Engineering Current through an Inductor in an RL circuit

[maxpound18]

Homework Statement

Find i₀(t) see attached picture

Homework Equations

i_o(t) = i_infinite + (i_0^- - i_infinite)e^-Rt/L

The Attempt at a Solution

I first re-drew the diagram for when the switch has NOT been closed, which is esssentially the same thing without the 37.5 V Source and the adjacent 10 ohm resistor.

I then found V(phi) using ohms law since the current through both the resistors should be the same. The current is (i=V/R) which is (250/50= 5 Amps)
So, V(phi) = iR which is (5amps*10ohms = 50 V)

I then applied Kirkchoff's Current Law to the node between the inductor, the dependent source, and the 10 ohm resistor.
0 = -(250/50) + 9V(phi) + i₀
0 = -5 + 9V(phi) + i₀
0 = -5 + 9(50) + i₀
i₀ = -445 Amps (This seems wrong) but moving on...then i drew the diagram for when t > 0

 

[maxpound18]

sorry i left out a resistor.

THIS is the correct drawing

 

[maxpound18]

For when t > 0

here are source conversions.

 

[maxpound18]

and more conversions

 

[maxpound18]

My question at this point is how do I reduce THIS circuit down to one resistor with the inductor? And what am i supposed to do with the Dependent Current Source.

The values I calculated for V(phi) and i₀ are 44.4V and -395.2A respectively.