Manipulating result of partial fractions

[aximus]

Homework Statement

Solve this IVP:

y'=(y-9x)^2 ; y(0)=1

Given a hint: Use the substitution v=y-9x and partial fractions.

Homework Equations

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The Attempt at a Solution

I was able to solve this DE through partial fractions, etc until I ended up at this point

ln (v-3/v+3) = 6x+C

or

(v-3)/(v+3) = Ae^6x (where A = e^C)

I won't show my entire working (unless requested) as the partial fractions and simplification is lengthy, but I know I am correct up to this point as the answers get to this point and then just tell me to use some simple algebraic rearranging to get:

y=9x-v=3(A*exp(6x)+1)/(1-A*exp(6x))

So yeah, I am stuck at manipulating this fraction about and to solve for v (and then y). I'm sure that this involves a very quick manipulation I should know, but it is really bothering me.

Thank you for your help.

 

[HallsofIvy]

It's not difficult to solve that equation for v: multiply both sides of
[tex]\frac{v-3}{v+3}= Ae^{6x}[/tex]
by v+ 3 to get
[tex]v- 3= Ae{6x}(v+ 3}= Ae^{6x}v+ 3Ae^{6x}[/tex]
Now, put everything involving v on the left, everything not involving v on the right:
[tex]v- Ae^{6x}v= 3Ae^{6x}+ 3[/tex]
[tex]v(1- Ae^{6x})= 3(Ae^{6x}+ 1)[/tex]
and divide both sides by [itex]1- Ae^{6x}[/itex]:
[tex]v= 3\frac{Ae^{6x}+ 1}{1- Ae^{6x}}[/tex]

Now, because v= y- 9x,
[tex]y= 9x+ v= 9x+ 3\frac{Ae^{6x}+ 1}{1- Ae^{6x}}[/tex]