How to solve $2y′′+8y′+80y=F(t), y(0)=0, y′(0)=0$?

Question

I am trying to solve the initial value problem, $2y′′+8y′+80y=F(t), y(0)=0, y′(0)=0$ where $F(t)=20e^{-t}$ But I am unable to do it. I got the answer of $2Ae^{-t}=20e^{-t}$ But this is wrong I think.

I solved the complementary equation and got: $y_{c}=e^{-2t}Acos(6t)+Bsin(6t))$

For the particular solution I guessed (correctly I think) $y_{p}=Ae^{-t}$ Then working some derivates I got: $2Ae^{-t}-8Ae^{-t}+80Ae^{-t}$ I plugged 0 into the corresponding initial values given but this does not produce the right answer. Thank you for you help I got a test tomorrow. This was the only part I neglected -.-

Answer 1

$$2y′′+8y′+80y=F(t), y(0)=0, y′(0)=0$$ For the homogeneous I got this $$2y′′+8y′+80y=0$$ The characteristic polynomial is $$r^2+4r+40=0$$ $$(r+2)^2-36i^2=0 \implies (r+2+6i)(r+2-6i)=0$$ $$\implies r=-2\pm 6i$$ Therefore the solution is $$y_h=e^{-2t}(c_1\cos(6t)+c_2\sin(6t))$$ For the particular solution your guess is correct $y_p=Ae^{-t}$ $$2Ae^{-t}-8Ae^{-t}+80Ae^{-t}=20e^{-t}$$ $$74Ae^{-t}=20e^{-t} \implies A=\frac {10}{37}$$ $$y_p=e^{-t}\frac {10}{37}$$ $$y(t)=e^{-2t}(c_1\cos(6t)+c_2\sin(6t))+\frac {10}{37}e^{-t}$$ Apply the initial conditions ...I got this $$(c_1,c_2)=(-\frac {10}{37},-\frac 5{111})$$ Finally, $$y(t)=e^{-2t}(-\frac {10}{37}\cos(6t)-\frac 5{111}\sin(6t))+\frac {10}{37}e^{-t}$$

Answer 2

$y_c= C_1 e^{-2t}\cos 6t + C_2 e^{-2t}\sin 6t\\ y_p = Ae^{-t}\\ 2Ae^{−t}−8Ae^{−t}+80Ae^{−t}=F(t)=20e^{−t}\\ 74A = 20\\ A = \frac {20}{74}\\ y(0) = C_1 + A = 0\\ C_1 = -\frac {20}{74}\\ y'(0) = -2C_1 + 6C_2 - A = 0$