Help solving y'=y(3-y)

just started ordinary differential equation class, and i have to find the general soln of the eqn: y'=y(3-y)... just started so i only know of the method of seperation of varibles.

Thought of multiplying the y through so it becomes:
y'=3y-y^2 then do..

dy/dx=3y-y^2 then... im stuck... I tried doing seperation of varibles by

Dividing 3y-y^2 to the other side so you get:

dy/3y-y^2=0? But that doesn't seem right.

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$$\frac{dy}{dx} = y(3-y)$$

$$\frac{dy}{y(3-y)} = dx$$

$$\int \frac{dy}{y(3-y)} = \int dx$$

Break up the left-hand side by the method of partial fractions.

after partial fractions and integrating both sides:

ln(y) - ln(y-3) = 3 x + C

by logarithmic identities:

ln(y/(y-3)) = 3 x + C

Taking the exponential of both sides:

y/(y-3) = C1 e^(3 x)

Where C1 = e^C

Solving for y:

y = 3 C1 e^(3 x)/(C1 e^(3 x)-1)

Well, geez... why don't ya just do the whole thing for him? Oh wait, you just did.

Way to encourage the joy of discovery. :uhh:

wurth_skidder_23 said:
after partial fractions and integrating both sides:

ln(y) - ln(y-3) = 3 x + C

by logarithmic identities:

ln(y/(y-3)) = 3 x + C

Taking the exponential of both sides:

y/(y-3) = C1 e^(3 x)

Where C1 = e^C

Solving for y:

y = 3 C1 e^(3 x)/(C1 e^(3 x)-1)

After my partial fraction decomposition I got:

-ln((y-3)/y)/3=x+c?

You have the same thing. Just change your - sign into an exponent and multiply both sides by 3.

how do you change a minus sigh into a exponent, im lost when you said that.

Use the property of logarithms: $$r log b = log b^{r}$$

where $$r = -1$$

and $$b = \frac{y-3}{y}$$

HallsofIvy