# Thread: Simple first order ordinary differential equation

1. ## Simple first order ordinary differential equation

Okay, so this question should be easy but for some reason I'm having major bother with it.

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

I got some sort of answer which I don't even think is right, and what I got I just couldn't simplify at all I know it's separable, hence why I should be able to do it, but I just can't.

2. I think the problem is the integration of $\frac1{1-x^2}.$ Any ideas how to tackle it?

3. Well it's difference of 2 squares, and I did that, then partial fractions and what not, but I really don't think my answer was right.

4. No partial fractions, just algebra: $\frac{1}{1-x^{2}}=\frac{(1+x)+(1-x)}{2(1+x)(1-x)}=\frac{1}{2}\left( \frac{1}{1-x}+\frac{1}{1+x} \right).$

5. Originally Posted by chella182
Okay, so this question should be easy but for some reason I'm having major bother with it.

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

I got some sort of answer which I don't even think is right, and what I got I just couldn't simplify at all I know it's separable, hence why I should be able to do it, but I just can't.
Yes it is separable...but I have a feeling we can't find an explicit solution (i.e. y=...)

The DE is equivalent to $\frac{\,dy}{y^2-1}=\frac{\,dx}{x^2-1}\implies \frac{\,dy}{(y+1)(y-1)}=\frac{\,dx}{(x+1)(x-1)}$

Now, we see that if we integrate, we have $\int\frac{\,dy}{(y+1)(y-1)}=\int\frac{\,dx}{(x+1)(x-1)}$.

Note that $(y+1)-(y-1)=2$ and $(x+1)-(x-1)=2$

Therefore, $\int\frac{\,dy}{(y+1)(y-1)}=\int\frac{\,dx}{(x+1)(x-1)}\implies\tfrac{1}{2}\int\frac{2\,dy}{(y+1)(y-1)}=\tfrac{1}{2}\int\frac{2\,dx}{(x+1)(x-1)}$ $\implies\tfrac{1}{2}\int\frac{\left[(y+1)-(y-1)\right]\,dy}{(y+1)(y-1)}=\tfrac{1}{2}\int\frac{\left[(x+1)-(x-1)\right]\,dx}{(x+1)(x-1)}$.

This now becomes $\int\frac{\,dy}{y-1}-\int\frac{\,dy}{y+1}=\int\frac{\,dx}{x-1}-\int\frac{\,dx}{x+1}$ $\implies\ln\left|y-1\right|-\ln\left|y+1\right|=\ln\left|x-1\right|-\ln\left|x+1\right|+C\implies \ln\left|\frac{y-1}{y+1}\right|=\ln\left|\frac{x-1}{x+1}\right|+C$

This format is fine...

Otherwise, you can say this is the same as

$\left|\frac{y-1}{y+1}\right|=C\left|\frac{x-1}{x+1}\right|\implies\frac{y-1}{y+1}=A\left|\frac{x-1}{x+1}\right|$ $\implies \left[1-A\left|\frac{x-1}{x+1}\right|\right]y=A\left|\frac{x-1}{x+1}\right|+1\implies y=\frac{A\left|\displaystyle\frac{x-1}{x+1}\right|+1}{1-A\left|\displaystyle\frac{x-1}{x+1}\right|}$ where $A=\pm C$

I hope this helps...

6. Partial fractions is algebra isn't it? Anyway that's what I got, but then integrating both sides and whatever went messy & I just couldn't do it.

Originally Posted by Chris L T521
Yes it is separable...but I have a feeling we can't find an explicit solution (i.e. y=...)

The DE is equivalent to $\frac{\,dy}{y^2-1}=\frac{\,dx}{x^2-1}\implies \frac{\,dy}{(y+1)(y-1)}=\frac{\,dx}{(x+1)(x-1)}$

Now, we see that if we integrate, we have $\int\frac{\,dy}{(y+1)(y-1)}=\int\frac{\,dx}{(x+1)(x-1)}$.

Note that $(y+1)-(y-1)=2$ and $(x+1)-(x-1)=2$

Therefore, $\int\frac{\,dy}{(y+1)(y-1)}=\int\frac{\,dx}{(x+1)(x-1)}\implies\tfrac{1}{2}\int\frac{2\,dy}{(y+1)(y-1)}=\tfrac{1}{2}\int\frac{2\,dx}{(x+1)(x-1)}$ $\implies\tfrac{1}{2}\int\frac{\left[(y+1)-(y-1)\right]\,dy}{(y+1)(y-1)}=\tfrac{1}{2}\int\frac{\left[(x+1)-(x-1)\right]\,dx}{(x+1)(x-1)}$.

This now becomes $\int\frac{\,dy}{y-1}-\int\frac{\,dy}{y+1}=\int\frac{\,dx}{x-1}-\int\frac{\,dx}{x+1}$ $\implies\ln\left|y-1\right|-\ln\left|y+1\right|=\ln\left|x-1\right|-\ln\left|x+1\right|+C\implies \ln\left|\frac{y-1}{y+1}\right|=\ln\left|\frac{x-1}{x+1}\right|+C$

This format is fine...

Otherwise, you can say this is the same as

$\left|\frac{y-1}{y+1}\right|=C\left|\frac{x-1}{x+1}\right|\implies\frac{y-1}{y+1}=A\left|\frac{x-1}{x+1}\right|$ $\implies \left[1-A\left|\frac{x-1}{x+1}\right|\right]y=A\left|\frac{x-1}{x+1}\right|+1\implies y=\frac{A\left|\displaystyle\frac{x-1}{x+1}\right|+1}{1-A\left|\displaystyle\frac{x-1}{x+1}\right|}$ where $A=\pm C$

I hope this helps...

Totally don't get that bit in the middle about things equalling 2. That's not how we've been taught to do stuff like that :\. I did get halves kicking about, but argh! Think I'm gonna have to go to the lecturer :\.

7. Originally Posted by chella182
Partial fractions is algebra isn't it? Anyway that's what I got, but then integrating both sides and whatever went messy & I just couldn't do it.
Chella - you're on the right track - it's partial fractions!

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

$A(x+1) + B(x-1)= 1$

$(A+B)x + (A-B) = 1\;\;\;\Rightarrow\;\;\;A+B=0,\;\;A-B=1$

so

$A = \frac{1}{2},\;\;\; b= - \frac{1}{2}$

and

$\frac{1}{2} \frac{1}{x-1} - \frac{1}{2} \frac{1}{x+1}$

Lot's of students don't see " that bit in the middle about things equalling 2" but the method above works fairly well!

8. They're both $\frac{1}{2}$ aren't they because it's $1-x$ in one of the brackets, not $x-1$

9. Originally Posted by chella182
They're both $\frac{1}{2}$ aren't they because it's $1-x$ in one of the brackets, not $x-1$
It depends what you want to split up

$\frac{1}{(1-x)(1+x)}$ or $\frac{1}{(x-1)(x+1)}$

the signs will change accordingly.

10. It's

Originally Posted by chella182

$(1-x^2)\frac{dy}{dx}=1-y^2$
Originally Posted by Krizalid
No partial fractions, just algebra: $\frac{1}{1-x^{2}}=\frac{(1+x)+(1-x)}{2(1+x)(1-x)}=\frac{1}{2}\left( \frac{1}{1-x}+\frac{1}{1+x} \right).$

11. Someone had a minus in between them up there I'm sure. Urgh, I'm leaving it for tonight.
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