To solve the given differential equation:

$y''' - 5y'' + 8y' - 4y = 0$

using the method of undetermined coefficients is not applicable, because this method is designed for non-homogeneous differential equations. However, this is a homogeneous differential equation, so the solution must be found by solving the characteristic equation. Below are the steps:

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Step 1: Write the characteristic equation

From the given differential equation: $y''' - 5y'' + 8y' - 4y = 0,$ the corresponding characteristic equation is: $r^3 - 5r^2 + 8r - 4 = 0.$

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Step 2: Solve the characteristic equation

To solve $r^3 - 5r^2 + 8r - 4 = 0$, we use factoring techniques.

1. Factorization by grouping: Group terms to factorize: $r^3 - 5r^2 + 8r - 4 = (r^2 - 4r)(r - 1).$

   Factor $r^2 - 4r$: $r^2 - 4r = r(r - 4),$ so the full factorization is: $(r - 2)^2(r - 1) = 0.$

2. Roots of the characteristic equation: The roots are: $r = 2 \quad (\text{repeated twice}), \quad r = 1 \quad (\text{simple root}).$

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Step 3: Write the general solution

Using the roots of the characteristic equation, the general solution of the differential equation is: $y(t) = C_1e^{2t} + C_2te^{2t} + C_3e^t,$ where:

• $C_1, C_2, C_3$ are arbitrary constants determined by initial or boundary conditions.

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This is the general solution. Let me know if you'd like to proceed with solving a non-homogeneous case or verifying the solution! Here are some related questions for practice:

1. How would the solution change if there were a non-homogeneous term such as $\sin(t)$?
2. What is the particular solution for the equation $y''' - 5y'' + 8y' - 4y = t^2e^t$?
3. How does the repeated root $r = 2$ affect the solution structure?
4. How can this solution be verified by substituting back into the original equation?
5. Can this problem be solved numerically if roots are complex or irrational?

Tip: Always verify roots of the characteristic equation carefully, especially if factoring is challenging.