Question

Solution of the differential equation $\frac{x+\frac{x^3}{3!}+\frac{x^5}{5!}+\dots }{1+\frac{x^2}{2!}+\frac{x^4}{4!}+\dots }=\frac{dx-dy}{dx+dy}$ is

• A. $2y{e}^{2x}=C\cdot e^{2x}+1$
• B. $2y{e}^{2x}=C\cdot e^{2x}-1$
• C. $y{e}^{2x}=C\cdot e^{2x}+2$
• D. none of these

Answer 1

Answer: (B)

Given equation can be rewritten as
$\frac{\frac{1}{2}\left(e^x-e^{-x}\right)}{\frac{1}{2}\left(e^x+e^{-x}\right)}=\frac{dx-dy}{dx+dy}$
Applying componendo and dividendo, we get
$\frac{dy}{dx}=\frac{e^{-x}}{e^x}=e^{-2x}$
$\Rightarrow 2y=-e^{-2x}+C$ (Integrating)
$\Rightarrow 2y{e}^{2x}=C\cdot e^{2x}-1$