Find a particular solution to y''+5y'+6y=-20te^(2t)

We are tasked with finding a particular solution to the non-homogeneous second-order differential equation:

\[
y'' + 5y' + 6y = -20te^{2t}
\]

### Step 1: Solve the homogeneous equation
First, solve the homogeneous equation:

\[
y'' + 5y' + 6y = 0
\]

The characteristic equation associated with this homogeneous equation is:

\[
r^2 + 5r + 6 = 0
\]

Factoring the quadratic equation:

\[
(r + 2)(r + 3) = 0
\]

So, the roots are \( r = -2 \) and \( r = -3 \).

Therefore, the general solution to the homogeneous equation is:

\[
y_h(t) = C_1 e^{-2t} + C_2 e^{-3t}
\]

### Step 2: Solve the non-homogeneous equation using the method of undetermined coefficients
We now seek a particular solution to the non-homogeneous equation:

\[
y_p'' + 5y_p' + 6y_p = -20te^{2t}
\]

Since the non-homogeneous term is of the form \( -20te^{2t} \), and \( e^{2t} \) is not a solution to the homogeneous equation, we propose a particular solution of the form:

\[
y_p(t) = (At + B) e^{2t}
\]

Now, find the derivatives of \( y_p(t) \):

\[
y_p'(t) = A e^{2t} + (At + B) 2e^{2t} = (A + 2At + 2B) e^{2t}
\]

\[
y_p''(t) = (2A + 2A + 4At + 4B) e^{2t} = (4At + 4A + 4B) e^{2t}
\]

Substitute \( y_p(t), y_p'(t), \) and \( y_p''(t) \) into the original equation \( y'' + 5y' + 6y = -20te^{2t} \):

\[
(4At + 4A + 4B) e^{2t} + 5(A + 2At + 2B)e^{2t} + 6(At + B) e^{2t} = -20te^{2t}
\]

Simplifying the left-hand side:

\[
(4At + 4A + 4B) + 5(A + 2At + 2B) + 6(At + B) = -20t
\]

Expand and group the terms:

\[
4At + 4A + 4B + 5A + 10At + 10B + 6At + 6B = -20t
\]

Now group the terms with \( t \) and the constants:

\[
(4A + 10A + 6A)t + (4A + 4B + 5A + 10B + 6B) = -20t
\]

This simplifies to:

\[
20At + (9A + 20B) = -20t
\]

### Step 3: Solve for \( A \) and \( B \)
For the equation to hold true, the coefficients of \( t \) and the constants on both sides must be equal. So, we set up the following system of equations:

\[
20A = -20 \quad \text{(from the \( t \)-terms)}
\]
\[
9A + 20B = 0 \quad \text{(from the constant terms)}
\]

Solving the first equation:

\[
A = -1
\]

Substitute \( A = -1 \) into the second equation:

\[
9(-1) + 20B = 0
\]
\[
-9 + 20B = 0
\]
\[
20B = 9
\]
\[
B = \frac{9}{20}
\]

### Step 4: Write the particular solution
Thus, the particular solution is:

\[
y_p(t) = (-t + \frac{9}{20}) e^{2t}
\]

### Step 5: General solution
The general solution to the differential equation is the sum of the homogeneous solution and the particular solution:

\[
y(t) = C_1 e^{-2t} + C_2 e^{-3t} + (-t + \frac{9}{20}) e^{2t}
\]

This is the general solution to the given non-homogeneous differential equation.

Do you have any questions or need further clarification?

---

Here are 5 related questions:
1. How would the method of variation of parameters differ in solving this equation?
2. What happens to the solution if the right-hand side of the equation were \( e^{2t} \) instead of \( te^{2t} \)?
3. How do undetermined coefficients differ from solving with the Laplace transform?
4. How does the solution change if the homogeneous equation has repeated roots?
5. What if the non-homogeneous term involved a sine or cosine function instead?

**Tip:** Always check if the non-homogeneous term is part of the homogeneous solution. If it is, modify your guess accordingly!

Discover how to solve the non-homogeneous differential equation y'' + 5y' + 6y = -20te^{2t} using the method of undetermined coefficients. The solution involves solving the characteristic equation, finding the general solution to the homogeneous equation, and identifying a particular solution for the given non-homogeneous term. Suitable for students at the undergraduate level.

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