https://file.aiquickdraw.com/mathbot/file/1728517938464gz7cea9d.jpg

The image shows a second-order linear differential equation:

\[
64y'' + 48y' + 25y = 0
\]

along with initial conditions:

\[
y(1) = 8 \quad \text{and} \quad y'(1) = 4.
\]

This is a homogeneous linear differential equation with constant coefficients. To solve it, follow these steps:

### 1. Solve the characteristic equation:
Start by finding the characteristic equation associated with the differential equation. For the general form \( ay'' + by' + cy = 0 \), the characteristic equation is given by:

\[
a r^2 + b r + c = 0
\]

In this case, \( a = 64 \), \( b = 48 \), and \( c = 25 \). Thus, the characteristic equation is:

\[
64r^2 + 48r + 25 = 0
\]

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

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

Substituting the values of \( a \), \( b \), and \( c \):

\[
r = \frac{-48 \pm \sqrt{48^2 - 4(64)(25)}}{2(64)}
\]
\[
r = \frac{-48 \pm \sqrt{2304 - 6400}}{128}
\]
\[
r = \frac{-48 \pm \sqrt{-4096}}{128}
\]
\[
r = \frac{-48 \pm 64i}{128}
\]
\[
r = \frac{-48}{128} \pm \frac{64i}{128}
\]
\[
r = -\frac{3}{8} \pm \frac{i}{2}
\]

Thus, the roots are complex: \( r = -\frac{3}{8} \pm \frac{1}{2}i \).

### 2. General solution:
For complex roots of the form \( r = \alpha \pm \beta i \), the general solution to the differential equation is:

\[
y(t) = e^{\alpha t} \left( C_1 \cos(\beta t) + C_2 \sin(\beta t) \right)
\]

Substituting \( \alpha = -\frac{3}{8} \) and \( \beta = \frac{1}{2} \):

\[
y(t) = e^{-\frac{3}{8} t} \left( C_1 \cos\left( \frac{1}{2} t \right) + C_2 \sin\left( \frac{1}{2} t \right) \right)
\]

### 3. Apply initial conditions:
Now, use the initial conditions \( y(1) = 8 \) and \( y'(1) = 4 \) to find \( C_1 \) and \( C_2 \).

#### Condition 1: \( y(1) = 8 \)

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

\[
8 = e^{-\frac{3}{8}} \left( C_1 \cos\left( \frac{1}{2} \right) + C_2 \sin\left( \frac{1}{2} \right) \right)
\]

This gives one equation in terms of \( C_1 \) and \( C_2 \).

#### Condition 2: \( y'(1) = 4 \)

To apply this, first compute \( y'(t) \) by differentiating the general solution:

\[
y'(t) = e^{-\frac{3}{8} t} \left[ -\frac{3}{8} \left( C_1 \cos\left( \frac{1}{2} t \right) + C_2 \sin\left( \frac{1}{2} t \right) \right) + \frac{1}{2} \left( -C_1 \sin\left( \frac{1}{2} t \right) + C_2 \cos\left( \frac{1}{2} t \right) \right) \right]
\]

Now substitute \( t = 1 \) into this expression and set it equal to 4:

\[
4 = e^{-\frac{3}{8}} \left[ -\frac{3}{8} \left( C_1 \cos\left( \frac{1}{2} \right) + C_2 \sin\left( \frac{1}{2} \right) \right) + \frac{1}{2} \left( -C_1 \sin\left( \frac{1}{2} \right) + C_2 \cos\left( \frac{1}{2} \right) \right) \right]
\]

This gives the second equation.

### 4. Solve for \( C_1 \) and \( C_2 \):
Solve the system of equations obtained from the initial conditions to determine the constants \( C_1 \) and \( C_2 \).

Let me know if you want help solving for \( C_1 \) and \( C_2 \) or if you would like further details!

---

### Questions:
1. How do we solve a characteristic equation when the roots are complex?
2. What is the significance of complex roots in the solution of differential equations?
3. How do we handle initial conditions in second-order differential equations?
4. What is the general form of the solution when the roots of the characteristic equation are complex conjugates?
5. Why does the solution involve both cosine and sine functions for complex roots?

### Tip:
When dealing with initial conditions, make sure to carefully differentiate the general solution and plug in the correct values at each step to avoid errors.

Solve the second-order linear differential equation: 64y'' + 48y' + 25y = 0, with initial conditions y(1) = 8, y'(1) = 4.

This page explains how to solve the second-order linear differential equation 64y'' + 48y' + 25y = 0, including steps to find the characteristic equation, solve for complex roots, and apply initial conditions y(1) = 8 and y'(1) = 4. Suitable for college-level calculus students.

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