- #1

Gyroscope

## Homework Statement

[tex]-kx=m\frac{d^2x}{dt^2}[/tex]

I don't know how to solve differential equations, can someone show me how to do it, with this example.

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- Thread starter Gyroscope
- Start date

- #1

Gyroscope

[tex]-kx=m\frac{d^2x}{dt^2}[/tex]

I don't know how to solve differential equations, can someone show me how to do it, with this example.

- #2

cristo

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This is an example of a second order differential equation; let's rewrite it as [tex]\frac{d^2x}{dt^2}+\frac{k}{m}x=0[/tex] This will have a general solution of the form x=Ae^{at}. Plug this into the equation and we obtain the equation a^{2}+k/m=0, which gives us the values for a; namely [itex]a=\pm \sqrt{-\frac{k}{m}}=\pm i\sqrt{\frac{k}{m}}[/itex]. Letting [itex]\sqrt{k/m}=\omega[/itex] we obtain [itex]x(t)=Ae^{i\omega t}+Be^{-i\omega t}[/itex]. Now, recognising that this is the form of a sum of cosine and sine functions, we obtain the general solution [itex]x(t)=C\cos(\omega t)+D\sin(\omega t)[/itex]

I hope you follow that; however, if you have not studied differential equations, then I would suggest first learning about first order differential equations, and then moving onto second order equations (although, I suspect you want this solution for a specific purpose).

[edit: I didn't notice that this was in the homework forum, and so a full solution shouldn't be given; however, I suspect from previous posts that you may be self-learning the subject, and so this is not a homework question]

I hope you follow that; however, if you have not studied differential equations, then I would suggest first learning about first order differential equations, and then moving onto second order equations (although, I suspect you want this solution for a specific purpose).

[edit: I didn't notice that this was in the homework forum, and so a full solution shouldn't be given; however, I suspect from previous posts that you may be self-learning the subject, and so this is not a homework question]

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- #3

Gyroscope

Why do you need both solutions?

- #4

cristo

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Why do you need both solutions?

Since both [itex]e^{i\omega t}[/itex] and [itex]e^{-i\omega t}[/itex] are solutions to the equation, then the general solution will be a linear combination of the two, with the constants determined by the boundary conditions.

- #5

Gyroscope

Thanks cristo. How can you pass from e^(something) to cosine and sine functions?

- #6

cristo

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Thanks cristo. How can you pass from e^(something) to cosine and sine functions?

Use the definition of complex exponential, namely [itex]e^{\pm i\theta}=\cos\theta \pm i\sin\theta[/itex]. With a bit of rearrangement, we find that [tex]\cos\theta =\frac{1}{2}(e^{i\theta}+e^{-i\theta})[/tex] and [tex]\sin\theta=\frac{1}{2i}(e^{i\theta}-e^{-i\theta})[/tex]

- #7

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## Homework Statement

[tex]-kx=m\frac{d^2x}{dt^2}[/tex]

I don't know how to solve differential equations, can someone show me how to do it, with this example.

While the method described above is very useful and practical, don't forget what the equation is asking:

Can you determine which functions x(t) are proportional to minus their second derivatives?

You don't need a [crash] course in differential equations to answer that question.

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