We are asked to solve the Cauchy problem for the differential equation $y'' - 3y' = 0$ with initial conditions $y(0) = 1$ and $y'(0) = -1$.

1. Problem Description

We are asked to solve the Cauchy problem for the differential equation $y'' - 3y' = 0$ with initial conditions $y(0) = 1$ and $y'(0) = -1$.

2. Solution Steps

The given differential equation is a second-order linear homogeneous differential equation with constant coefficients.

First, we find the characteristic equation. Let $y = e^{rx}$. Then $y' = re^{rx}$ and $y'' = r^2e^{rx}$. Substituting these into the differential equation, we get:

$r^2e^{rx} - 3re^{rx} = 0$

$e^{rx}(r^2 - 3r) = 0$

Since $e^{rx}$ is never zero, we have the characteristic equation:

$r^2 - 3r = 0$

$r(r - 3) = 0$

This gives us two distinct real roots: $r_1 = 0$ and $r_2 = 3$.

The general solution is then given by:

$y(x) = c_1e^{r_1x} + c_2e^{r_2x}$

$y(x) = c_1e^{0x} + c_2e^{3x}$

$y(x) = c_1 + c_2e^{3x}$

Now we apply the initial conditions.

$y(0) = 1$

$y(0) = c_1 + c_2e^{3(0)} = c_1 + c_2e^0 = c_1 + c_2 = 1$

So, $c_1 + c_2 = 1$.

Next, we find the first derivative:

$y'(x) = 0 + 3c_2e^{3x} = 3c_2e^{3x}$

Now apply the second initial condition:

$y'(0) = -1$

$y'(0) = 3c_2e^{3(0)} = 3c_2e^0 = 3c_2 = -1$

So, $3c_2 = -1$, which implies $c_2 = -\frac{1}{3}$.

Now we can find $c_1$ using the equation $c_1 + c_2 = 1$:

$c_1 - \frac{1}{3} = 1$

$c_1 = 1 + \frac{1}{3} = \frac{4}{3}$

Substituting the values of $c_1$ and $c_2$ into the general solution:

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

3. Final Answer

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