Question

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

• A. $2x{e}^{ta{n}^{-1}y},=e^{2ta{n}^{-1}y}+k$
• B. $x{e}^{ta{n}^{-1}y},=e^{ta{n}^{-1}y}+k$
• C. $x{e}^{2ta{n}^{-1}y},=e^{-ta{n}^{-1}y}+k$
• D. $(x-2)k{e}^{ta{n}^{-1}y}$

Answer 1

The correct option is A $2x{e}^{ta{n}^{-1}y},=e^{2ta{n}^{-1}y}+k$

$\frac{dx}{dy}+\frac{1}{1+y^2}x=\frac{1}{1+y^2}{e}^{ta{n}^{-1}y}$
$I.F=e^{\int \frac{1}{1+y^2}dy}=e^{ta{n}^{-1}y}$
$\therefore$ Solution is $x.e^{ta{n}^{-1}}y$
$=\int e^{ta{n}^{-1}y}.\frac{1}{1+y^2}{e}^{ta{n}^{-1}}dy=\frac{1}{2}{e}^{2ta{n}^{-1}y}+\frac{1}{2}k\Rightarrow 2x{e}^{ta{n}^{-1}y}=e^{2ta{n}^{-1}y}+k$