Question

How to solve laplace transform differential equations?

Answer 1

Are you working on a specific type of differential equation, or do you need a general overview of how to solve them using Laplace transforms?

Answer 2

To solve differential equations using Laplace Transforms, follow these steps:

### 1. **Take the Laplace Transform of the Entire Equation:**
   - Use the linearity of the Laplace Transform to apply it to each term of the differential equation.
   - Remember to apply the following Laplace Transforms for derivatives:
     \[
     \mathcal{L}\{y'(t)\} = sY(s) - y(0)
     \]
     \[
     \mathcal{L}\{y''(t)\} = s^2Y(s) - sy(0) - y'(0)
     \]
     where \( Y(s) \) is the Laplace Transform of \( y(t) \).

### 2. **Substitute the Initial Conditions:**
   - Plug in the given initial conditions (e.g., \( y(0) \), \( y'(0) \)) into the transformed equation.

### 3. **Solve the Algebraic Equation in Terms of \( Y(s) \):**
   - After applying the Laplace Transform, you will get an algebraic equation in terms of \( Y(s) \). Solve for \( Y(s) \).

### 4. **Take the Inverse Laplace Transform:**
   - Once you have \( Y(s) \), use the inverse Laplace Transform to find \( y(t) \).
   - Use a table of common Laplace Transforms to match the terms in \( Y(s) \) with known inverse transforms.

### Example Problem:
Solve the differential equation:
\[
y'' + 3y' + 2y = 0, \quad y(0) = 1, \quad y'(0) = 0
\]

#### Step 1: Apply the Laplace Transform
\[
\mathcal{L}\{y''\} + 3\mathcal{L}\{y'\} + 2\mathcal{L}\{y\} = 0
\]
\[
(s^2Y(s) - sy(0) - y'(0)) + 3(sY(s) - y(0)) + 2Y(s) = 0
\]

#### Step 2: Substitute Initial Conditions
Given \( y(0) = 1 \) and \( y'(0) = 0 \), substitute them:
\[
(s^2Y(s) - s - 0) + 3(sY(s) - 1) + 2Y(s) = 0
\]
\[
s^2Y(s) - s + 3sY(s) - 3 + 2Y(s) = 0
\]

#### Step 3: Solve for \( Y(s) \)
Combine terms:
\[
(s^2 + 3s + 2)Y(s) = s + 3
\]
\[
Y(s) = \frac{s + 3}{s^2 + 3s + 2}
\]
Factor the denominator:
\[
Y(s) = \frac{s + 3}{(s + 1)(s + 2)}
\]

#### Step 4: Inverse Laplace Transform
Decompose the fraction using partial fraction decomposition:
\[
Y(s) = \frac{A}{s + 1} + \frac{B}{s + 2}
\]
Solving for \( A \) and \( B \), you get:
\[
Y(s) = \frac{1}{s + 1}
\]
Taking the inverse Laplace transform:
\[
y(t) = e^{-t}
\]

The solution is \( y(t) = e^{-t} \).