0
$\begingroup$

I need to write down the second order linear equation which has v(t) as its solution. $A>1$ and$g <\dfrac{ 4(A-1)}{A^2} $ . $v(t)$ is an underdamped harmonic oscillation.

Given: $u' = -g(Au+v) \\ v' = (A-1)u $

I know how to make a second order equation into a first order one, but no clue how to do the opposite.

Thanks. Any help is appreciated.

1 Answers 1

0

hint :$$\\\begin{cases}u' = -g(Au+v)\\v' = (A-1)u\end{cases}\\ \to \begin{cases}u' = -g(Au+v) &\to u''=-g(Au'+v') \\v' = (A-1)u\end{cases}\\$$now put $v'$ into first one $$ u''=-g(Au'+\color{red} {v'}) \to u''=-g(Au'+\color{red} {(A-1)u}) $$

  • 0
    Wouldn't that be an equation where u(t) is a solution, not v(t)? For v(t), I coudl do something similar to what you did by taking the derivative of v' and plugging in u' into the resulting v'' expression. But even in this case, v'' would be a statement with the variable u in it. – 2017-02-19
  • 0
    @JaneDoe :when you obtain $u(t)$ ,you can obtain $v(t)$ (by second equation) – 2017-02-20