What is the solution to the differential equation dy/dx=(y-y2)/x for x≠0?

[mmekosh]

Homework Statement

Solve the differential equation: dy/dx=(y-y²)/x , for all x\neq 0

Homework Equations

Integration by Parts: \int u dv = u v - \int v du

\intlnx= 1/x + C
\int (1/x) = lnx + C
dy/dx lnx = 1/x
dy/dx 1/x = lnx

The Attempt at a Solution

dy/(y-y²)=dx/x

\int 1/(y-y²) dy = \int 1/x dx

\int (1/y)(1/(1-y))dy = lnx + C

(Integration by parts)
u=1/x dv=(1/(1-y))dy
du=lnydy v= -ln(1-y)dy

-lny / y + \int lny ln(1-y) dy

And then if I continue and do integration by parts again, it just goes back to the original integral.

 

[Mark44]

Instead of trying to do this by integration by parts, use the technique of partial fractions. In a nutshell you want to rewrite your integral on the left side:
\int \frac{dy}{y(1 - y)}

with the fraction broken up into two separate fractions, like so:
\int \left[\frac{A}{y} + \frac{B}{1 - y}\right]dy

What you need to do is to find constants A and B so that 1/(y(1 - y)) is identically equal to A/y + B/(1 - y).

 

[nobahar]

\intlnx= 1/x + C
\int (1/x) = lnx + C
dy/dx lnx = 1/x
dy/dx 1/x = lnx

Hello!
I don't mean to butt in, but can someone verify that:
\frac{d}{dx}(\frac{1}{x}) = \ln{x}?
Also, the integral:
\int{\ln{x}} \left \left dx = \frac{1}{x}?
I'm confused...
Many thanks.

 

[Altabeh]

Hello!
I don't mean to butt in, but can someone verify that:
\frac{d}{dx}(\frac{1}{x}) = \ln{x}?
Also, the integral:
\int{\ln{x}} \left \left dx = \frac{1}{x}?
I'm confused...
Many thanks.

I think you've confused the concept of integral with that of derivative as

\frac{d}{dx}(\ln(x)) = \frac{1}{x}
\int{\frac{1}{x}} \left \left dx = \ln(x)+C.

And yeah both are correct.

AB

 

[HallsofIvy]

Hello!
I don't mean to butt in, but can someone verify that:
\frac{d}{dx}(\frac{1}{x}) = \ln{x}?

No, no one can verify that- it's not true. What is true is that
\frac{d}{dx}(\frac{1}{x})= -\frac{1}{x^2}
because \frac{1}{x}= x^{-1} and the derivative of x^n is nx^{n-1}.

Also, the integral:
\int{\ln{x}} \left \left dx = \frac{1}{x}?
I'm confused...
Many thanks.

No, that's also not true. \int ln(x) dx= ln(x)(x- 1)+ C

You have these both backwards: d(ln(x))/dx= 1/x and \int 1/x dx= ln(x)+ C.

 

[nobahar]

Thanks both Altabeh and Halls.
I didn't consider them to be true, hence why I said I was confused. It was from the OP's relevant equations section.