We are tasked with finding a particular solution to the differential equation:

$9 y''' + 6 y'' + y' - 3y = e^{-t}$

Step 1: Solve the Homogeneous Equation

The homogeneous version of the differential equation is:

$9 y''' + 6 y'' + y' - 3y = 0$

To solve this, we first find the characteristic equation:

$9r^3 + 6r^2 + r - 3 = 0$

Step 2: Find a Particular Solution Using Undetermined Coefficients

We will now focus on finding a particular solution to the non-homogeneous equation:

$9 y''' + 6 y'' + y' - 3y = e^{-t}$

Since the right-hand side of the equation is $e^{-t}$, we assume a particular solution of the form:

$y_p(t) = Ae^{-t}$

where $A$ is a constant to be determined.

Step 3: Substitute into the Equation

Now, we compute the derivatives of $y_p(t)$:

$y_p'(t) = -Ae^{-t}$ $y_p''(t) = Ae^{-t}$ $y_p'''(t) = -Ae^{-t}$

Substitute these into the original differential equation:

$9(-Ae^{-t}) + 6(Ae^{-t}) + (-Ae^{-t}) - 3(Ae^{-t}) = e^{-t}$

Simplifying:

$-9Ae^{-t} + 6Ae^{-t} - Ae^{-t} - 3Ae^{-t} = e^{-t}$

Combine like terms:

$(-9A + 6A - A - 3A)e^{-t} = e^{-t}$ $-7Ae^{-t} = e^{-t}$

Step 4: Solve for $A$

Equating the coefficients of $e^{-t}$:

$-7A = 1$

Thus, solving for $A$:

$A = -\frac{1}{7}$

Step 5: Write the Particular Solution

Therefore, the particular solution is:

$y_p(t) = -\frac{1}{7} e^{-t}$

---

Would you like more details or clarification on any step?

Here are 5 related questions you may find helpful:

1. How do you find the general solution to a higher-order linear differential equation?
2. Can you explain the method of undetermined coefficients in more detail?
3. What happens when the right-hand side of the equation is not $e^{-t}$?
4. How do you solve a characteristic equation for higher-order linear equations?
5. What are some common forms for the particular solution in the method of undetermined coefficients?

Tip: Always check if any terms in the particular solution form are already present in the homogeneous solution, as this would require a modification (e.g., multiplying by $t$) to avoid linear dependence.