Question on showing general formula of solution

[RoboNerd]

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!

 

[Ray Vickson]

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.

 

[pasmith]

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
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 a and b the quantities e^{a + ib} and e^{a - ib} are complex conjugates, so it follows that the linearly independent combinations <br /> \frac12(e^{a + ib} + e^{a - ib}) and \frac1{2i}(e^{a + ib} - e^{a - ib}) are real, and you can easily establish that they are equal to e^{a}\cos b and e^a \sin b respectively.

 

[RoboNerd]

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

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

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!