Solve Swimmer's Speed to Drift 1 Mile: 4 mi/h

[JJRKnights]

I found an answer, but I am skeptical.

What must the swimmer's speed, vs, be in order that he drifts only 1 mile downstream as he crosses the river?

The width of the river = 2 miles
The velocity of the river, vr = 3(1-x^2)

Relevant equations:

tanα=vr/vs
dy/dx = tanα
vr = v0(1-x^2/a^2)

Image:

So I started off with:

tanα=vr/vs
plugged in dy/dx for tanα, and vr = 3(1-x^2)

dy/dx = (3/vs)(1-x^2)
I assumed that vs was a constant.
∫dy/dx = (3/vs)∫(1-x^2)dx

y = (3/vs)(x - 1/3 x^3) + C

plugged in initial values y(-1) = 0,

0 = (3/vs)((-1) - 1/3(-1)^3) + C
((-1) - 1/3(-1)^3) = -2/3
0 = (3/vs)(-2/3) + C
C = 2/vs

now I have the equation all with 1 unknown:

y = (3/vs)(x - 1/3 x^3) + 2/vs

here is the part where I just improvised not really knowing what to do...
I figured that if I used the initial values, I wouldn't get my vs, so I used the final values:
y(1)=1

1 = (3/vs)(1 - 1/3) + 2/vs
and for this one I get that:
vs = 4 mi/h

 

[LCKurtz]

Looks OK to me. The $a$ in the picture is evidently 1.