Solving Differential Equations with Laplace Transforms
Solving Laplace Using Differential Equations
Intro to Laplace »
Inverse »
Heaviside »
Differential Equations »
Laplace Transforms
First-Order ODE Solver: $ay' + by = K$
Solve $ay'(t) + by(t) = K$ with $y(0) = y_0$
Alternatively, fill in the coefficients below directly:
Solution steps will appear here...
Introduction
The Laplace transform is a powerful mathematical tool used to solve linear ordinary differential equations (ODEs) with constant coefficients. It transforms a differential equation in the time domain (typically involving functions of $t$) into an algebraic equation in the complex frequency domain (involving functions of $s$).
The key advantages of using Laplace transforms are:
- It converts differentiation and integration operations in the time domain into algebraic operations (multiplication and division by $s$) in the $s$-domain.
- Initial conditions are directly incorporated into the transformation process, simplifying the solution for initial value problems.
- It can readily handle discontinuous or impulsive forcing functions (like step functions or Dirac delta functions), which are common in engineering systems.
After solving the algebraic equation in the $s$-domain for $Y(s)$ (the Laplace transform of the solution $y(t)$), the solution $y(t)$ is found by applying the inverse Laplace transform.
Procedure for Solving ODEs
The general procedure to solve an $n$-th order linear ODE with constant coefficients using Laplace transforms is as follows:
- Take the Laplace Transform: Apply the Laplace transform to both sides of the given differential equation. Use the linearity property and the transform rules for derivatives. This will convert the ODE in $y(t)$ into an algebraic equation in $Y(s) = \mathcal{L}\{y(t)\}$.
- Incorporate Initial Conditions: The Laplace transforms of derivatives will introduce the initial conditions of the problem (e.g., $y(0)$, $y'(0)$, etc.) directly into the $s$-domain equation.
- Solve for $Y(s)$ algebraically: Rearrange the transformed equation to solve for $Y(s)$. This typically involves basic algebraic manipulation.
- Perform Partial Fraction Expansion (if necessary): The expression for $Y(s)$ is often a rational function of $s$. To find its inverse transform, it's usually necessary to decompose $Y(s)$ into simpler terms using partial fraction expansion.
- Take the Inverse Laplace Transform: Apply the inverse Laplace transform to $Y(s)$ (often term by term after partial fraction expansion) to obtain the solution $y(t)$ in the time domain. This involves recognizing standard Laplace transform pairs from a table.
Common Laplace Transforms and Properties
Basic Transforms
| $f(t)$ | $F(s) = \mathcal{L}\{f(t)\}$ |
|---|---|
| $1$ | $$ \frac{1}{s} $$ |
| $t$ | $$ \frac{1}{s^2} $$ |
| $t^n$ ($n \ge 0$ integer) | $$ \frac{n!}{s^{n+1}} $$ |
| $e^{at}$ | $$ \frac{1}{s-a} $$ |
| $\sin(kt)$ | $$ \frac{k}{s^2+k^2} $$ |
| $\cos(kt)$ | $$ \frac{s}{s^2+k^2} $$ |
| $t e^{at}$ | $$ \frac{1}{(s-a)^2} $$ |
| $e^{at}\sin(kt)$ | $$ \frac{k}{(s-a)^2+k^2} $$ |
| $e^{at}\cos(kt)$ | $$ \frac{s-a}{(s-a)^2+k^2} $$ |
| Unit step function $u(t-c)$ | $$ \frac{e^{-cs}}{s} $$ |
| Dirac delta function $\delta(t-c)$ | $$ e^{-cs} $$ |
Properties (especially for derivatives)
| Property | Time Domain $f(t)$ | Frequency Domain $F(s)$ |
|---|---|---|
| Linearity | $af(t) + bg(t)$ | $aF(s) + bG(s)$ |
| First Derivative | $f'(t)$ | $sF(s) - f(0)$ |
| Second Derivative | $f''(t)$ | $s^2F(s) - sf(0) - f'(0)$ |
| $n$-th Derivative | $f^{(n)}(t)$ | $s^nF(s) - s^{n-1}f(0) - \dots - f^{(n-1)}(0)$ |
| Time Shift | $f(t-c)u(t-c)$ | $e^{-cs}F(s)$ |
| Frequency Shift | $e^{at}f(t)$ | $F(s-a)$ |