https://file.aiquickdraw.com/mathbot/file/1727924777294n0bi59cp.jpg

Let's solve both differential equations from the image using the method of undetermined coefficients:

### Question 5:
The differential equation is:
\[
y'' - 12y' + 36y = 2te^{6t}
\]

To solve this using the method of undetermined coefficients, first, we need the general solution for the homogeneous part \(y'' - 12y' + 36y = 0\). The characteristic equation is:
\[
r^2 - 12r + 36 = 0
\]
Factoring this gives:
\[
(r - 6)^2 = 0
\]
Thus, the solution to the homogeneous equation is:
\[
y_h = (C_1 + C_2 t)e^{6t}
\]

Next, for the non-homogeneous part, the right-hand side is \(2te^{6t}\), which suggests that we try a particular solution of the form:
\[
y_p = At^2 e^{6t}
\]
because the term \(t e^{6t}\) already appears in the homogeneous solution. 

Now, compute the first and second derivatives of \(y_p\):
\[
y_p' = A(2t e^{6t} + t^2 6 e^{6t}) = A e^{6t}(2t + 6t^2)
\]
\[
y_p'' = A(2e^{6t} + 12t e^{6t} + 6 e^{6t} 6t^2) = A e^{6t}(2 + 12t + 36t^2)
\]

Substitute \(y_p, y_p'\), and \(y_p''\) into the differential equation \(y'' - 12y' + 36y = 2te^{6t}\):
\[
A e^{6t}(2 + 12t + 36t^2) - 12A e^{6t}(2t + 6t^2) + 36A t^2 e^{6t} = 2te^{6t}
\]
Simplify:
\[
A e^{6t}(2 + 12t + 36t^2 - 24t - 72t^2 + 36t^2) = 2te^{6t}
\]
\[
A e^{6t}(2 - 12t) = 2te^{6t}
\]
Equating coefficients of like powers of \(t\), we find:
\[
A(-12) = 2 \quad \Rightarrow \quad A = -\frac{1}{6}
\]

Thus, the particular solution is:
\[
y_p = -\frac{1}{6}t^2 e^{6t}
\]
The correct answer is \( \frac{1}{6} t e^{6t} \).

### Question 6:
The differential equation is:
\[
y'' - 6y' + 9y = 6t^2 e^{3t}
\]

The characteristic equation for the homogeneous part is:
\[
r^2 - 6r + 9 = 0
\]
This factors to:
\[
(r - 3)^2 = 0
\]
So, the homogeneous solution is:
\[
y_h = (C_1 + C_2 t) e^{3t}
\]

For the non-homogeneous part, we try a particular solution of the form:
\[
y_p = At^3 e^{3t}
\]
because \(t^2 e^{3t}\) already appears in the homogeneous solution.

Now, differentiate \(y_p\):
\[
y_p' = A(3t^2 e^{3t} + t^3 3 e^{3t}) = A e^{3t}(3t^2 + 3t^3)
\]
\[
y_p'' = A(6t e^{3t} + 9t^2 e^{3t} + 9t^3 e^{3t}) = A e^{3t}(6t + 9t^2 + 9t^3)
\]

Substitute these into the differential equation:
\[
A e^{3t}(6t + 9t^2 + 9t^3) - 6A e^{3t}(3t^2 + 3t^3) + 9A e^{3t}t^3 = 6t^2 e^{3t}
\]
Simplify:
\[
A e^{3t}(6t + 9t^2 + 9t^3 - 18t^2 - 18t^3 + 9t^3) = 6t^2 e^{3t}
\]
\[
A e^{3t}(6t - 9t^2) = 6t^2 e^{3t}
\]
Equating coefficients of like powers of \(t\):
\[
A(-9) = 6 \quad \Rightarrow \quad A = -\frac{2}{3}
\]

Thus, the particular solution is:
\[
y_p = -\frac{2}{3} t^3 e^{3t}
\]
None of the provided options exactly match this result, so the correct answer seems to be "none of the above."

---

Let me know if you want details on any part of this solution or further clarification. Here are 5 follow-up questions for practice:

1. How do you solve homogeneous linear differential equations with constant coefficients?
2. What is the method of undetermined coefficients used for?
3. Why is it necessary to choose a higher power of \(t\) when solving using undetermined coefficients?
4. How do you identify the form of the particular solution based on the right-hand side of the DE?
5. What happens when the right-hand side of the DE is already part of the homogeneous solution?

**Tip:** When choosing the form of the particular solution, always check for duplication with the homogeneous solution to adjust the guess accordingly (by multiplying by a higher power of \(t\)).

Find a particular solution using the method of undetermined coefficients for the DE: y'' - 12y' + 36y = 2te^{6t} and y'' - 6y' + 9y = 6t^2e^{3t}.

Learn how to find particular solutions for second-order differential equations like y'' - 12y' + 36y = 2te^{6t} and y'' - 6y' + 9y = 6t^2e^{3t} using the method of undetermined coefficients. This solution explores the steps to solve homogeneous and non-homogeneous equations and provides examples of key formulas.

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