The solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1+y^2}{1+x^2}$ is
The solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1+y^2}{1+x^2}$ is
$x+y=\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
$x-y=\mathrm{c}(x y)$, where $\mathrm{c}$ is a constant of integration.
$x+y=\mathrm{c}(1+x y)$, where $\mathrm{c}$ is a constant of integration.
$y-x=\mathrm{c}(1+x y)$, where $\mathrm{c}$ is a constant of integration.
Solution
Given differential equation is \(d y / d x=\left(1+y^2\right) /\left(1+x^2\right)\)
or, \(\left(1 /\left(1+y^2\right)\right) d y=\left(1 /\left(1+x^2\right)\right) d x\)
on, integrating,
\(\int\left(1 /\left(1+y^2\right)\right) d x=\int\left(1 /\left(1+x^2\right)\right) d y\)
or, \(\tan ^{-1} \mathrm{y}=\tan ^{-1} \mathrm{x}+\mathrm{c}\)
or \(\tan ^{-1} \mathrm{y}-\tan ^{-1} \mathrm{x}=\mathrm{c}\)
or, \(\tan ^{-1}((y-x) /(1+y x))=c\)
or, \((y-x) /(1+y x)=\tan c\)
\(\therefore(y-x) /(1+y x)=c\) (constant) is, the require solution