Question

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

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

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