Question

The solution of $\frac{d^{2}y}{d{x}^2}+2\frac{dy}{dx}+17y=0$; $y\left(0\right)=1$,$\frac{dy}{dx}\left(\frac{\pi }{4}\right)=0$ in the range $0<x<\frac{\pi }{4}$ is given by

• A. $e^{-x}(cos4x+\frac{1}{4}sin4x)$
• B. $e^x(cos4x-\frac{1}{4}sin4x)$
• C. $e^{-4x}(cos4x-\frac{1}{4}sin4x)$
• D. $e^{-4x}(cos4x-\frac{1}{4}sin4x)$

Answer 1

The correct option is A $e^{-x}(cos4x+\frac{1}{4}sin4x)$

Given the differential equation
$\frac{d^{2}y}{d{x}^2}+2\frac{dy}{dx}+17y=0$ ... (i)
Its auxiliary equation is
$m^2+2m+17=0$
$m=\frac{-2\pm \sqrt{4-4\times 17}}{2}$
$=-1\pm 4i$
Solution of (i) is
$y=e^{-x}(C_{1}cos4x+C_{2}sin4x)$ ...(ii)
Given $y\left(0\right)=1$
$\Rightarrow 1=1(C_1+C_2\times 0$)
$C_1=1$
Also$\frac{dy}{dx}\left(\pi /4\right)=0$ ... (iii)
Differentiating (ii) w.r.t $x$
$\frac{dy}{dx}=e^{-x}[-4C_{1}sin4x+4C_{2}cos4x-C_{1}cos4x-C_{2}sin4x]$
Using (iii),
$\Rightarrow 0=e^{-\pi /4}[0-4C_2+C_1-0]$
$C_1-4C_2=0$
$\Rightarrow 1-4C_2=0$
$\Rightarrow C_2=\frac{1}{4}$
$\therefore y=e^{-x}(cos4x+\frac{1}{4}sin4x$)