Question

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

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

Answer 1

Answer: (B)

$2ye^{2x}=C\cdot e^{2x}-1$

Answer 2

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 $\frac{dy}{dx}=\frac{e^{-x}}{e^{x}}=e^{-2x}$ $\Rightarrow 2y=-e^{-2x}+C\quad$ (Integrating) $\Rightarrow 2ye^{2x}=C\cdot e^{2x}-1$