Question

Consider the differential equation $\frac{x+\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}+\ldots}{1+\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}+\ldots}=\frac{d x-d y}{d x+d y}$. If $y(0)=1$ and the solution of the differential equation is of the form $2 y f(x)=m e^{2 x}+n .$ Evaluate $(m+n) f(0)$

Answer 1

$\frac{x+\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}+\ldots}{1+\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}+\ldots}=\frac{d x-d y}{d x+d y}$
$\frac{e^{x}}{e^{-x}}=\frac{(d x+d y)+(d x-d y)}{(d x+d y)-(d x-d y)}$
...[By Componendo dividendo]
$\Leftrightarrow e^{2 x}=\frac{d x}{d y}$
Integrating, we get
$y=\frac{-e^{-2 x}}{2}+k$
$\Leftrightarrow 2 y e^{2 x}=2 k e^{2 x}-1$
$y(0)=1$
$ \Rightarrow k=\frac{3}{2}$
$\Rightarrow 2 y e^{2 x}=3 e^{2 x}-1$
$\Rightarrow f(x)=e^{2 x}, m=3, n=-1$
$\Rightarrow(m+n) f(0)=(3-1)(1)=2$