# $\tan^{-1}x+\tan^{-1}y=c$ is the general solution of the differential equation:

\begin{array}{1 1}(A)\;\frac{dy}{dx}=\frac{1+y^2}{1+x^2} & (B)\;\frac{dy}{dx}=\frac{1+x^2}{1+y^2}\\(C)\;(1+x^2)dy+(1+y^2)dx=0 & (D)\;(1+x^2)dx+(1+y^2)dy=0\end{array}

Toolbox:
• A linear differential equation of the form $\large\frac{dy}{dx}$$=f(x) can be solved by seperating the variables and then integrating it. • \int \large\frac{dx}{x^2+a^2}=\large\frac {1}{a}$$ \tan ^{-1} \bigg(\large\frac{x}{a}\bigg)+c$
We are asked to find to which of the following equation is $\tan ^{-1}x+\tan ^{-1}y=c$ is the general solution
$(A)\;\large\frac{dy}{dx}=\frac{1+y^2}{1+x^2}$
$(B)\;\large\frac{dy}{dx}=\frac{1+x^2}{1+y^2}$
$(C)\;(1+x^2)dy+(1+y^2)dx=0$
$(D)\;(1+x^2)dx+(1+y^2)dy=0$
Consider equ $(A) \large\frac{dy}{dx}=\frac{1+y^2}{1+x^2}$
If we seperate the variables, we get,
$\large\frac{dy}{1+y^2}=\frac{dx}{1+x^2}$
On integrating we get,
$\int \large\frac{dy}{1+y^2}=\int \frac{dx}{1+x^2}$
$=>\tan ^{-1}(y)=\tan ^{-1}(x)+c$
$=>\tan ^{-1}(y)-\tan ^{-1}(x)=c$
Hence this is not correct
Consider equ $(B)\;\large\frac{dy}{dx}=\frac{1+x^2}{1+y^2}$
On seperating the variables we get,
$dy(1+y^2)=dx(1+x^2)$
On integrating we get,
$\int (1+y^2)dy=\int (1+x^2)dx$
$=y+\large\frac{y^3}{3}$$=x+\large\frac{x^3}{3}$$+c$
Hence this is also not correct.
Consider equ $(C)\;(1+x^2)dy+(1+y^2)dx=0$
This can be written as
$(1+x^2)dy=-(1+y^2)dx$
On seperating the variables we get,
$\large\frac{dy}{1+y^2}=\frac{-dx}{1+x^2}$
On integrating we get,
$\int \large\frac{dy}{1+y^2}=-\int \frac{dx}{1+x^2}$
$=>\tan ^{-1}(y)=-\tan ^{-1}(x)+c$
$=>\tan ^{-1}(y)+\tan ^{-1}(x)=c$
Hence $C$ is the correct option
answered May 23, 2013 by