Question

Solve the Bernoulli equation y'+y=y^2.

Answer 1

Here's the work:
n=2
v=1/y
y=1/v
-1/v^2*dv/dx+1/v=1/v^2
dv/dx-v=-1
u(x)=e^-x -->the integrating factor
(e^-x)v'-ve^-x=-e^-x
e^-x*v=e^-x+C
v=(e^-x+C)/(e^-x)
1/y=(e^-x+C)/(e^-x)
y=e^(-x)/(e^(-x)+C)
But that's not the answer. The answer is y=1/(1-ce^x) in the book. What's wrong?

Answer 2

No, you're right. That *is* the answer. Try multiplying by \(\dfrac{e^x}{e^x}\).

Answer 3

Wait a minute.

Answer 4

I got 1/(1+ce^x) instead of 1/(1-ce^x).

Answer 5

Well \(C\) is an arbitrary constant. I don't think it matters. Or maybe it's a typo.

Wolfram Alpha has the same answer as you: http://www.wolframalpha.com/input/?i=y%27%2By%3Dy%5E2

Answer 6

Does anyone know the input of this problem on Ti-89?

Answer 7

Refer to the attachment from WolframAlpha.

Answer 8

Thank you guys for the help.