in sin(d*t) and cos(d*t) into the differential equation but are stuck. If anyone can help, that would be much appreciated.
  • #1
RoboNerd
410
11

Homework Statement


show that the general solution of the differential equation d^2/dt^2 + 2 *alpha * dr/dt + omega^2 * r = 0,

where alpha and w are constant and R is a function of time "t" is R = e^(-alpha * t) * [ C1*sin( sqrt(omega^2 - alpha^2) * t) + C2*cos( sqrt(omega^2 - alpha^2) * t)

IF alpha^2 - omega^2 < 0.I was struggling over how to solve this problem and I would be very grateful if someone have me a badly needed hint.

Homework Equations



None come to mind as of now[/B]

The Attempt at a Solution



I said let R = f(t) = e^(d * t).

I then had the expression when I plugged it into the differential equation

d^2 * e^(d * t) + 2 *alpha * d * e^(d * t) + omega^2 * e^(d * t) = 0

I canceled the e^(d * t) term and I got the characteristic equation:
d^2 + 2*alpha*d + omega^2 = 0

I then applied the quadratic formula to solve for d:
d = [ -2 *alpha +/- sqrt( (2*alpha)^2 - 4*omega^2 ) ] / 2.

I thus get d = -alpha +/- sqrt (alpha^2 - omega^2 )

Now I have a huge problem. If alpha^2 - omega^2 < 0, then my d is non-real and I can not proceed from here.
I tried substituting in the functions sin(d*t) and cos(d*t) instead of e^(d*t) into the differential equation and solving, but then I become stuck as I can not eliminate the plugged in functions at all from the general expression.

Could anyone please be kind enough to help?

Thanks, and have a great day!
 
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  • #2
RoboNerd said:

Homework Statement


show that the general solution of the differential equation d^2/dt^2 + 2 *alpha * dr/dt + omega^2 * r = 0,

where alpha and w are constant and R is a function of time "t" is R = e^(-alpha * t) * [ C1*sin( sqrt(omega^2 - alpha^2) * t) + C2*cos( sqrt(omega^2 - alpha^2) * t)

IF alpha^2 - omega^2 < 0.I was struggling over how to solve this problem and I would be very grateful if someone have me a badly needed hint.

Homework Equations



None come to mind as of now[/B]

The Attempt at a Solution



I said let R = f(t) = e^(d * t).

I then had the expression when I plugged it into the differential equation

d^2 * e^(d * t) + 2 *alpha * d * e^(d * t) + omega^2 * e^(d * t) = 0

I canceled the e^(d * t) term and I got the characteristic equation:
d^2 + 2*alpha*d + omega^2 = 0

I then applied the quadratic formula to solve for d:
d = [ -2 *alpha +/- sqrt( (2*alpha)^2 - 4*omega^2 ) ] / 2.

I thus get d = -alpha +/- sqrt (alpha^2 - omega^2 )

Now I have a huge problem. If alpha^2 - omega^2 < 0, then my d is non-real and I can not proceed from here.
I tried substituting in the functions sin(d*t) and cos(d*t) instead of e^(d*t) into the differential equation and solving, but then I become stuck as I can not eliminate the plugged in functions at all from the general expression.

Could anyone please be kind enough to help?

Thanks, and have a great day!

The absolutely 100% standard method of showing something is a solution is to substitute it into the equation to see if it "works". you have an explicit formula for r(t), so nothing prevents you from taking the derivatives, etc.
 
  • #3
RoboNerd said:

Homework Statement


show that the general solution of the differential equation d^2/dt^2 + 2 *alpha * dr/dt + omega^2 * r = 0,

where alpha and w are constant and R is a function of time "t" is R = e^(-alpha * t) * [ C1*sin( sqrt(omega^2 - alpha^2) * t) + C2*cos( sqrt(omega^2 - alpha^2) * t)

IF alpha^2 - omega^2 < 0.I was struggling over how to solve this problem and I would be very grateful if someone have me a badly needed hint.

Homework Equations



None come to mind as of now[/B]

The Attempt at a Solution



I said let R = f(t) = e^(d * t).

