- #1

thaiqi

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- TL;DR Summary
- How to solve this vibration equation?

[tex] \ddot{x} + \omega_{0}^2 x = {e \over m} E_{x} [/tex]

Hello,everyone:

I got an equation:

[tex]

\ddot{x} + \omega_{0}^2 x = {e \over m} E_{x}

[/tex]

I know the solution is:

[tex]

x(t) = {e \over {\omega_{0} m}} \int_{0}^{t} E_{x}(\xi) \sin{ \omega_{0} } (t - \xi) d \xi \\

[/tex]

[tex] x(0) = \dot{x} (0) = 0[/tex]

My attempt to verify this solution is by using formula:

[tex]

{d \over dx } \int_{\alpha (x)}^{\beta (x)} f(x,y)dy

= \int_{\alpha (x)}^{\beta (x)} { {\partial f(x,y)} \over {\partial x}} dy

+ f[x, \beta (x)] \beta^{\prime}(x)

- f[x, \alpha (x)] \alpha^{\prime}(x)

[/tex]

But I don't know how the physicist obtained it. Can anyone give an answer? Thanks.

I got an equation:

[tex]

\ddot{x} + \omega_{0}^2 x = {e \over m} E_{x}

[/tex]

I know the solution is:

[tex]

x(t) = {e \over {\omega_{0} m}} \int_{0}^{t} E_{x}(\xi) \sin{ \omega_{0} } (t - \xi) d \xi \\

[/tex]

[tex] x(0) = \dot{x} (0) = 0[/tex]

My attempt to verify this solution is by using formula:

[tex]

{d \over dx } \int_{\alpha (x)}^{\beta (x)} f(x,y)dy

= \int_{\alpha (x)}^{\beta (x)} { {\partial f(x,y)} \over {\partial x}} dy

+ f[x, \beta (x)] \beta^{\prime}(x)

- f[x, \alpha (x)] \alpha^{\prime}(x)

[/tex]

But I don't know how the physicist obtained it. Can anyone give an answer? Thanks.