Question

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

Answer 1

Answer: 2

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