We are given the general solution of a differential equation: $y = C_1e^x + C_2e^{2x} + 3e^{3x}$ and the initial conditions $y(0) = 0$ and $y'(0) = 0$. We need to find the particular solution by determining the values of the constants $C_1$ and $C_2$.

1. Problem Description

We are given the general solution of a differential equation:

$y = C_1e^x + C_2e^{2x} + 3e^{3x}$

and the initial conditions $y(0) = 0$ and $y'(0) = 0$. We need to find the particular solution by determining the values of the constants $C_1$ and $C_2$.

2. Solution Steps

First, we apply the initial condition $y(0) = 0$ to the general solution:

$y(0) = C_1e^0 + C_2e^{2(0)} + 3e^{3(0)} = 0$

$C_1(1) + C_2(1) + 3(1) = 0$

$C_1 + C_2 + 3 = 0$

$C_1 + C_2 = -3$ (Equation 1)

Next, we need to find the derivative of the general solution:

$y' = \frac{d}{dx}(C_1e^x + C_2e^{2x} + 3e^{3x})$

$y' = C_1e^x + 2C_2e^{2x} + 9e^{3x}$

Now, we apply the second initial condition $y'(0) = 0$:

$y'(0) = C_1e^0 + 2C_2e^{2(0)} + 9e^{3(0)} = 0$

$C_1(1) + 2C_2(1) + 9(1) = 0$

$C_1 + 2C_2 + 9 = 0$

$C_1 + 2C_2 = -9$ (Equation 2)

We now have a system of two linear equations with two unknowns:

$C_1 + C_2 = -3$ (Equation 1)

$C_1 + 2C_2 = -9$ (Equation 2)

Subtract Equation 1 from Equation 2:

$(C_1 + 2C_2) - (C_1 + C_2) = -9 - (-3)$

$C_2 = -6$

Substitute $C_2 = -6$ into Equation 1:

$C_1 + (-6) = -3$

$C_1 = -3 + 6$

$C_1 = 3$

Now we have the values for $C_1$ and $C_2$. Substituting these values into the general solution:

$y = 3e^x + (-6)e^{2x} + 3e^{3x}$

$y = 3e^x - 6e^{2x} + 3e^{3x}$

3. Final Answer

The particular solution is $y = 3e^x - 6e^{2x} + 3e^{3x}$.