Question

Find the solution of the differential equation:
$(1+y^2)+(x-e^{ta{n}^{-1}y})\frac{dy}{dx}=0$

Answer 1

Given,

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

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

$\frac{1}{1+y^2}=\frac{1}{e^{{\tan }^{-1}y}-x}\frac{dx}{dy}$

$\frac{e^{{\tan }^{-1}y}}{1+y^2}-\frac{x}{1+y^2}=\frac{dx}{dy}$

$P=\frac{1}{1+y^2},Q=\frac{e^{{\tan }^{-1}y}}{1+y^2}$

$x\times I.F=\int Q\times I.F\times dy$

$I.F=e^{\int Pdx}=e^{\int \frac{1}{1+y^2}dx}=e^{{\tan }^{-1}y}$

$x\times e^{{\tan }^{-1}y}=\int \frac{e^{{\tan }^{-1}y}}{1+y^2}{e}^{{\tan }^{-1}y}dy$

substitute ${\tan }^{-1}y=t$

$x\times e^t=\int e^t\times e^{t}dt$

$x\times e^t=\int e^{2t}dt$

$x\times e^t=\frac{e^{2t}}{2}+c$

$x=\frac{e^t}{2}+c{e}^{-t}$

$x=\frac{e^{{\tan }^{-1}y}}{2}+c{e}^{-{\tan }^{-1}y}$