4.4: Solving ODEs with Laplace Transforms

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Reading: Notes on Diffy Q’s Section 6.2

Steps for Solving ODE’s with Laplace Transforms:

Let’s focus our exploration with Laplace transforms on their application to differential equations. We can apply the Laplace transform to solve differential equations with a frequently used problem solving strategy:

Step 1: Transform a difficult problem into an easier one.

Step 2: Solve the easier problem.

Step 3: Use the previous solution to obtain a solution to the original problem.

We will walk through each step to solve the differential equation

\[\color{dodgerblue}{y''-2y'+5y=0 \quad \mbox{with} \quad y(0)=2 \mbox{ and } y'(0)=4}.\]

Step 1: Transform a difficult problem into an easier one.

\[\begin{array}{rcll} \mathscr{L} \left\{ y''-2y'+5y \right\} &=& \mathscr{L} \left\{ 0 \right\} & {\color{tomato}{\mbox{Apply $\mathscr{L}$ to both sides.}}} \\ \mathscr{L} \left\{ y'' \right\} -2\mathscr{L} \left\{y'\right\} + 5 \mathscr{L} \left\{ y \right\} &=& \mathscr{L} \left\{ 0 \right\} & {\color{tomato}{\mbox{Applying properties 1 and 2 on left side.}}}\\ \mathscr{L} \left\{ y'' \right\} -2\mathscr{L} \left\{y'\right\} + 5 Y(s) &=& 0 & {\color{tomato}{\mbox{a. justification ??}}}\\ \big( s^2Y(s)-sy(0)-y'(0) \big) - 2 \big( sY(s) - y(0) \big) + 5 Y(s) &=& 0 & {\color{tomato}{\mbox{b. justification ??}}}\\ \big( s^2Y(s)-2s-4 \big) - 2 \big( sY(s) - 2 \big) + 5 Y(s) &=& 0 & {\color{tomato}{\mbox{c. justification ??}}}\\ \end{array} \]

Question 1:

Justify each of the steps labeled a, b and c in the work above.

Solution to Question 1:


  1. ??

  2. ??

  3. ??




Step 2: Solve the easier problem.

  • Rearrange and group like terms.
  • Then solve for \(Y(s)\).

\[\begin{array}{rcll} \big( s^2Y(s)-2s-4 \big) - 2 \big( sY(s) - 2 \big) + 5 Y(s) &=& 0 & {\color{tomato}{\mbox{Result from Step 1}}}\\ \big( s^2Y(s) - 2sY(s) +5Y(s) \big) - 2s &=& 0 & {\color{tomato}{\mbox{a. justification ??}}}\\ (s^2 - 2s +5)Y(s) &=& 2s & {\color{tomato}{\mbox{b. justification ??}}}\\ Y(s) &=& \dfrac{2s}{s^2-2s+5} & {\color{tomato}{\mbox{c. justification ??}}}\\ \end{array}\]

Question 2:

Justify each of the steps labeled a, b and c in the work above.

Solution to Question 2:


  1. ??

  2. ??

  3. ??



Step 3: Use the previous solution to obtain a solution of the original problem.

  • Take the inverse Laplace transform and solve for \(y(t) = \mathscr{L}^{-1}\{Y(x) \}\).
  • The function sympy.inverse_laplace_transform(F,s,t) will be helpful!

A solution to the original differential equation is a function \(y(t)\) whose Laplace transform is \(Y(s)\) from Step 2. Thus we apply the inverse Laplace transfrom to the result from Step 2 and solve for \(y(t)\).

\[\begin{array}{rcll} \mathscr{L}^{-1} \left\{ Y(s) \right\} &=& \mathscr{L}^{-1} \left\{ \dfrac{2s}{s^2-2s+5} \right\} & {\color{tomato}{\mbox{Applying $\mathscr{L}^{-1}$ to both sides.}}}\\ y(t) &=& \mathscr{L}^{-1} \left\{ \dfrac{2s}{(s-1)^2 + 4} \right\} & {\color{tomato}{\mbox{a. justification ??}}}\\ y(t) &=& 2 \mathscr{L}^{-1} \left\{ \dfrac{s}{(s-1)^2 + 4} \right\} & {\color{tomato}{\mbox{b. justification ??}}}\\ y(t) &=& 2 \mathscr{L}^{-1} \left\{ \dfrac{s-1+1}{(s-1)^2 + 4} \right\} & {\color{tomato}{\mbox{c. justification ??}}}\\ y(t) &=& 2 \mathscr{L}^{-1} \left\{ \dfrac{s-1}{(s-1)^2 + 4} + \dfrac{1}{(s-1)^2 + 4} \right\} & {\color{tomato}{\mbox{d. justification ??}}}\\ y(t) &=& 2 \mathscr{L}^{-1} \left\{ \dfrac{s-1}{(s-1)^2 + 4} \right\} + 2 \mathscr{L}^{-1} \left\{ \dfrac{1}{(s-1)^2 + 4} \right\} & {\color{tomato}{\mbox{e. justification ??}}}\\ y(t) &=& 2 \mathscr{L}^{-1} \left\{ \dfrac{s-1}{(s-1)^2 + 4} \right\} + \mathscr{L}^{-1} \left\{ \dfrac{2}{(s-1)^2 + 4} \right\} & {\color{tomato}{\mbox{f. justification ??}}}\\ y(t) &=& 2e^t\cos{(2t)}+ e^t\sin{(2t)} & {\color{tomato}{\mbox{g. justification ??}}}\\ \end{array}\]

