https://file.aiquickdraw.com/mathbot/file/1728262882847cxgsx9bo.jpg

We are given the second-order linear homogeneous differential equation:

\[
9y'' + 6y' + y = 0
\]

with the initial conditions \( y(1) = 1 \) and \( y'(1) = 3 \).

### Step 1: Solve the characteristic equation
To solve this, we first solve the associated characteristic equation by assuming \( y = e^{rt} \). Substituting this into the differential equation yields:

\[
9r^2 e^{rt} + 6r e^{rt} + e^{rt} = 0
\]

Since \( e^{rt} \) is never zero, we can divide through by \( e^{rt} \):

\[
9r^2 + 6r + 1 = 0
\]

This is a quadratic equation, and we can solve it using the quadratic formula:

\[
r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \( 9r^2 + 6r + 1 = 0 \), we have \( a = 9 \), \( b = 6 \), and \( c = 1 \). Substituting these into the quadratic formula gives:

\[
r = \frac{-6 \pm \sqrt{6^2 - 4(9)(1)}}{2(9)} = \frac{-6 \pm \sqrt{36 - 36}}{18} = \frac{-6 \pm 0}{18} = \frac{-6}{18} = -\frac{1}{3}
\]

Thus, the characteristic equation has a repeated root \( r = -\frac{1}{3} \).

### Step 2: General solution
Since we have a repeated root, the general solution to the differential equation is:

\[
y(t) = (C_1 + C_2 t)e^{-\frac{1}{3}t}
\]

### Step 3: Apply the initial conditions
We now use the given initial conditions \( y(1) = 1 \) and \( y'(1) = 3 \) to determine the constants \( C_1 \) and \( C_2 \).

#### Condition 1: \( y(1) = 1 \)
Substitute \( t = 1 \) into the general solution:

\[
y(1) = (C_1 + C_2 \cdot 1)e^{-\frac{1}{3}} = 1
\]

This simplifies to:

\[
(C_1 + C_2)e^{-\frac{1}{3}} = 1
\]

or

\[
C_1 + C_2 = e^{\frac{1}{3}}
\quad \text{(equation 1)}
\]

#### Condition 2: \( y'(1) = 3 \)
First, find the derivative of \( y(t) \):

\[
y'(t) = \frac{d}{dt}\left[(C_1 + C_2 t)e^{-\frac{1}{3}t}\right]
\]

Using the product rule:

\[
y'(t) = (C_1 + C_2 t)\left(-\frac{1}{3}e^{-\frac{1}{3}t}\right) + C_2 e^{-\frac{1}{3}t}
\]

Simplifying:

\[
y'(t) = \left[-\frac{1}{3}(C_1 + C_2 t) + C_2\right]e^{-\frac{1}{3}t}
\]

Substitute \( t = 1 \) and \( y'(1) = 3 \):

\[
\left[-\frac{1}{3}(C_1 + C_2) + C_2\right]e^{-\frac{1}{3}} = 3
\]

Simplify:

\[
\left[-\frac{1}{3}(C_1 + C_2) + C_2\right] = 3e^{\frac{1}{3}}
\]

Substitute \( C_1 + C_2 = e^{\frac{1}{3}} \) from equation 1:

\[
\left[-\frac{1}{3}(e^{\frac{1}{3}}) + C_2\right] = 3e^{\frac{1}{3}}
\]

Simplifying:

\[
-\frac{1}{3}e^{\frac{1}{3}} + C_2 = 3e^{\frac{1}{3}}
\]

Solve for \( C_2 \):

\[
C_2 = 3e^{\frac{1}{3}} + \frac{1}{3}e^{\frac{1}{3}} = \frac{10}{3}e^{\frac{1}{3}}
\]

### Step 4: Solve for \( C_1 \)
Substitute \( C_2 = \frac{10}{3}e^{\frac{1}{3}} \) into equation 1:

\[
C_1 + \frac{10}{3}e^{\frac{1}{3}} = e^{\frac{1}{3}}
\]

Solving for \( C_1 \):

\[
C_1 = e^{\frac{1}{3}} - \frac{10}{3}e^{\frac{1}{3}} = -\frac{7}{3}e^{\frac{1}{3}}
\]

### Final Answer:
Thus, the solution to the initial value problem is:

\[
y(t) = \left(-\frac{7}{3} + \frac{10}{3}t\right)e^{-\frac{1}{3}t}
\]

Do you want more details or have any questions?

Here are 5 related questions:
1. How do repeated roots affect the general solution of a second-order linear differential equation?
2. What is the method for solving non-homogeneous linear differential equations?
3. How does the product rule apply in differentiating terms like \( (C_1 + C_2 t)e^{-\frac{1}{3}t} \)?
4. Can the method of undetermined coefficients be used for non-homogeneous differential equations?
5. How would the solution change if the initial conditions were different?

**Tip:** For higher-order differential equations, always start by solving the characteristic equation first to identify the general solution form.

Solve the initial value problem: 9y'' + 6y' + y = 0, y(1) = 1, y'(1) = 3.

Learn how to solve the second-order linear differential equation 9y'' + 6y' + y = 0 with initial conditions y(1) = 1 and y'(1) = 3. The solution involves solving the characteristic equation and applying the initial values.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation