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:
<br /> z = x^{1 - n} = x^{-1}<br />
The derivative is:
<br /> x&#039; = -x^{-2} \, x&#039;<br />
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:
<br /> z = x^{1 - n} = x^{-1}<br />
The derivative is:
<br /> x&#039; = -x^{-2} \, x&#039;<br />
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&#039; and y&#039; 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:
<br /> y&#039; = \frac{(1 + y)^2}{x (y + 1) - x^2} = \frac{(1 + y)^2}{x (y + 1 - x)}<br />
The sign of y&#039; 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&#039; diverges, but x&#039; = 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!