Solve Mutual Inductance Homework: V_0cosωt, L, M, R, i_1 & i_2

[papkin]

Homework Statement

A coil of self inductance $$L_{1}$$ is coupled to a second coil of self inductance $$L_{2}$$ and resitance $$R$$. The coefficient of mutual inductance is $$M$$. The second coil is short circuited whereas the first coil is connected via a switch, to a voltage source $$V_{0}cos \omega t$$

Homework Equations

The switch is closed at time t = 0. Find an expression for the current which then flows in the second coil, as a function of time.

The Attempt at a Solution

$$V_{0}cos \omega t = L \frac {di_{1}}{dt}$$
$$M \frac {di_{1}}{dt}= - i_{2}R$$
Transferring into Laplace domain:
$$I_{1}(s) = \frac {v_{0}}{L_{1}} \frac {1}{s^{2}+\omega^{2}}$$
$$I_{2}(s) = \frac {-Mv_{0}}{L_{1}R} \frac {1}{s^{2}+\omega^{2}}$$
and back to the real world:
$$i_{2} = \frac {-Mv_{0}}{L_{1}R\omega} sin \omega t$$
and that just doesn't feel right.

Thanks, in advance.
EDIT: Sorry, lost an s somewhere along the way, got $$\frac {-Mv_{0}}{L_{1}R} cos \omega t$$ now.
I think that's correct, but wouldn't mind somebody reassurance by someone more experienced.

 

[AvroArrow]

The Attempt at a Solution V_{0}cos \omega t = L_{1} \frac {di_{1}}{dt} M \frac {di_{1}}{dt}= - i_{2}R Transferring into Laplace domain: I_{1}(s) = \frac {v_{0}}{L_{1}} \frac {1}{s^{2}+\omega^{2}} I_{2}(s) = \frac {-Mv_{0}}{L_{1}R} \frac {1}{s^{2}+\omega^{2}} and back to the real world: i_{2} = \frac {-Mv_{0}}{L_{1}R\omega} sin \omega t and that just doesn't feel right.Thanks, in advance.EDIT: Sorry, lost an s somewhere along the way, got \frac {-Mv_{0}}{L_{1}R} cos \omega t now.I think that's correct, but wouldn't mind somebody reassurance by someone more experienced.