Find y as a function of t if 9y''+31y=0, y(0)=5,y'(0)=6

The given differential equation is:

\[
9y'' + 31y = 0
\]

This is a second-order linear homogeneous differential equation with constant coefficients. Let's solve it step by step.

### Step 1: Characteristic Equation

To solve this, assume the solution is of the form \( y(t) = e^{rt} \), where \( r \) is a constant to be determined. Substituting \( y(t) = e^{rt} \) into the differential equation:

\[
9r^2 e^{rt} + 31 e^{rt} = 0
\]

Since \( e^{rt} \neq 0 \), we can divide through by \( e^{rt} \) to obtain the characteristic equation:

\[
9r^2 + 31 = 0
\]

### Step 2: Solve the Characteristic Equation

Solve for \( r \):

\[
9r^2 = -31 \quad \Rightarrow \quad r^2 = -\frac{31}{9}
\]

Taking the square root of both sides gives:

\[
r = \pm i \frac{\sqrt{31}}{3}
\]

Thus, the roots are complex numbers \( r = \pm i \frac{\sqrt{31}}{3} \).

### Step 3: General Solution

For complex roots \( r = \pm i \omega \), the general solution to the differential equation is of the form:

\[
y(t) = c_1 \cos(\omega t) + c_2 \sin(\omega t)
\]

where \( \omega = \frac{\sqrt{31}}{3} \). Substituting this value for \( \omega \), we get:

\[
y(t) = c_1 \cos\left(\frac{\sqrt{31}}{3} t\right) + c_2 \sin\left(\frac{\sqrt{31}}{3} t\right)
\]

### Step 4: Apply Initial Conditions

We are given the initial conditions:

- \( y(0) = 5 \)
- \( y'(0) = 6 \)

#### Apply \( y(0) = 5 \):

Substitute \( t = 0 \) into the general solution:

\[
y(0) = c_1 \cos\left(0\right) + c_2 \sin\left(0\right) = c_1
\]

Thus, \( c_1 = 5 \).

#### Apply \( y'(0) = 6 \):

Now, differentiate the general solution:

\[
y'(t) = -c_1 \cdot \frac{\sqrt{31}}{3} \sin\left(\frac{\sqrt{31}}{3} t\right) + c_2 \cdot \frac{\sqrt{31}}{3} \cos\left(\frac{\sqrt{31}}{3} t\right)
\]

Substitute \( t = 0 \) and use \( y'(0) = 6 \):

\[
y'(0) = -c_1 \cdot \frac{\sqrt{31}}{3} \sin(0) + c_2 \cdot \frac{\sqrt{31}}{3} \cos(0)
\]

This simplifies to:

\[
y'(0) = c_2 \cdot \frac{\sqrt{31}}{3} = 6
\]

Solve for \( c_2 \):

\[
c_2 = 6 \cdot \frac{3}{\sqrt{31}} = \frac{18}{\sqrt{31}}
\]

### Final Solution:

The solution to the differential equation is:

\[
y(t) = 5 \cos\left(\frac{\sqrt{31}}{3} t\right) + \frac{18}{\sqrt{31}} \sin\left(\frac{\sqrt{31}}{3} t\right)
\]

Would you like further details on any of the steps or have questions?

Here are some related questions to expand your understanding:

1. How does the nature of the roots affect the form of the solution to second-order ODEs?
2. What are the general forms of solutions for real vs complex roots in homogeneous differential equations?
3. How would the solution differ if the coefficients of the differential equation were different?
4. Can you explain the physical significance of the solution in a context like harmonic oscillators?
5. What would change if we had a non-homogeneous equation?

**Tip:** For second-order linear differential equations, the form of the solution depends crucially on the discriminant of the characteristic equation.

Find y as a function of t if 9y'' + 31y = 0, y(0) = 5, y'(0) = 6

Learn how to solve the second-order linear differential equation 9y'' + 31y = 0 with given initial conditions y(0) = 5 and y'(0) = 6. This detailed step-by-step solution explains the characteristic equation and complex root method, suitable for college-level calculus students.

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