Solving y'=\frac{(1+y)^2} {x(y+1)-x^2}: An Alternative Approach

[estro]

I want to solve:
$y'=\frac{(1+y)^2} {x(y+1)-x^2}$

What I tried:
I have no basis to think that y' is positive or negative in some domain, but if I do, I can write:
$x'(y)=\frac {x(y+1)-x^2}{(1+y)^2}=\frac{x}{(1+y)} +\frac{x^2}{(1+y)^2}$

and then I can substitute $z=\frac{x}{(1+y)}$

And I get the following ODE: $(y+1)\frac{dz}{dy}+z=z-z^2$.

So the solution is $\frac{1}{z}=\frac{(y+1)}{x}=\ln|y+1|+C$.

Then I can mark all this as "draft" and write: Let's notice that if $F(x,y)=\frac{(y+1)}{x}-\ln|y+1|=C$ is a potential, so this is indeed the solution.

Is this solution legit?
How can I solve this problem alternatively?

Thanks!

 

[haruspex]

And I get the following ODE: $(y+1)\frac{dz}{dy}+z=z-z^2$.

You might want to check that. I get something a little simpler.

Then I can mark all this as "draft" and write: Let's notice that if $F(x,y)=\frac{(y+1)}{x}-\ln|y+1|=C$ is a potential, so this is indeed the solution.

Sorry, I don't follow your logic there. Did you check that your solution satisfies the original equation?
Try substituting z = (y+1)/x straight off.

 

[estro]

This is what I've got: attached pdf document.
Can you help me to spot what I did wrong?

 

[STEMucator]

Hmm I only looked at this for a moment, but it looks like a POSSIBLE candidate for the method of successive approximations.

 

[Dickfore]

I want to solve:
$y'=\frac{(1+y)^2} {x(y+1)-x^2}$

What I tried:
I have no basis to think that y' is positive or negative in some domain, but if I do, I can write:
$x'(y)=\frac {x(y+1)-x^2}{(1+y)^2}=\frac{x}{(1+y)} +\frac{x^2}{(1+y)^2}$

This is a Bernoulli ODE with n = 2, with respect to $x(y)$ and is transformed to a linear ODE with the substitution:
$$z = x^{1 - n} = x^{-1}$$
The derivative is:
$$x' = -x^{-2} \, x'$$
So, multiply the equation by $(-x^{-2})$, and see what you get.

 

[estro]

This is a Bernoulli ODE with n = 2, with respect to $x(y)$ and is transformed to a linear ODE with the substitution:
$$z = x^{1 - n} = x^{-1}$$
The derivative is:
$$x' = -x^{-2} \, x'$$
So, multiply the equation by $(-x^{-2})$, and see what you get.

Thanks!

But, can I assume that y' is positive or negative in some domain?
How can I explain this?

 

[Dickfore]

$x'$ and $y'$ have the same sign.

 

[estro]

I understand this. but how I can understand that any of one of the tho preserve sign in any domain?

 

[Dickfore]

I understand this. but how I can understand that any of one of the tho preserve sign in any domain?

Ok, so:
$$y' = \frac{(1 + y)^2}{x (y + 1) - x^2} = \frac{(1 + y)^2}{x (y + 1 - x)}$$
The sign of $y'$ is the same as the sign of $x (y + 1 - x)$. You may draw the regions in the x-y plane where this is positive, negative, or zero. When it is zero, $y'$ diverges, but $x' = 0$ (unless $y = -1$).

When you find the general solution, try and see if any particular solution crosses from one region to another.

 

[estro]

Thank you very much!