Question

Solve $(x^2+1)dy+(2y-1)dx=0$

• A. $y=-c{e}^{-2{\tan }^{-1}x}+\frac{1}{2}$
• B. $y=c{e}^{-2{\tan }^{-1}x}-\frac{1}{2}$
• C. $y=c{e}^{-2{\tan }^{-1}x}+\frac{1}{2}$
• D. None of these.

Answer 1

Answer: (C)

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

Given differential eqn
$(x^2+1)dy+(2y-1)dx=0$
$\Rightarrow \frac{dy}{dx}=\frac{(1-2y)}{(x^2+1)}$
$\Rightarrow \frac{dy}{dx}+\frac{2y}{x^2+1}=\frac{1}{x^2+1}$
which is a linear differential equation
Here $P=\frac{2}{x^2+1};Q=\frac{1}{x^2+1}$
Integrating Factor $I.F.=e^{\int Pdx}$
$=e^{\int \frac{2dx}{x^2+1}}$
$\Rightarrow I.F.=e^{2{\tan }^{-1}x}$
Solution is given by
$y{e}^{2{\tan }^{-1}x}=\int \frac{e^{2{\tan }^{-1}x}}{x^2+1}dx$
Put ${\tan }^{-1}x=t$
$\Rightarrow \frac{1}{1+x^2}dx=dt$
$y{e}^{2{\tan }^{-1}x}=\int e^{2t}dt$
$\Rightarrow y{e}^{2{\tan }^{-1}x}=\frac{1}{2}{e}^{2t}+C$
$y{e}^{2{\tan }^{-1}x}=\frac{1}{2}{e}^{2{\tan }^{-1}x}+C$
$\Rightarrow y=c{e}^{-2{\tan }^{-1}x}+\frac{1}{2}$