Laplace Transform and Initial Value Problems

Transformations of Derivatives

• The transformation of the first derivative of a function is given by:
    - $L y'(t) = s Y_s - y(0)$ where $Y_s = L y(t)$

• For the second derivative, the transformation is:
    - $L y''(t) = s^2 Y_s - s y(0) - y'(0)$

Solving Initial Value Problems (IVPs)

General Steps:

• Start with a differential equation (DE) in $y(t)$ and initial conditions
• Produce an algebraic equation in $Y_s$ (Laplace transformed variable)
• Apply the Laplace transform to convert the differential equation to an algebraic equation
• Solve the transformed equation for $Y_s$
• Find the solution $y(t)$ of the original initial value problem (IVP)
• Apply the inverse Laplace transform to derive $y(t)$

Example IVPs Using Laplace Transform:

Problem 1

• Equation: $2y' + y = 0$

• Initial Condition: $y(0) = -3$
  

• Steps to Solve:
    1. Set $Y = L y$.
    2. Take the Laplace transform of each side:
       - $L(2y') + L(y) = 0$
    3. Solve the transformed algebraic equation for $Y$.
    4. Find the inverse Laplace transform of $Y$.
    5. Write the resulting equation for $y(t)$.

• Solution: The solution for the given IVP is:
    - $y(t) = -\frac{35}{2} e^{0t} + 0$ with $y(0) = -3$

Problem 2

• Equation: $y' - y = 1$

• Initial Condition: $y(0) = 0$
  

• Steps to Solve:
    1. Set $Y = L y$.
    2. Take the Laplace transform of both sides:
       - $L(y') - L(y) = L(1)$
    3. Solve the transformed algebraic equation for $Y$.
    4. Find the inverse Laplace transform of $Y$.
    5. Express in terms of $y(t)$.

• Solution: The unique solution for the IVP is:
    - $y(t) = e^{t} - 1$ with $y(0) = 0$

Problem 3

• Equation: $y'' + 9y = e^{t}$

• Initial Conditions: $y(0) = 0, y'(0) = 0$

• Steps to Solve:
    1. Set $Y = L y$.
    2. Take the Laplace transform of both sides:
       - $L(y'') + 9 L(y) = L(e^{t})$
    3. Solve the transformed algebraic equation for $Y$.

• Finding Inverse:
    - Use partial fraction decomposition to break down the expression:
      - For instance, consider the rational expression given by:
  $\frac{1}{s(s^2 + 9)}$
      - Decompose this:
  $\frac{1}{s} - \frac{1}{s^2 + 9}$
        and apply LT tables to find the inverse of each part:
        - $\frac{0.1}{s} - \frac{0.1 s}{s^2 + 9} - \frac{0.1}{s^2 + 9}$
    4. After finding individual inverses, compile to find the solution:

• Solution: The unique solution for IVP:
    - $y(t) = 0.1 e^t - 0.1 ext{cos}(3t) - 0.1\frac{1}{3} ext{sin}(3t)$

Solving Boundary Value Problems (BVPs)

Problem 4

• Equation: $y'' + 4y = 0$
    - Initial Conditions: $y(0) = 2, y(\frac{ heta}{4}) = -2$

• Steps to Solve:
    1. Set $Y = L y$ and let $C = y'(0)$.
    2. Take the Laplace transform:
       - $L(y'') + 4L(y) = 0$
    3. Solve the resulting equation for $Y$.
    4. Use inverse Laplace transform to recover $y(t)$.

Combined Exponential and Polynomial Terms IVP

Problem 5

• Equation: $y'' - 3y' + 2y = e^{4t}$
    - Initial Conditions: $y(0) = 1, y'(0) = 5$

• Steps to Solve:
    1. Set $Y = L y$.
    2. Take Laplace transform:
       - $L(y'') - 3L(y') + 2L(y) = L(e^{4t})$
    3. Solve the algebraic equation for $Y$.
       - $Y = \frac{s^2 + 6s + 9}{s^3} + \frac{1}{s - 4}$
    4. Use partial fraction decomposition:
       - Decompose it into individual fractions:
  $-\frac{16}{5(s - 1)} + \frac{25}{6(s - 2)} + \frac{1}{30(s + 4)}$
    5. Write the equation for $y(t)$:

• Final Result:
    - $y(t) = -\frac{16}{5} e^t + \frac{25}{6} e^{2t} + \frac{1}{30} e^{4t}$