Given the differential equation
y''-8y'+32y=0, y(0)=-1, y'(0)=1
Apply the Laplace Transform and solve for Y(s)
Find the inverse laplace transform y(t) = L^-1{Y(s)}

To solve the differential equation 
\[
y'' - 8y' + 32y = 0, \quad y(0) = -1, \quad y'(0) = 1
\]
using the Laplace Transform, let's proceed step by step.

### Step 1: Apply the Laplace Transform to the Differential Equation
The Laplace Transform of \( y(t) \), \( y'(t) \), and \( y''(t) \) are:
1. \( \mathcal{L}\{y(t)\} = Y(s) \)
2. \( \mathcal{L}\{y'(t)\} = sY(s) - y(0) \)
3. \( \mathcal{L}\{y''(t)\} = s^2 Y(s) - s y(0) - y'(0) \)

Now, applying the Laplace Transform to each term in the differential equation:
\[
\mathcal{L}\{y'' - 8y' + 32y\} = \mathcal{L}\{0\}
\]

Substituting the transforms:
\[
s^2 Y(s) - s \cdot y(0) - y'(0) - 8(s Y(s) - y(0)) + 32Y(s) = 0
\]

Using \( y(0) = -1 \) and \( y'(0) = 1 \):
\[
s^2 Y(s) - s(-1) - 1 - 8(s Y(s) - (-1)) + 32 Y(s) = 0
\]

Simplifying each term:
\[
s^2 Y(s) + s - 1 - 8s Y(s) - 8 + 32 Y(s) = 0
\]

Combine like terms in \( Y(s) \):
\[
(s^2 - 8s + 32) Y(s) = -s + 9
\]

Thus, we have:
\[
Y(s) = \frac{-s + 9}{s^2 - 8s + 32}
\]

### Step 2: Simplify \( Y(s) \)
To proceed, we’ll factor the denominator:
\[
s^2 - 8s + 32 = (s - 4)^2 + 16
\]
Thus, the expression for \( Y(s) \) becomes:
\[
Y(s) = \frac{-s + 9}{(s - 4)^2 + 16}
\]

### Step 3: Partial Fraction Decomposition
Rewrite \( -s + 9 \) in terms of \( s - 4 \):
\[
-s + 9 = -(s - 4) + 5
\]
Substitute this back into \( Y(s) \):
\[
Y(s) = \frac{-(s - 4) + 5}{(s - 4)^2 + 16} = \frac{-(s - 4)}{(s - 4)^2 + 16} + \frac{5}{(s - 4)^2 + 16}
\]

This separates \( Y(s) \) into two terms:
\[
Y(s) = \frac{-(s - 4)}{(s - 4)^2 + 16} + \frac{5}{(s - 4)^2 + 16}
\]

### Step 4: Take the Inverse Laplace Transform
Now we can find \( y(t) \) by taking the inverse Laplace transform of each term separately.

1. For the first term:
   \[
   \mathcal{L}^{-1} \left( \frac{-(s - 4)}{(s - 4)^2 + 16} \right) = -\cos(4t) \cdot e^{4t}
   \]

2. For the second term:
   \[
   \mathcal{L}^{-1} \left( \frac{5}{(s - 4)^2 + 16} \right) = 5 \cdot \frac{\sin(4t)}{4} \cdot e^{4t} = \frac{5}{4} e^{4t} \sin(4t)
   \]

### Final Solution
Combining these results, the solution \( y(t) \) is:
\[
y(t) = -e^{4t} \cos(4t) + \frac{5}{4} e^{4t} \sin(4t)
\]

Let me know if you need more details on any of these steps or have any questions!

---

#### Additional Related Questions:
1. How can the Laplace transform be used to solve non-homogeneous differential equations?
2. What is the role of initial conditions in solving differential equations using Laplace transforms?
3. How would the solution change if the differential equation had different initial conditions?
4. How can we interpret the behavior of \( y(t) \) based on the exponential and sinusoidal terms?
5. How would you handle complex roots in characteristic equations for differential equations?

#### Tip:
For differential equations with repeated or complex roots, using the exponential and sinusoidal functions derived from the Laplace domain is a powerful technique.

Given the differential equation y''-8y'+32y=0, y(0)=-1, y'(0)=1, apply the Laplace Transform and solve for Y(s). Find the inverse laplace transform y(t) = L^-1{Y(s)}.

Explore the solution to the differential equation y''-8y'+32y=0 with initial conditions y(0)=-1 and y'(0)=1 using Laplace Transform techniques. This detailed guide walks you through the application of Laplace transforms, providing the solution Y(s) and the inverse transform y(t). Ideal for students in Grades 11-12.

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