 #1
thaiqi
 161
 8
 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.