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Marizza181 [45]
3 years ago
12

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
Mathematics
1 answer:
pogonyaev3 years ago
3 0
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}
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