Question 3:

Justify each of the steps labeled a to g in the work above.

Solution to Question 3:


  1. ??

  2. ??

  3. ??

  4. ??

  5. ??

  6. ??

  7. ??




Question 4:

In Questions 1, 2, and 3 we apply the Laplace transform to solve the differential equation

\[y''-2y'+5y=0 \quad \mbox{with} \quad y(0)=2 \mbox{ and } y'(0)=4,\]

and we discovered the solution

\[y(t) = 2e^t\cos{(2t)}+ e^t\sin{(2t)}.\]

Question 4a:

What other method(s) for solving differential equations could we use to solve this differential equation?

Solution to Question 4a:






Question 4b:

Based on your answer to part a, explain why the form of the solution is not surprising.

Solution to Question 4b:






Question 4c:

What are some of the pros and cons of using the Laplace transforms to solve the differential equation compared to the method you identified in part a?

Solution to Question 4c:






Solving ODE’s with Laplace Transforms and SymPy

As you may have gathered, using the Laplace transform to solve differential equations may present some challenges at each step. In particular, finding the inverse Laplace transform of \(Y(s)\) in the last step involved the most work. We can use the built-in Laplace transform functions in SymPy to help with some of the work.

Below are a couple of additional resources for help with Laplace transforms with SymPy.

Defining Laplace and Inverse Laplace Transform Functions:

Run the code cell below to define the Laplace transform function L(f) and inverse Laplace transform function invL(F).

import sympy as sym
from sympy.abc import s,t

# Just run this code cell. Do Not Edit.

# Define the Laplace transform function
def L(f):
    return sym.laplace_transform(f, s, t)

# Define the inverse Laplace transform function
def invL(F):
    return sym.inverse_laplace_transform(F, s, t)

Question 5:

In Question 3, you explain the algebra and properties of inverse Laplace transforms applied in Step 3 of solving a differential equation with the Laplace transform. Below is a recap of the result.

What is a function \(y(t)\) whose Laplace tranform is \(Y(s) = \dfrac{2s}{s^2-2s+5}\)?

\[\begin{array}{rcl} \mathscr{L}^{-1} \left\{ Y(s) \right\} &=& \mathscr{L}^{-1} \left\{ \dfrac{2s}{s^2-2s+5} \right\} \\ \vdots & = & \vdots \\ y(t) &=& 2e^t\cos{(2t)}+ e^t\sin{(2t)}\\ \end{array}\]

Edit the code cell below to use SymPy to solve for \(y(t)\) and verify the previous answer. Be sure you first run the previous code cell to import SymPy and Laplace transform functions.

Solution to Question 5:




Edit the code cell below. 


#############################################################
# STUDENT TO DO: 
# Replace the ?? with an approrpriate expression
#############################################################

F =   # define your function Y(s) with respect to s

invL(F)  # Solve for y(t)

Question 6: Checking Your Answer in Python

We applied Laplace transforms to solve the differential equation

\[y''-2y'+5y=0 \quad \mbox{with} \quad y(0)=2 \mbox{ and } y'(0)=4,\]

and we discovered the solution

\[y(t) = 2e^t\cos{(2t)}+ e^t\sin{(2t)}.\]

Edit the code cell below to use SymPy to plug \(y(t)\) into the differential equation to verify it satisfies the differential equation.

Solution to Question 6:





Edit the code cell below. 


#############################################################
# STUDENT TO DO: 
# Replace the ?? with an approrpriate expression
#############################################################

y = ??

dy = y.diff(t, 1)  # compute y'
ddy = y.diff(t, 2)  # compute y''

#############################################################
# STUDENT TO DO: 
# Replace the ?? with an approrpriate expression
#############################################################

left = sym.simplify(??)
right = 0

#############################################################
# Check initial conditions
#############################################################

y0 = y.evalf(subs={t:0})
prime0 = dy.evalf(subs={t:0})

print("The left side is ", left, 
      "\n \n The right side is ", right,
      "\n \n y(0) =  ", y0,
      "\n \n y'(0) =  ", prime0)

Question 7:

Solve the initial value problem using the Laplace Transform. You may use Python as much (or as little) as you like.

\[y''-y'-2y=0 \quad \mbox{with} \quad y(0)=-2 \mbox{ and } y'(0)=5.\]

Solution to Question 7:




It is recommended to do some work by hand and use Python to help with part of the process. Feel free to add some Python code cells below.


