Question

Solve differential equation $(1+y^2)dx=({\tan }^{-1}y-x)dy$.

Answer 1

$(1+y^2)dx=({\tan }^{-1}y-x)dx$
$\therefore \frac{dx}{dy}=\frac{{\tan }^{-1}y-x}{1+y^2}$
$\therefore \frac{dx}{dy}=\frac{{\tan }^{-1}y}{1+y^2}-\frac{x}{1+y^2}$
$\therefore \frac{dx}{dy}+\frac{x}{1+y^2}=\frac{{\tan }^{-1}y}{1+y^2}$ $\left(1\right)$
$\frac{dx}{dy}+P\left(y\right)x=Q\left(y\right)$ which is linear differential equation of the type
$\therefore P=1+y^2$ $Q\left(A\right)=\frac{{\tan }^{-1}y}{1+y^2}$
$\therefore$ Integrating factor $=e^{\int P\left(y\right)dy}$
$=e^{\int \frac{1}{1+y^2}dy}$
$\therefore$ Multiplying equation $\left(1\right)$ by $e^{{\tan }^{-1}}y$
$\therefore e^{{\tan }^{-1}y}\frac{dx}{dy}+\frac{e^{{\tan }^{-1}y}}{1+y^{-}}-=e^{{\tan }^{-1}y}\left(\frac{{\tan }^{-1}y}{1+y^2}\right)$
$\therefore \frac{d}{dx}(e^{{\tan }^{-1}y}\cdot x)=e^{{\tan }^{-1}y}\cdot \left(\frac{{\tan }^{-1}y}{1+y^2}\right)$
$\therefore$ Integrating both sides pairwise
$\therefore x{e}^{{\tan }^{-1}y}=\int e^{{\tan }^{-1}y}\cdot \left(\frac{{\tan }^{-1}y}{1+y^2}\right)dy$ .$\left(2\right)$
$\therefore$ Now, finding $\int e^{{\tan }^{-1}y}\cdot \left(\frac{{\tan }^{-1}y}{1+y^2}\right)dy$
Taking ${\tan }^{-1}y=t$
$\therefore \frac{1}{1+y^2}=dy=dt$
$\therefore x{e}^{{\tan }^{-1}y}=\int e^{t}dt$
$\therefore \int e^t\left[\right(t-1)+1]dt$
$\therefore$ Here taking $f\left(t\right)=1$
$\therefore x{e}^{{\tan }^{-1}y}=\int e^t\left(f\right(t)+f^{\prime }(t\left)\right)dt$
$e^{t}f\left(t\right)+c$
$\therefore x{e}^{{\tan }^{-1}y}=e^t(t-1)+c$
$\therefore$ Result $\left(2\right)$ becames as follows
$\therefore x{e}^{{\tan }^{-1}y}=e^{{\tan }^{-1}y}({\tan }^{-1}y-1)+c$
$\therefore x=({\tan }^{-1}y-1)+c\cdot e^{-{\tan }^{-1}y}$
$\therefore$ Which is required solution of given differential equation.