I then had the expression when I plugged it into the differential equation

d^2 * e^(d * t) + 2 *alpha * d * e^(d * t) + omega^2 * e^(d * t) = 0

I canceled the e^(d * t) term and I got the characteristic equation:
d^2 + 2*alpha*d + omega^2 = 0

I then applied the quadratic formula to solve for d:
d = [ -2 *alpha +/- sqrt( (2*alpha)^2 - 4*omega^2 ) ] / 2.

I thus get d = -alpha +/- sqrt (alpha^2 - omega^2 )

Now I have a huge problem. If alpha^2 - omega^2 < 0, then my d is non-real and I can not proceed from here.

Have you seen the identity
[tex]e^{ix} = \cos x + i \sin x[/tex] before? It's very important to the theory of second-order linear constant coefficient ODEs. Now for real [itex]a[/itex] and [itex]b[/itex] the quantities [itex]e^{a + ib}[/itex] and [itex]e^{a - ib}[/itex] are complex conjugates, so it follows that the linearly independent combinations [tex]
\frac12(e^{a + ib} + e^{a - ib})[/tex] and [tex]\frac1{2i}(e^{a + ib} - e^{a - ib})[/tex] are real, and you can easily establish that they are equal to [itex]e^{a}\cos b[/itex] and [itex]e^a \sin b[/itex] respectively.
 
  • #4
pasmith said:
Have you seen the identity
eix=cosx+isinxeix=cos⁡x+isin⁡x​
e^{ix} = \cos x + i \sin x before? It's very important to the theory of second-order linear constant coefficient ODEs. Now for real aaa and bbb the quantities ea+ibea+ibe^{a + ib} and ea−ibea−ibe^{a - ib} are complex conjugates, so it follows that the linearly independent combinations
12(ea+ib+ea−ib)12(ea+ib+ea−ib)​
\frac12(e^{a + ib} + e^{a - ib}) and
12i(ea+ib−ea−ib)12i(ea+ib−ea−ib)​
\frac1{2i}(e^{a + ib} - e^{a - ib}) are real, and you can easily establish that they are equal to eacosbeacos⁡be^{a}\cos b and easinbeasin⁡be^a \sin b respectively.

Hi. No, I have not seen this identity before. Thanks for drawing it to my attention.

I understand how one might be able to simplify the expression with the given identity, but I do not understand why one would after finding two complex conjugate expressions do the linear combinations
pasmith said:
12(ea+ib+ea−ib)12(ea+ib+ea−ib)​
\frac12(e^{a + ib} + e^{a - ib}) and
12i(ea+ib−ea−ib)​

which you say are equal to the final results
pasmith said:
eacosbeacos⁡be^{a}\cos b and easinbeasin⁡be^a \sin b
.

Could you please explain why you feel the need to take the linear combinations by adding/subtracting the two conjugate pairs and then dividing by 2 or 2i respectively?
What is the theory behind your actions? How does taking these linear combinations help?

Thanks so much for your help, and have a great day!
 

Related to Question on showing general formula of solution

1. What is the general formula for a solution?

The general formula for a solution is C = m/V, where C is the concentration of the solution, m is the mass of the solute, and V is the volume of the solution.

2. How do you calculate the concentration of a solution?

The concentration of a solution can be calculated by dividing the mass of the solute (in grams) by the volume of the solution (in liters). The resulting value is usually expressed in units of molarity (M).

3. Why is it important to know the general formula for a solution?

Knowing the general formula for a solution allows scientists to accurately calculate the concentration of a solution and make precise measurements in experiments. It also helps in understanding the properties and behavior of different solutions.

4. Are there any exceptions to the general formula for a solution?

Yes, there are exceptions to the general formula for a solution. For example, in some cases, the solute may not completely dissolve in the solvent, resulting in a different formula for the concentration. Additionally, some solutions may require different units of measurement, such as molality or percent concentration.

5. Can the general formula for a solution be used for all types of solutions?

The general formula for a solution can be used for most types of solutions, including aqueous solutions, solutions in non-aqueous solvents, and solutions with multiple solutes. However, it may not be applicable to solutions that involve reactions or significant changes in volume or mass.

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