The given differential equation is:

\[
5y'' + 10y' - 15y = 0
\]

We can start by solving this second-order linear homogeneous differential equation with constant coefficients using the characteristic equation method.

### Step 1: Rewrite the equation in standard form

Divide the whole equation by 5 to simplify:

\[
y'' + 2y' - 3y = 0
\]

### Step 2: Find the characteristic equation

The characteristic equation for a second-order equation \( y'' + 2y' - 3y = 0 \) is given by:

\[
r^2 + 2r - 3 = 0
\]

### Step 3: Solve the characteristic equation

To solve \( r^2 + 2r - 3 = 0 \), we can factor it:

\[
(r + 3)(r - 1) = 0
\]

So, the roots are:

\[
r_1 = -3 \quad \text{and} \quad r_2 = 1
\]

### Step 4: General solution

Since the roots are real and distinct, the general solution to the differential equation is:

\[
y(t) = C_1 e^{-3t} + C_2 e^{t}
\]

where \( C_1 \) and \( C_2 \) are constants determined by initial conditions.

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Would you like more details on this, or do you have any further questions? Here are five related questions for deeper understanding:

1. How do we solve a second-order differential equation with complex roots?
2. What if the characteristic equation has repeated roots?
3. How do we apply initial conditions to find the particular solution?
4. How do we solve non-homogeneous differential equations?
5. Can we use matrix methods to solve systems of differential equations?

**Tip:** Always check the discriminant of the characteristic equation to determine the nature of the roots (real, distinct, repeated, or complex).

Learn how to solve the second-order differential equation 5y'' + 10y' - 15y = 0 using the characteristic equation method. This step-by-step solution involves solving the characteristic equation and finding the general solution. Suitable for advanced high school and college-level calculus students.

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