Find the particular solution to: y’’+4y’+3y=-18te^(4t)

To find the particular solution of the non-homogeneous differential equation:

\[
y'' + 4y' + 3y = -18t e^{4t}
\]

We will use the **method of undetermined coefficients**. Here’s how we proceed:

### Step 1: Solve the homogeneous equation

The homogeneous part of the equation is:

\[
y'' + 4y' + 3y = 0
\]

The characteristic equation for this is:

\[
r^2 + 4r + 3 = 0
\]

Factoring the characteristic equation:

\[
(r + 3)(r + 1) = 0
\]

So, the roots are \( r = -3 \) and \( r = -1 \).

Thus, the general solution to the homogeneous equation is:

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

### Step 2: Find the particular solution

The non-homogeneous term is \( -18t e^{4t} \). Since the non-homogeneous term involves a product of a polynomial \( t \) and an exponential \( e^{4t} \), we assume the particular solution to be of the form:

\[
y_p(t) = (At + B) e^{4t}
\]

Here, \( A \) and \( B \) are constants to be determined.

### Step 3: Differentiate \( y_p(t) \)

First derivative of \( y_p(t) \):

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

Simplify:

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

Second derivative of \( y_p(t) \):

\[
y_p''(t) = (0 + A + 4A + 16At + 16B) e^{4t}
\]

Simplify:

\[
y_p''(t) = (5A + 16At + 16B) e^{4t}
\]

### Step 4: Substitute into the original equation

Substitute \( y_p(t) \), \( y_p'(t) \), and \( y_p''(t) \) into the original equation:

\[
(5A + 16At + 16B) e^{4t} + 4(A + 4At + 4B) e^{4t} + 3(At + B) e^{4t} = -18t e^{4t}
\]

Factor out \( e^{4t} \):

\[
e^{4t} \left[ (5A + 16At + 16B) + 4(A + 4At + 4B) + 3(At + B) \right] = -18t e^{4t}
\]

Now, simplify the terms inside the brackets:

\[
(5A + 16At + 16B) + (4A + 16At + 16B) + (3At + 3B)
\]

This simplifies to:

\[
(5A + 4A) + (16At + 16At + 3At) + (16B + 16B + 3B)
\]

\[
9A + 35At + 35B
\]

Thus, the equation becomes:

\[
e^{4t} \left[ 9A + 35At + 35B \right] = -18t e^{4t}
\]

Cancel \( e^{4t} \) on both sides:

\[
9A + 35At + 35B = -18t
\]

### Step 5: Solve for \( A \) and \( B \)

Now, equate the coefficients of \( t \) and the constant terms:

- For the \( t \)-terms: \( 35A = -18 \), so \( A = \frac{-18}{35} \).
- For the constant terms: \( 9A + 35B = 0 \). Substituting \( A = \frac{-18}{35} \):

\[
9\left(\frac{-18}{35}\right) + 35B = 0
\]

\[
\frac{-162}{35} + 35B = 0
\]

\[
35B = \frac{162}{35}
\]

\[
B = \frac{162}{35 \times 35} = \frac{162}{1225}
\]

Thus, \( A = \frac{-18}{35} \) and \( B = \frac{162}{1225} \).

### Step 6: Write the particular solution

The particular solution is:

\[
y_p(t) = \left( \frac{-18}{35} t + \frac{162}{1225} \right) e^{4t}
\]

### Step 7: General solution

The general solution to the differential equation is:

\[
y(t) = C_1 e^{-3t} + C_2 e^{-t} + \left( \frac{-18}{35} t + \frac{162}{1225} \right) e^{4t}
\]

---

Would you like more details or have any questions?

Here are 5 related questions:
1. How would you solve a second-order differential equation with a different non-homogeneous term?
2. What if the exponential function in the non-homogeneous term had a different base, like \( e^{kt} \)?
3. How would you handle repeated roots in the characteristic equation?
4. Can we solve this problem using variation of parameters?
5. What is the role of the characteristic equation in solving homogeneous equations?

**Tip:** Always factor the characteristic equation carefully, as it helps determine the form of the homogeneous solution.

Explore the step-by-step solution for finding the particular solution of the non-homogeneous differential equation y’’ + 4y’ + 3y = -18t e^(4t) using the method of undetermined coefficients. This guide covers solving the characteristic equation, finding the homogeneous solution, and determining the particular solution form. Suitable for undergraduate students studying differential equations.

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