simplification/possible error

**schteve** · February 25th 2010, 08:28 AM

Hi,

I have the differential equation (1-x^2)y' = 1 - y^2.
Using partial fractions and exponentials, i now have the equation in the following form:

sqrt(1+y)/sqrt(1-y) = A(sqrt(1+x)/sqrt(1-x))

I am not sure if i have simplified this enough, as the square roots are causing me problems.
Also i am not sure if I have used the right method. I have used seperation, but I'm not sure if i could have used a homogeneous equation method.

Any thoughts or comments would be appreciated.
Thanks

stephen

**Chris L T521** · February 25th 2010, 09:00 AM

Originally Posted by schteve

Hi,

I have the differential equation (1-x^2)y' = 1 - y^2.
Using partial fractions and exponentials, i now have the equation in the following form:

sqrt(1+y)/sqrt(1-y) = A(sqrt(1+x)/sqrt(1-x))

I am not sure if i have simplified this enough, as the square roots are causing me problems.
Also i am not sure if I have used the right method. I have used seperation, but I'm not sure if i could have used a homogeneous equation method.

Any thoughts or comments would be appreciated.
Thanks

stephen

Separation is the way to go about this, since this equation is not homogeneous (do you know why?).

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

Using partial fractions, you should have

$\frac{1}{2}\left[\frac1{1+y}+\frac1{1-y}\right]\,dy=\frac{1}{2}\left[\frac1{1+x}+\frac1{1-x}\right]\,dx$ $\implies \ln|1+y|-\ln|1-y|=\ln|1+x|-\ln|1-x|+C\implies \left|\frac{1+y}{1-y}\right|=C\left|\frac{1+x}{1-x}\right|$

At this stage, let's get rid of absolute values on the LHS:

$\frac{1+y}{1-y}=\pm C\left|\frac{1+x}{1-x}\right|\implies 1+y=C\left|\frac{1+x}{1-x}\right|-C\left|\frac{1+x}{1-x}\right|y$ $\implies \left(C\left|\frac{1+x}{1-x}\right|+1\right)y=C\left|\frac{1+x}{1-x}\right|-1\implies y=\frac{C\left|\dfrac{1+x}{1-x}\right|-1}{C\left|\dfrac{1+x}{1-x}\right|+1}$

This looks quite complicated, but at least we have a value for y. Does this part make sense?

**schteve** · February 25th 2010, 09:04 AM

Hi

Thanks, I think it's not homogeneous because all the terms are not of the same degree - i think that's right?

Also, I forgot the halves can cancel, hence why i have sqaure roots in my example.
Another thought has also occured to me, and that is that you could convert the integrals into arctan expressions? I don't know which is the best way to approach the integral now.

Again thanks fo the help

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