The given differential equation is:

\[
y'' - y' - 20y = 20\sin(t) - 15\cos(t)
\]

This is a non-homogeneous second-order linear differential equation. We can solve it using the method of **undetermined coefficients**. The general solution \( y(t) \) will consist of two parts:

1. **The complementary solution** \( y_c(t) \), which solves the homogeneous equation:
   \[
   y'' - y' - 20y = 0
   \]
   
2. **The particular solution** \( y_p(t) \), which solves the non-homogeneous equation:
   \[
   y'' - y' - 20y = 20\sin(t) - 15\cos(t)
   \]

### Step 1: Find the complementary solution \( y_c(t) \)

First, solve the homogeneous equation:
\[
y'' - y' - 20y = 0
\]
The characteristic equation for this is:
\[
r^2 - r - 20 = 0
\]
Solving this quadratic equation:
\[
r = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-20)}}{2(1)}
\]
\[
r = \frac{1 \pm \sqrt{1 + 80}}{2} = \frac{1 \pm \sqrt{81}}{2} = \frac{1 \pm 9}{2}
\]
Thus, the roots are:
\[
r = 5 \quad \text{and} \quad r = -4
\]
So, the general solution to the homogeneous equation is:
\[
y_c(t) = C_1 e^{5t} + C_2 e^{-4t}
\]

### Step 2: Find the particular solution \( y_p(t) \)

For the non-homogeneous part, the right-hand side is \( 20\sin(t) - 15\cos(t) \), which suggests we try a particular solution of the form:
\[
y_p(t) = A\sin(t) + B\cos(t)
\]
Now, differentiate \( y_p(t) \):
\[
y_p'(t) = A\cos(t) - B\sin(t)
\]
\[
y_p''(t) = -A\sin(t) - B\cos(t)
\]

Substitute \( y_p(t) \), \( y_p'(t) \), and \( y_p''(t) \) into the original differential equation:
\[
(-A\sin(t) - B\cos(t)) - (A\cos(t) - B\sin(t)) - 20(A\sin(t) + B\cos(t)) = 20\sin(t) - 15\cos(t)
\]
Simplify the terms:
\[
(-A - 20A)\sin(t) + (-B - 20B)\cos(t) - A\cos(t) + B\sin(t) = 20\sin(t) - 15\cos(t)
\]
Group the terms involving \(\sin(t)\) and \(\cos(t)\):
\[
(-21A + B)\sin(t) + (-21B - A)\cos(t) = 20\sin(t) - 15\cos(t)
\]

Now, equate the coefficients of \(\sin(t)\) and \(\cos(t)\) on both sides:
1. \(-21A + B = 20\)
2. \(-21B - A = -15\)

Solve this system of equations. From equation (1):
\[
B = 20 + 21A
\]
Substitute this into equation (2):
\[
-21(20 + 21A) - A = -15
\]
Simplify:
\[
-420 - 441A - A = -15
\]
\[
-442A = 405
\]
\[
A = -\frac{405}{442}
\]

Now, substitute \( A = -\frac{405}{442} \) into \( B = 20 + 21A \):
\[
B = 20 + 21\left(-\frac{405}{442}\right)
\]
\[
B = 20 - \frac{8505}{442}
\]
\[
B = \frac{8840 - 8505}{442} = \frac{335}{442}
\]

Thus, the particular solution is:
\[
y_p(t) = -\frac{405}{442}\sin(t) + \frac{335}{442}\cos(t)
\]

### Step 3: General solution

The general solution to the differential equation is:
\[
y(t) = y_c(t) + y_p(t) = C_1 e^{5t} + C_2 e^{-4t} - \frac{405}{442}\sin(t) + \frac{335}{442}\cos(t)
\]

---

Would you like more details on any part of this process or have any questions?

Here are five related questions for further exploration:
1. How can we determine the values of \( C_1 \) and \( C_2 \) for specific initial conditions?
2. How does the method of undetermined coefficients change if the right-hand side was \( e^t \)?
3. How do we solve a similar non-homogeneous equation using the method of variation of parameters?
4. What happens if the characteristic equation has complex roots?
5. How does damping affect the solution in a physical system modeled by such an equation?

**Tip:** Always check the form of the non-homogeneous term to choose the appropriate particular solution form (e.g., polynomial, exponential, sine/cosine).

This page explains how to solve the second-order differential equation y'' - y' - 20y = 20sin(t) - 15cos(t) using the method of undetermined coefficients. It covers finding the complementary and particular solutions and provides a detailed step-by-step breakdown of the process. Ideal for undergraduate students studying differential equations.

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