Question 8:

Solve the initial value problem using Laplace Transforms. You may use Python as much (or as little) as you like.

\[y''-4y'-5y=4e^{3t} \quad \mbox{with} \quad y(0)=2 \mbox{ and } y'(0)=7.\]

Solution to Question 8:




It is recommended to do some work by hand and use Python to help with part of the process. Feel free to add some Python code cells below.


Question 9:

Solve the initial value problem using Laplace Transforms. You may use Python as much (or as little) as you like.

\[ty''-ty'+y=2 \quad \mbox{with} \quad y(0)=2 \mbox{ and } y'(0)=-1.\]

Solution to Question 9:




It is recommended to do some work by hand and use Python to help with part of the process. Feel free to add some Python code cells below.


Question 10:

Solve the initial value problem using Laplace Transforms. You may use Python as much (or as little) as you like.

\[y''+ty'-y=0 \quad \mbox{with} \quad y(0)=0 \mbox{ and } y'(0)=3.\]

Solution to Question 10:




It is recommended to do some work by hand and use Python to help with part of the process. Feel free to add some Python code cells below.


Appendix A: Common Laplace Transforms

\(\displaystyle \large f(t)\) \(\displaystyle \large F(s) = \mathscr{L} \left\{ f(t) \right\}\)
\(\displaystyle \large f(t)=1\) \(\displaystyle \large F(s)=\frac{1}{s}, \ s > 0\)
\(\displaystyle\large f(t)=e^{at}\) \(\displaystyle \large F(s) = \frac{1}{s-a}, \ s > a\)
\(\displaystyle \large f(t)=t^n, \ n=1,2, \ldots\) \(\displaystyle \large F(s) = \frac{n!}{s^{n+1}}, \ s > 0\)
\(\displaystyle \large f(t)=\sin{(bt)}\) \(\displaystyle \large F(s) = \frac{b}{s^2+b^2}, \ s > 0\)
\(\displaystyle \large f(t)=\cos{(bt)}\) \(\displaystyle \large F(s) = \frac{s}{s^2+b^2}, \ s > 0\)
\(\displaystyle \large e^{at}t^n, \ n=1,2, \ldots\) \(\displaystyle \large F(s) = \frac{n!}{(s-a)^{n+1}}, \ s > a\)
\(\displaystyle \large e^{at}\sin{(bt)}\) \(\displaystyle \large F(s) = \frac{b}{(s-a)^2+b^2}, \ s > a\)
\(\displaystyle \large e^{at}\cos{(bt)}\) \(\displaystyle \large F(s) = \frac{s-a}{(s-a)^2+b^2}, \ s > a\)

Appendix B: Properties of Laplace Transforms

1. \(\color{dodgerblue}{\mathscr{L} \left\{ cf(t) \right\} = c \mathscr{L} \left\{ f(t) \right\}}\), where \(c\) is a constant.

2. \(\color{dodgerblue}{\mathscr{L} \left\{ f_1(t) + f_2(t) \right\} = \mathscr{L} \left\{ f_1(t) \right\} + \mathscr{L} \left\{ f_2(t)\right\}}\).

3. If \(F(s) = \mathscr{L} \left\{ f(t) \right\}\) exists for all \(s > \alpha\), then \(\color{dodgerblue}{\displaystyle \mathscr{L} \left\{ e^{at} f(t) \right\} = F(s-a)}\) for all \(s>\alpha + a\).

4. If \(F(s) =\mathscr{L} \left\{ f(t) \right\}\) exists for all \(s > \alpha\), then for all \(s>\alpha\),

\[\color{dodgerblue}{\mathscr{L} \left\{ f^{(n)}(t) \right\} = s^n \mathscr{L} \{ f(t) \}-s^{n-1} f(0)- s^{n-2} f'(0) - \ldots - f^{(n-1)}(0)}.\]

5. If \(F(s) =\mathscr{L} \left\{ f(t) \right\}\) exists for all \(s > \alpha\), then

\[\color{dodgerblue}{\mathscr{L} \left\{ t^n f(t) \right\} = (-1)^n \frac{d^nF}{ds^n} \mbox{ for all } s > \alpha}.\]

Creative Commons License Information

Creative Commons License
Exploring Differential Equations by Adam Spiegler is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
Based on a work at https://github.com/CU-Denver-MathStats-OER/ODEs and original content created by Rasmussen, C., Keene, K. A., Dunmyre, J., & Fortune, N. (2018). Inquiry oriented differential equations: Course materials. Available at https://iode.sdsu.edu.