Question

For the differential equation $(1+y^2)+(x-e^{-{\tan }^{-1}y})\frac{dy}{dx}=0;y\left(0\right)=\frac{\pi }{4}$ then find the negative of the only constant in the solution to the given system.

Answer 1

Given, $(1+y^2)+(x-e^{-{\tan }^{-1}y})\frac{dy}{dx}=0$

$\Rightarrow \frac{dx}{dy}+\frac{x}{1+y^2}=\frac{e^{-{\tan }^{-1}y}}{1+y^2}$ ...(1)
Here $P=\frac{1}{1+y^2}\Rightarrow \int Pdy=\int \frac{1}{1+y^2}dy={\tan }^{-1}y$
$\therefore I.F.=e^{{\tan }^{-1}y}$
Multiplying (1) by $I.F.$ we get
$e^{{\tan }^{-1}y}\frac{dx}{dy}+\frac{x{e}^{{\tan }^{-1}y}}{1+y^2}=\frac{1}{1+y^2}$
Integrating both sides we get
$x{e}^{{\tan }^{-1}y}=\int \frac{1}{1+y^2}dy+c={\tan }^{-1}y+c$
As $y\left(0\right)=\frac{\pi }{4}\Rightarrow 0=1+c\Rightarrow c=-1$

Hence the negative of the only constant is $-(-1)=1$