To solve for $a$ and $b$, we start by analyzing the given differential equation and the solution.

The differential equation is: $y''' + a y'' + b y' - 4y = 0$ and we are told that $t e^{2t}$ is a solution.

Step 1: Identifying the Form of the Solution

Since $t e^{2t}$ is a solution, it suggests that $e^{2t}$ is a root of the characteristic polynomial with multiplicity of at least two (because of the $t$ term). This leads us to suspect the characteristic polynomial has a root at 2 with multiplicity at least 2.

Step 2: Form of the Characteristic Polynomial

Let's assume the characteristic polynomial $P(r)$ of the differential equation is of the form: $P(r) = r^3 + a r^2 + b r - 4$ Since $t e^{2t}$ is a solution, 2 must be a root with multiplicity at least 2. This means $r = 2$ is a double root, and possibly a triple root if $t^2 e^{2t}$ were also a solution.

Step 3: Substitute $r = 2$

Since $r = 2$ is a root, we substitute $r = 2$ into the polynomial: $2^3 + a \cdot 2^2 + b \cdot 2 - 4 = 0$ which simplifies to: $8 + 4a + 2b - 4 = 0$ or: $4 + 4a + 2b = 0$ Dividing through by 2 gives: $2 + 2a + b = 0$ or: $b = -2 - 2a$

Step 4: Ensuring a Double Root

To ensure $r = 2$ is a double root, the derivative of the characteristic polynomial $P(r)$ with respect to $r$, evaluated at $r = 2$, should also be zero: $P'(r) = 3r^2 + 2a r + b$ Substituting $r = 2$: $P'(2) = 3 \cdot 2^2 + 2a \cdot 2 + b = 0$ This simplifies to: $12 + 4a + b = 0$ We already found that $b = -2 - 2a$, so substituting gives: $12 + 4a - 2 - 2a = 0$ $10 + 2a = 0$ $a = -5$

Step 5: Substitute $a = -5$ to Find $b$

Using $a = -5$ in the equation $b = -2 - 2a$: $b = -2 - 2(-5) = -2 + 10 = 8$

Solution

Thus, the values of $a$ and $b$ are: $a = -5 \quad \text{and} \quad b = 8$

Let me know if you would like further details on any part of this solution.

Here are some questions to deepen your understanding:

1. Why does $t e^{2t}$ imply that $r = 2$ is a double root?
2. How would the solution change if we were given a different solution form, such as $e^{3t}$?
3. Can this method be used to solve for higher-order polynomials with more roots?
4. How would you verify that $t e^{2t}$ is indeed a solution for these values of $a$ and $b$?
5. What is the general form of the solution if all roots were distinct?

Tip: When dealing with characteristic polynomials, identifying the multiplicity of roots can provide essential clues for forming the correct polynomial structure.