Differential Equations: Bernoulli Equation

[dmoney123]

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!

 

[LCKurtz]

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)

which is equal to $-\frac 1 3 + Ce^{-3x}$

the solution given to me is

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

Any help is appreciated. thanks!

Isn't that the reciprocal of what you have? Remember $v = \frac 1 y$.

 

[dmoney123]

i always feel so stupid when i come on the site. yes it is the reciprocal. thank you for your help