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# The solution of the differential equation $\Large \frac{dy}{dx}=\frac{1-y^2}{1-x^2}$ is:\begin{array}{1 1}(A)\;y=\tan^{-1}x & (B)\;y-x=k(1+xy)\\(C)\;x=tan^{-1}y & (D)\;\tan (xy)=k\end{array}

Toolbox:
• If a linear differential equation is of the form $\large\frac{dy}{dx}$$=f(x), then ot can be solved by seperating the variables and then integrating. • \int \large\frac{dx}{x+a}$$=\log |x+a|$
Given $\large\frac{dy}{dx}=\frac{1-y^2}{1-x^2}$
Seperating the variables we get
$\large\frac{dy}{(1-y^2)}=\frac{dx}{1-x^2}$
$\int \large\frac{dx}{a^2-x^2}=\frac{1}{2a} $$\log \large\frac{a+x}{a-x} integrating on both sides, \int \large\frac{dy}{(1-y^2)}=\int \frac{dx}{1-x^2} \large\frac{1}{2}$$\log \large\bigg|\frac{1+y}{1-y}\bigg|=\frac{1}{2} $$\log \large\bigg|\frac{1+x}{1-x}\bigg|$$+\log c$