Question

The particular solution of the initial value problem given below is $\frac{d^{2}y}{d{x}^2}+12\frac{dy}{dx}+36y=0$ with $y\left(0\right)=3$ and ${\begin{array}{c}\frac{dy}{dx}\end{array}|}_{x=0}=-36$

• A. $(3-18x)e^{-6x}$
• B. $(3+25x)e^{-6x}$
• C. $(3+20x)e^{-6x}$
• D. $(3-12x)e^{-6x}$

Answer 1

The correct option is A $(3-18x)e^{-6x}$

$\frac{d^{2}y}{d{x}^2}+12\frac{dy}{dx}+36y=0;$
$y\left(0\right)=3$ & ${\begin{array}{c}\frac{dy}{dx}\end{array}|}_{x=0}=-36$
$\Rightarrow$ $D^2+12D+36=0$
$\Rightarrow$ $(D+6)(D+6)=0$
$\Rightarrow$ $D=-6,-6$
$\therefore$ $y=(C_1+C_{2}x)e^{-6x}$ ....(i)
at $x=0,y=3$
$\Rightarrow$ $3=c_1{e}^0+C_2\left(0\right)e^0$
$\Rightarrow$ $c_1=3$
also, $\frac{dy}{dx}=-36$ at $x=0$
$\Rightarrow$ $\frac{dy}{dx}=c_2{e}^{-6x}+(c_1+c_{2}x)e^{-6x}(-6)$
[using (i)]
$\Rightarrow$ $-36=c_2{e}^0+(c_1+c_2(0\left)\right)e^0(-6)$
$\Rightarrow$ $-36=c_2+(3+0)(-6)$
$\Rightarrow$ $c_2=-18$
Hence, $y=(3-18x)e^{-6x}$