# Study Notes: Solving ODEs with the Laplace Transform The Laplace transform is a powerful tool for solving ordinary differential equations (ODEs), especially when initial conditions are involved. This study note will guide you through key concepts, properties, and examples to help you master the Laplace transform and its inverse. --- ## Key Concepts ### What is the Laplace Transform? The **Laplace transform** converts a time-domain function $f(t)$ into a complex frequency-domain function $F(s)$. It is defined as: $$ L[f(t)] = F(s) = \int_0^\infty e^{-st} f(t) \, dt $$ - **$t$**: Time variable (independent variable in the time domain) - **$s$**: Complex frequency variable (independent variable in the Laplace domain) The Laplace transform simplifies differential equations by converting derivatives into algebraic terms. --- ### Solving ODEs Using the Laplace Transform To solve an ODE using the Laplace transform: 1. Take the Laplace transform of both sides of the ODE. 2. Use initial conditions to simplify the transformed equation. 3. Solve for $F(s)$ (the Laplace transform of the solution). 4. Apply the **inverse Laplace transform** to find $f(t)$. --- ## Example: Solving an ODE with the Laplace Transform Solve the ODE: $$ y' - y = e^{2t}, \quad y(0) = 1 $$ ### Step 1: Take the Laplace Transform Using the property $L[y'(t)] = sY(s) - y(0)$, where $Y(s)$ is the Laplace transform of $y(t)$: $$ L[y'(t)] - L[y(t)] = L[e^{2t}] $$ Substitute the initial condition $y(0) = 1$: $$ (sY(s) - 1) - Y(s) = \frac{1}{s - 2} $$ ### Step 2: Simplify the Equation Combine terms: $$ sY(s) - Y(s) - 1 = \frac{1}{s - 2} $$ Factor $Y(s)$: $$ Y(s)(s - 1) = \frac{1}{s - 2} + 1 $$ Simplify the right-hand side: $$ Y(s)(s - 1) = \frac{1}{s - 2} + \frac{s - 2}{s - 2} = \frac{s - 1}{s - 2} $$ Solve for $Y(s)$: $$ Y(s) = \frac{s - 1}{(s - 1)(s - 2)} = \frac{1}{s - 2} $$ ### Step 3: Apply the Inverse Laplace Transform The inverse Laplace transform of $\frac{1}{s - 2}$ is $e^{2t}$. Thus: $$ y(t) = e^{2t} $$ Let’s verify the solution by substituting $y(t) = e^{2t}$ back into the original ODE. The derivative $y'(t) = 2e^{2t}$, and substituting into $y' - y = e^{2t}$ confirms the solution is correct. --- ## Properties of the Laplace Transform ### Linearity The Laplace transform is **linear**, meaning: $$ L[A f(t) + B g(t)] = A L[f(t)] + B L[g(t)] $$ - **$A$ and $B$**: Constants - **$f(t)$ and $g(t)$**: Functions of $t$ To find $L[\sin(kt) + \cos(kt)]$, use linearity: $$ L[\sin(kt)] + L[\cos(kt)] = \frac{k}{s^2 + k^2} + \frac{s}{s^2 + k^2} $$ --- ### Common Laplace Transforms Here are some frequently used Laplace transforms: | Function $f(t)$ | Laplace Transform $F(s)$ | |------------------------|----------------------------------| | $1$ | $\frac{1}{s}$ | | $t$ | $\frac{1}{s^2}$ | | $e^{at}$ | $\frac{1}{s - a}$ | | $\sin(kt)$ | $\frac{k}{s^2 + k^2}$ | | $\cos(kt)$ | $\frac{s}{s^2 + k^2}$ | Memorizing these common transforms will save time when solving problems. --- ## Inverse Laplace Transform The **inverse Laplace transform** converts a function $F(s)$ back into the time domain $f(t)$. It is denoted as: $$ L^{-1}[F(s)] = f(t) $$ ### Example: Find the Inverse Laplace Transform Given: $$ F(s) = \frac{1}{s + 2} + \frac{4}{s^2} $$ 1. Break into separate terms: - $\frac{1}{s + 2}$ and $\frac{4}{s^2}$ 2. Apply the inverse transform: - $L^{-1}\left[\frac{1}{s + 2}\right] = e^{-2t}$ - $L^{-1}\left[\frac{4}{s^2}\right] = 4t$ 3. Combine results: - $f(t) = e^{-2t} + 4t$ Don’t forget to handle each term separately when applying the inverse Laplace transform. --- ## Practice Problems 1. Compute the Laplace transform of $t^2$. 2. Solve the ODE $y'' + 3y' + 2y = 0$, with $y(0) = 1$ and $y'(0) = 0$. 3. Find the inverse Laplace transform of $\frac{2s + 3}{s^2 + 4s + 5}$. Use partial fraction decomposition when necessary to simplify $F(s)$ before applying the inverse Laplace transform. --- ## Key Takeaways - The Laplace transform converts differential equations into algebraic equations, making them easier to solve. - Use initial conditions to simplify the transformed equation. - The inverse Laplace transform is essential for converting solutions back to the time domain. - Memorize common Laplace transforms and properties like linearity for efficiency. Mastering the Laplace transform requires practice. Work through examples and verify your solutions to build confidence.