Question

Solve the initial value problem frac%7Bdy%7D%7Bdx%7D%20%2B%205y%3D0" id="TexFormula1" title=" \frac{d^{2}y}{dx^{2}} + 2 \frac{dy}{dx} + 5y=0" alt=" \frac{d^{2}y}{dx^{2}} + 2 \frac{dy}{dx} + 5y=0" align="absmiddle" class="latex-formula">
$y(0)=2$
$y'(0)=2$

Answer 1

The characteristic equation for this ODE is

$r^2+2r+5=0$

which has roots at $r=-1\pm2i$, so the general solution is

$y_c=(C_1\cos2x+C_2\sin2x)e^{-x}$

Given $y(0)=2$, we have

$2=(C_1\cos0+C_2\sin0)e^{-0}\implies C_1=2$

and $y'(0)=2$, we have (upon differentiating $y_c$)

${y_c}'=((2C_2-C_1)\cos2x-(2C_1+C_2)\sin2x)e^{-x}$
$2=((2C_2-2)\cos0-(4+C_2)\sin0)e^{-0}$
$2=2C_2-2\implies C_2=2$

So the particular solution is

$y_c=(2\cos2x+2\sin2x)e^{-x}$