Solving differential equations (circular motion)

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Gatsby88
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Homework Statement



I have a differential equation of the form

[itex]\frac{dZ}{d\theta} + cZ = a cos \theta + b sin \theta[/itex]

Where [itex]Z = \frac{1}{2}\dot{\theta}^{2}[/itex]

I need to find the general solution of this equation. a, b and c are all constants.

Homework Equations



The questions suggests using this to help:

[itex]\int e^{\lambda x} (a cos x + bsin x ) = \frac{1}{1+\lambda^2}e^{\lambda x}(\lambda (a cos x + b sin x) a sin x - b cos x) + C[/itex]

The Attempt at a Solution



I just don't know how that integral is supposed to help me solve the equation. How does e become relevant to this function?

Im also a bit unsure about this.. If I integrate

[itex]\frac{1}{2}C \dot{\theta}^2[/itex]

with respect to θ, do I get

[itex]\frac{1}{2}C \theta ^2[/itex] ?
 
on Phys.org
Gatsby88 said:

Homework Statement



I have a differential equation of the form

[itex]\frac{dZ}{d\theta} + cZ = a cos \theta + b sin \theta[/itex]

Where [itex]Z = \frac{1}{2}\dot{\theta}^{2}[/itex]

I need to find the general solution of this equation. a, b and c are all constants.

Homework Equations



The questions suggests using this to help:

[itex]\int e^{\lambda x} (a cos x + bsin x ) = \frac{1}{1+\lambda^2}e^{\lambda x}(\lambda (a cos x + b sin x) a sin x - b cos x) + C[/itex]

The Attempt at a Solution



I just don't know how that integral is supposed to help me solve the equation. How does e become relevant to this function?

The first step involves solving for ##Z## in terms of ##\theta##. Are you familiar with the technique of using an integrating factor? You can read more about it here: http://en.wikipedia.org/wiki/Integrating_factor

Multiply by the appropriate integrating factor, and the reason for the hint should become very clear.

Im also a bit unsure about this.. If I integrate

[itex]\frac{1}{2}C \dot{\theta}^2[/itex]

with respect to θ, do I get

[itex]\frac{1}{2}C \theta ^2[/itex] ?

No. Remember that ##\dot \theta## signifies a derivative wrt time. There's no (simple) relationship between the two expressions.
 
Thank you.

Ah. Yes, we've done integrating factors earlier in the course.

The integrating factor will be

[itex]e^{\int {c}} = e^{c\theta}[/itex]

multiplying by this gives

[itex]e^{c\theta} \frac{dZ}{d\theta} + C e^{c\theta} Z = e^{c\theta}(a cos \theta + b sin \theta)[/itex]

And then

[itex]\frac{d}{d\theta}(e^{c\theta}Z) = e^{c\theta}(a cos \theta + b sin \theta)[/itex]

And then I can integrate both sides and use the integral given as well as substituting back in for Z at the end and everything is finished.

However, I largely just followed a 'process solution' for this from my textbook. I am unsure how the LHS goes from

[itex]e^{c\theta} \frac{dZ}{d\theta} + C e^{c\theta} Z[/itex]

to

[itex]\frac{d}{d\theta}(e^{c\theta}Z)[/itex]

Can someone explain that to me please?