<s>[INST] How do I solve Cauchy problem for $y''-2y'-3y = e^{4x}$? 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.
 [/INST] 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
$$y(x) = \frac{1}{5} \, [ 4 \, e^{-x} + e^{4 x} ].$$
</s>