Node Volt Analysis: Find \frac{V_o}{V_i}

[Cursed]

Homework Statement

Find $$\frac{V_o}{V_i}$$

for

http://img685.imageshack.us/img685/8658/nodevoltage.png

Homework Equations

$$s=σ+jω$$, where σ=real component & jω=imaginary component

The Attempt at a Solution

Solution is
$$\frac{V_o}{V_i}=\frac{1}{s+2}$$

Solution manual says:
$$\frac{V_o-V_i}{s}+\frac{V_o}{s}+V_o=0$$

I don't know where that comes from. I don't understand why there are 2 s's
Shouldn't it be;

$$\frac{V_L-V_i}{R}+\frac{V_L-0}{Ls}+\frac{V_L-V_o}{R}=0$$

where $$V_L=V_o$$

so

$$\frac{V_o-V_i}{R}+\frac{V_o-0}{Ls}+\frac{V_o-V_o}{R}=0$$
$$\frac{V_o-V_i}{R}+\frac{V_o-0}{Ls}=0$$

That won't give me an s+2 in the denominator...

 

[rude man]

Homework Statement

Find $$\frac{V_o}{V_i}$$

for

http://img685.imageshack.us/img685/8658/nodevoltage.png

Homework Equations

$$s=σ+jω$$, where σ=real component & jω=imaginary component

The Attempt at a Solution

Solution is
$$\frac{V_o}{V_i}=\frac{1}{s+2}$$

Solution manual says:
$$\frac{V_o-V_i}{s}+\frac{V_o}{s}+V_o=0$$

I don't know where that comes from. I don't understand why there are 2 s's
Shouldn't it be;

$$\frac{V_L-V_i}{R}+\frac{V_L-0}{Ls}+\frac{V_L-V_o}{R}=0$$

where $$V_L=V_o$$

...

You seem to be using V_o for ground, whereas the diagram shows V_o = V_L.
You're close, fix that up, & shoot again.
You should call ground (the bottom node voltage) zero like everybody else does!