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## Homework Statement

Find the general solution:

y'-3y=(y^2)

## Homework Equations

## The Attempt at a Solution

divide both sides by y^2

y'(y^-2) -3(y^-1) = 1

we know v=y^(n-1)

v=y^-1

v'=d/dx(y^-1)

v'=-(y^-2) y'

plug it back into

y'(y^-2) -3(y^-1) = 1

-v'-3v=1

this is where I think I am making a mistake

im putting it into a 1st ODE by making v' positive

v'+3v=-1

then u(x)=e^(3x)

=-{[e^(3x)]/3+C}/e^(3x)

=-1/3+C/e^(3x)

the solution given to me is

1/((Ce^(-3x))-1/3)

Any help is appreciated. thanks!