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?
On
$$y'=\frac{cx+1-c(x+c)}{(cx+1)^2}=\frac{1-c^2}{(cx+1)^2},$$ $$\frac{1-y^2}{1-x^2}=\frac{\frac{(cx+1)^2-(x+c)^2}{(cx+1)^2}}{(1-x^2)}=\frac{(c^2-1)x^2+(1-c^2)}{(cx+1)^2(1-x^2)}=\frac{1-c^2}{(cx+1)^2}.$$
On
\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*}
On
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 $c=\dfrac{k_2}{k_1}$.
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 $\frac{1-y^2}{1-x^2}$.