Question

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

• A. $(x-2)=K{e}^{{\tan }^{-1}y}$
• B. $2x{e}^{{\tan }^{-1}y}=e^{2{\tan }^{-1}y}+K$
• C. $x{e}^{{\tan }^{-1}y}={\tan }^{-1}y+K$
• D. $x{e}^{2{\tan }^{-1}y}=e^{{\tan }^{-1}y}+K$

Answer 1

Answer: (B)

$\left(1+y^2\right)+\left(x-e^{{\tan }^{-1}y}\right)\frac{dy}{dx}=0$
$\left(1+y^2\right)\frac{dx}{dy}+x=e^{{\tan }^{-1}y}$
$\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{e^{{\tan }^{-1}y}}{\left(1+y^2\right)}$
$\Rightarrow IF=e^{\int \frac{1}{1+y^2}dy}=e^{{\tan }^{-1}y}$
$\Rightarrow x\cdot e^{{\tan }^{-1}y}=\int \frac{e^{{\tan }^{-1}y}}{1+y^2}\cdot e^{{\tan }^{-1}y}\cdot dy$
$\Rightarrow x\left(e^{{\tan }^{-1}y}\right)=\frac{e^{2{\tan }^{-1}y}}{2}+c$
$\therefore 2x{e}^{{\tan }^{-1}y}=e^{2{\tan }^{-1}y}+K$