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The differential equation given is:

\[
y'' + 5y = -180x^2 e^{5x}
\]

To solve this using the method of undetermined coefficients, we follow these steps:

### Step 1: Solve the homogeneous equation
The homogeneous version of the differential equation is:

\[
y'' + 5y = 0
\]

The characteristic equation for this is:

\[
r^2 + 5 = 0 \implies r^2 = -5 \implies r = \pm i\sqrt{5}
\]

Thus, the general solution for the homogeneous equation is:

\[
y_h(x) = c_1 \cos(\sqrt{5}x) + c_2 \sin(\sqrt{5}x)
\]

### Step 2: Solve the particular solution
The non-homogeneous term is \(-180x^2 e^{5x}\). Since the term involves both a polynomial in \(x\) and an exponential, we try a particular solution of the form:

\[
y_p(x) = (Ax^2 + Bx + C)e^{5x}
\]

We will need to find \(A\), \(B\), and \(C\) by plugging this guess into the original equation and solving for these constants.

#### First derivative of \(y_p(x)\):
\[
y_p'(x) = \left(2Ax + B\right)e^{5x} + 5(Ax^2 + Bx + C)e^{5x}
\]
Simplifying:
\[
y_p'(x) = \left(2Ax + B + 5Ax^2 + 5Bx + 5C\right)e^{5x}
\]

#### Second derivative of \(y_p(x)\):
\[
y_p''(x) = \left(2A + 5(2Ax + B)\right)e^{5x} + 5\left(2Ax + B + 5Ax^2 + 5Bx + 5C\right)e^{5x}
\]
Simplifying further:
\[
y_p''(x) = \left(2A + 10Ax + 5B + 25Ax^2 + 25Bx + 25C\right)e^{5x}
\]

### Step 3: Substitute into the original equation
Now, substitute \(y_p(x)\) and \(y_p''(x)\) into the original equation:

\[
y_p'' + 5y_p = -180x^2 e^{5x}
\]

This gives:

\[
\left(2A + 10Ax + 5B + 25Ax^2 + 25Bx + 25C\right)e^{5x} + 5\left(Ax^2 + Bx + C\right)e^{5x} = -180x^2 e^{5x}
\]

Simplify and combine like terms:

\[
\left(30Ax^2 + (50A + 25B)x + (2A + 5B + 25C)\right)e^{5x} = -180x^2 e^{5x}
\]

Equating the coefficients of like powers of \(x\):

- For \(x^2\): \(30A = -180 \implies A = -6\)
- For \(x^1\): \(50A + 25B = 0 \implies -300 + 25B = 0 \implies B = 12\)
- For the constant: \(2A + 5B + 25C = 0 \implies 2(-6) + 5(12) + 25C = 0 \implies -12 + 60 + 25C = 0 \implies C = -48/25\)

Thus, the particular solution is:

\[
y_p(x) = \left(-6x^2 + 12x - \frac{48}{25}\right)e^{5x}
\]

### Step 4: General solution
The general solution to the differential equation is:

\[
y(x) = y_h(x) + y_p(x)
\]

\[
y(x) = c_1 \cos(\sqrt{5}x) + c_2 \sin(\sqrt{5}x) + \left(-6x^2 + 12x - \frac{48}{25}\right)e^{5x}
\]

This is the solution to the given differential equation.

Would you like further clarification or details on any step?

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Here are some related questions:
1. What is the method of undetermined coefficients, and when is it used?
2. Can this approach be applied to other forms of non-homogeneous differential equations?
3. What is the significance of the characteristic equation in solving differential equations?
4. How does the presence of the exponential term \(e^{5x}\) influence the choice of the particular solution?
5. What other methods exist to solve non-homogeneous differential equations?

**Tip:** When solving differential equations, always solve the homogeneous equation first to understand the complementary solution before tackling the particular solution.

Solve the differential equation y'' + 5y = -180x^2 e^{5x} using the method of undetermined coefficients.

Learn how to solve the second-order non-homogeneous differential equation y'' + 5y = -180x^2 e^{5x} using the method of undetermined coefficients. Detailed steps involve solving the homogeneous equation, determining a particular solution, and obtaining the general solution. Suitable for college-level calculus students.

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