Solving a separable differential equation

Solve the differential equation:

$$y'=\frac{1-y^2}{1-x^2}$$

My book says the solution is: $$y=\frac{x+c}{cx+1},$$ where $c$ is a constant. It's been ten minutes I tried to verify if it was correct but I'm pretty sure the book is wrong. Can someone confirm it?

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Ten minutes is not much. Try harder. –  Yves Daoust Jun 6 '14 at 9:03
The integral of the differential equation is: $\frac{1+y}{1-y}=\frac{1+x}{1-x}c$ I think it's correct... So how to find the explicit equation: $$y=\frac{x+c}{cx+1}$$ –  user155542 Jun 6 '14 at 9:06
Solve $\frac{1+y}{1-y}=A$. –  Yves Daoust Jun 6 '14 at 9:08
$y=\frac{A-1}{A+1}$ –  user155542 Jun 6 '14 at 9:10
Now you have it. What's $A$ ? –  Yves Daoust Jun 6 '14 at 9:11

We have $$\frac{dy}{dx}=\frac{1-y^2}{1-x^2}.$$ Using separation variable we obtain $$\frac{dy}{1-y^2}=\frac{dx}{1-x^2}.$$ Integrating both sides yields \eqalign { \int\frac{dy}{1-y^2}&=\int\frac{dx}{1-x^2}\\ \frac12\int\left[\frac{1}{1+y}+\frac{1}{1-y}\right]\ dy&=\frac12\int\left[\frac{1}{1+x}+\frac{1}{1-x}\right]\ dx\\ \ln(1+y)-\ln(1-y)&=\ln(1+x)-\ln(1-x)+C_1\\ \ln\frac{1+y}{1-y}&=\ln\frac{1+x}{1-x}+C_1\\ \frac{1+y}{1-y}&=C_2\cdot\frac{1+x}{1-x}\\ y&=\frac{k_1x+k_2}{k_2x+k_1}\quad\Rightarrow\quad\text{dividing by}\ k_1\\ \color{blue}{y(x)}&\color{blue}{=\frac{x+c}{c\, x+1}}. } where $C_2=e^{C_1}$, $k_1=C_2+1$, $k_2=C_2-1$, and $c=\dfrac{k_2}{k_1}$.

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answer is not wrong @tunk-fey. It is just well arranged –  tattwamasi amrutam Jun 6 '14 at 9:13
@Dmoreno Thanks for your editing. –  Tunk-Fey Jun 6 '14 at 13:28
You are welcome @Tunk-Fey! –  Dmoreno Jun 6 '14 at 13:30
"Using separation variable we obtain": I've been calling that "cross-multiplying" my entire life. Is it not called that when working with derivatives? –  Cole Johnson Jun 6 '14 at 15:48
@Cole No, because $dy/dx$ is not a fraction. Of course, the effect is the same because of the chain rule. –  Ryan Reich Jun 6 '14 at 22:20

$$y'=\frac{cx+1-c(x+c)}{(cx+1)^2}=\frac{1-c^2}{(cx+1)^2},$$ $$\frac{1-y^2}{1-x^2}=\frac{\frac{(cx+1)^2-(x+c)^2}{(cx+1)^2}}{(1-x^2)}=\frac{(c^2-1)x^2+(1-c^2)}{(cx+1)^2(1-x^2)}=\frac{1-c^2}{(cx+1)^2}.$$

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\begin{align*} \text{Separate:} && \frac{dy}{1-y^2} & =\frac{dx}{1-x^2} \\ \text{Partial fractions:} && \frac{1}{2}\left(\int\frac{1}{1+y}dy+\int\frac{1}{1-y}dy\right) & =\frac{1}{2}\left(\int\frac{1}{1+x}dx+\int\frac{1}{1-x}dx\right) \\ \text{Integrate:} && \log\frac{1+y}{1-y} & = \log c+\log\left(\frac{1+x}{1-x}\right) \\ \text{Solve for $y$:} && y & =\frac{c+cx-1+x}{c+cx+1-x} \\ \text{Factor:} && y & =\frac{x(c+1)+c-1}{x(c-1)+c+1} \\ \text{Divide by $c+1$:} && y & =\frac{x+\frac{c-1}{c+1}}{\frac{c-1}{c+1}x+1} \\ \text{Let $k=\frac{c-1}{c+1}$:} && y & =\frac{x+k}{kx+1}. \end{align*}

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The book is correct. If you will show how you calculated that it isn't, I can show you where you made a mistake. Show us how you calculated $y'$ and how you calculated $\frac{1-y^2}{1-x^2}$.

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By integration from a table of primitives, $argth(y)=argth(x)+c$, so that $$y=\tanh(argth(x)+c).$$ (Can be shown to be the same as the given solution.)

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