Does anyone know how to solve this?

Solve the differential equation by variation of parameters, subject to the initial conditions
y(0) = 1, y'(0) = 0.
y'' + 2y' − 8y = 5e^(−3x) − e^(−x)

I get y = c1e^(-4x)+c2e^(2x)-e^(-3x)+e^(-x)/9 but idk how to carry on with the initial condition part.

Question

Does anyone know how to solve this? 

Solve the differential equation by variation of parameters, subject to the initial conditions 
y(0) = 1, y'(0) = 0.
y'' + 2y' − 8y = 5e^(−3x) − e^(−x)

I get y = c1e^(-4x)+c2e^(2x)-e^(-3x)+e^(-x)/9 but idk how to carry on with the initial condition part.

sirm3d · Answer

$$\Large {y=c_1e^{-4x}+c_2e^{2x}-e^{-3x}+e^{-x}/9\1=c_1+c_2-1+1/9}$$

$$\Large {y'=-4c_1e^{-4x}+2c_2e^{2x}+3e^{-3x}-e^{-x}/9\0=-4c_1+2c_2+3-1/9}$$

solve the unknowns $\large c_1,\;c_2$

anonymous · Answer

Thanks. I was able to solve the unknowns. $$c_{1} = \frac{ 121 }{ 54 } , c_{2} = -\frac{ 7 }{54  }$$ 
Is that what you got as well?