How do I solve Cauchy problem for $y''-2y'-3y = e^{4x}$?

Question

I have one:

$$y''-2y'-3y = e^{4x} \quad y(0) = 1 \quad y'(0) = 0$$

I've found the solution as a sum of general solution and particular one:

$$y(x) = C_1e^{-x}+C_2e^{3x}+\frac{1}{5}e^{4x}$$

Applying the first condition I got:

$$C_1+C_2+ \frac{1}{5} = 1$$

but I do not know how to proceed, techically.

Answer 1

Using $$y(x) = C_1e^{-x}+C_2e^{3x}+\frac{1}{5}e^{4x}$$ then differentiate to obtain $$y'(x) = - C_1 e^{-x} + 3 C_2 e^{3x} + \frac{4}{5} e^{4x}.$$ Now set $x=0$ in these equations to obtain \begin{align} C_{1} + C_{2} + \frac{1}{5} &= 1 \\ - C_{1} + 3 C_{2} + \frac{4}{5} &= 0. \end{align} Solving this set for $C_{1}$ and $C_{2}$ yields $C_{1} = 4/5$ and $C_{2} = 0$ and leads to $$y(x) = \frac{1}{5} \, [ 4 \, e^{-x} + e^{4 x} ].$$