Solve Separable Diff. Eqn.: (y-1)dx+x(x+1)dy=0

[Arman777]

Homework Statement

$(y-1)dx+x(x+1)dy=0$

Homework Equations

The Attempt at a Solution

[/B]
I multiplied both equation with, $\frac {1} {(y-1)x(x+1)}$ so I get
$\frac {dx} {x(x+1)}+\frac {dy} {y-1}=0$
taking integral for both sides
then I get
$ln(x)-ln(x+1)+ln(y-1)=ln(c)$
so
$ln(\frac {x(y-1)} {x+1})=ln(c)$
then I get $xy-x=c(x+1)$

Anyone who can notice where I am doing wrong ?

 

[Dick]

Homework Statement

$(y-1)dx+x(x+1)dy=0$

Homework Equations

The Attempt at a Solution

[/B]
I multiplied both equation with, $\frac {1} {(y-1)x(x+1)}$ so I get
$\frac {dx} {x(x+1)}+\frac {dy} {y-1}=0$
taking integral for both sides
then I get
$ln(x)-ln(x+1)+ln(y-1)=ln(c)$
so
$ln(\frac {x(y-1)} {x+1})=ln(c)$
then I get $xy-x=c(x+1)$

Anyone who can notice where I am doing wrong ?

What makes you think something is wrong?

 

[Arman777]

What makes you think something is wrong?

Im my book it says $xy+1=c(x+1)$ also in wolfram

 

[Dick]

Im my book it says $xy+1=c(x+1)$ also in wolfram

It's the same solution with a different choice of $c$. Substitute $c-1$ for $c$ in your solution and you'll get the book solution.

 

[Dick]

The multiplication by $\frac {1} {(y-1)x(x+1)}$ affects the differentials. You cannot treat it as a constant.

Hmm? In what way does it affect the differentials?

 

[fresh_42]

Hmm? In what way does it affect the differentials?

I made a mistake, sorry.

 

[Dick]

I made a mistake, sorry.

It happens. No problem.

 

[Arman777]

It's the same solution with a different choice of $c$. Substitute $c-1$ for $c$ in your solution and you'll get the book solution.

So, I did right then Since $c-1$ and $c$ are same.

 

[Dick]

So, I did right then Since $c-1$ and $c$ is same.

RIght.