Find the particular solution to y’’+7y’+10y=5te^(5t)

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

\[
y'' + 7y' + 10y = 5te^{5t}
\]

### Step 1: Solve the homogeneous equation
The corresponding homogeneous equation is:

\[
y_h'' + 7y_h' + 10y_h = 0
\]

We begin by solving the characteristic equation:

\[
r^2 + 7r + 10 = 0
\]

Factoring the quadratic:

\[
(r + 5)(r + 2) = 0
\]

Thus, the solutions to the characteristic equation are \( r = -5 \) and \( r = -2 \). Therefore, the general solution to the homogeneous equation is:

\[
y_h(t) = C_1 e^{-5t} + C_2 e^{-2t}
\]

### Step 2: Form of the particular solution
Now, we need to find a particular solution to the non-homogeneous equation:

\[
y_p'' + 7y_p' + 10y_p = 5te^{5t}
\]

The right-hand side \( 5te^{5t} \) suggests we try a particular solution of the form:

\[
y_p(t) = (At + B) e^{5t}
\]

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

### Step 3: Differentiate \( y_p(t) \)
We compute the first and second derivatives of \( y_p(t) \):

1. First derivative \( y_p'(t) \):
\[
y_p'(t) = \frac{d}{dt} \left( (At + B) e^{5t} \right) = A e^{5t} + (At + B) \cdot 5e^{5t} = (5At + 5B + A) e^{5t}
\]

2. Second derivative \( y_p''(t) \):
\[
y_p''(t) = \frac{d}{dt} \left( (5At + 5B + A) e^{5t} \right) = (5A e^{5t}) + (5At + 5B + A) \cdot 5e^{5t} = (25At + 25B + 6A) e^{5t}
\]

### Step 4: Substitute into the original equation
Substitute \( y_p(t) \), \( y_p'(t) \), and \( y_p''(t) \) into the original equation:

\[
(25At + 25B + 6A) e^{5t} + 7(5At + 5B + A) e^{5t} + 10(At + B) e^{5t} = 5te^{5t}
\]

Factor out \( e^{5t} \):

\[
e^{5t} \left[ (25At + 25B + 6A) + 7(5At + 5B + A) + 10(At + B) \right] = 5te^{5t}
\]

Simplify the expression inside the brackets:

\[
(25A + 35A + 10A)t + (25B + 35B + 6A + 7A + 10B) = 5t
\]
\[
(70A)t + (70B + 13A) = 5t
\]

### Step 5: Solve for \( A \) and \( B \)
To satisfy the equation for all \( t \), equate the coefficients of \( t \) and the constant terms:

1. Coefficient of \( t \):
\[
70A = 5 \quad \Rightarrow \quad A = \frac{1}{14}
\]

2. Constant term:
\[
70B + 13A = 0 \quad \Rightarrow \quad 70B + 13 \times \frac{1}{14} = 0
\]
\[
70B + \frac{13}{14} = 0 \quad \Rightarrow \quad B = -\frac{13}{980}
\]

### Step 6: Write the particular solution
The particular solution is:

\[
y_p(t) = \left( \frac{1}{14}t - \frac{13}{980} \right) e^{5t}
\]

### Step 7: General solution
The general solution to the differential equation is the sum of the homogeneous and particular solutions:

\[
y(t) = C_1 e^{-5t} + C_2 e^{-2t} + \left( \frac{1}{14}t - \frac{13}{980} \right) e^{5t}
\]

Let me know if you'd like more details on any part of this process or further clarifications!

### Further exploration:
1. How do you solve differential equations with repeated roots in the characteristic equation?
2. What is the method of undetermined coefficients and its applications?
3. Can you apply variation of parameters to this problem? How would the solution change?
4. What are some common forms for the particular solution in non-homogeneous equations?
5. How do the properties of the homogeneous solution affect the particular solution choice?

**Tip:** When using the method of undetermined coefficients, always check if your trial solution conflicts with the homogeneous solution and adjust accordingly!

Learn how to solve the non-homogeneous second-order differential equation y'' + 7y' + 10y = 5te^(5t) using the method of undetermined coefficients. This step-by-step solution includes solving the homogeneous equation, applying the method of undetermined coefficients, and finding the general solution.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation