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Q.
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
Differential Equations
Solution:
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)
$\Rightarrow 2ye^{2x}=C\cdot e^{2x}-1$