Separating Variables in a Nonlinear Differential Equation

[prace]

Problem: (y-yx^2)\frac{dy}{dx} = (y+1)^2

So, the first thing I tried was just dividing the whole equation by (y-yx^2) and then factored out the y to get \frac{dy}{dx} = \frac{(y+1)^2}{y(1-x^2)}. Next I expanded the numerator on the right side of the equation and then split them all into idividual fractions to get:

\frac{dy}{dx} = \frac{y}{1-x^2}+\frac{2}{1-x^2}+\frac{1}{y(1-x^2)}.

I just don't see how I can go from here?

 

[prace]

Ok, so I worked it out some more and got \frac{ydy}{(y+1)^2} = \frac{dx}{(1-x^2)}

Is this correct?

 

[cristo]

Ok, so I worked it out some more and got \frac{ydy}{(y+1)^2} = \frac{dx}{(1-x^2)}

Is this correct?

Yes, that looks right to me

 

[prace]

Cool Thanks. So I worked this one out and got the left hand side of the equation, but I can seem to get the integration on the right.

\frac{ydy}{(y+1)^2} => \frac{dx}{(1-x^2)} = \int{\frac{ydy}{(y+1)^2}} = \int{\frac{dx}{(1-x^2)}}

so, ln|y+1|+\frac{1}{y+1} = \int{\frac{dx}{(1-x^2)}}

what can I do with this one?

* For some reason on my screen the y integral keeps showing up in the post even though it is not in the LaTex. So, if you see it too, please disregard the middle term that reads \int{\frac{ydy}{(y+1)^2}} in the last part of the problem. Thanks

 

[cristo]

Cool Thanks. So I worked this one out and got the left hand side of the equation, but I can seem to get the integration on the right.

\frac{ydy}{(y+1)^2} => \frac{dx}{(1-x^2)} = \int{\frac{ydy}{(y+1)^2}} = \int{\frac{dx}{(1-x^2)}}

so, ln|y+1|+\frac{1}{y+1} = \int{\frac{dx}{(1-x^2)}}

what can I do with this one?

Thats not the correct integral of the LHS. Have you come across the method of partial fractions?

 

[prace]

Yes I have worked with partial fractions a little bit, but I don't see how it applies to this one. I don't mean to second guess you, but are you sure the LHS is not correct? I worked it out again and got the same answer and my TI-89 gives the same answer as well.

 

[cristo]

Yes I have worked with partial fractions a little bit, but I don't see how it applies to this one. I don't mean to second guess you, but are you sure the LHS is not correct? I worked it out again and got the same answer and my TI-89 gives the same answer as well.

I get a factor 1/2, but it might just be me. How did you integrate that by hand without using partial fraction, just out of curiosity?! EDIT: the factor 1/2 is just me! sorry

For the x integral, consider the denominator (1-x^2). This can be written as
(1+x)(1-x). So, if you can express \frac{1}{1-x^2} as a sum\frac{a}{1+x}+\frac{b}{1-x} then this can be integrated directly

 

[prace]

Thanks, I will try that out. To get the LHS w/o partial fractions:

\int{\frac{ydy}{(y+1)^2}}

let u = y+1, du = dy, and y = u-1

So, \int{\frac{(u-1)}{(u)^2}du}

=

\int{\frac{u}{u^2}-\frac{1}{u^2}du}

=

ln|u| + \frac{1}{u}

Substitute y+1 for u and get ln|y+1|+\frac{1}{y+1}

 

[cristo]

Thanks, I will try that out. To get the LHS w/o partial fractions:

\int{\frac{ydy}{(y+1)^2}}

let u = y+1, du = dy, and y = u-1

So, \int{\frac{(u-1)}{(u)^2}du} = \int{\frac{u}{u^2}-\frac{1}{u^2}du} = ln|u| + \frac{1}{u}

Substitute y+1 for u and get ln|y+1|+\frac{1}{y+1}

Ahh, ok, nice. I didn't spot that! I would generally use partial fractions for expressions like that, but looks like the sub'n is simpler there!

 

[prace]

Oh man, haha.. I like the substitution method, but I am not too keen on the trig substitution. Although, this one seems pretty nice for that. Do you mind taking a look at my partial fraction work? I need some practice with it so I was wondering if I did it correctly.

\int{\frac{dx}{(1-x^2)}}

\frac{1}{(1+x)(1-x)}=\frac{A}{1+x}+\frac{B}{1-x}

So, B and A both = (1/2)

Then the solution to the integral is:
\frac{ln|1+x|}{2}+\frac{ln|1-x|}{2}

Thanks!