\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\units}[1]{\,\text{#1}} \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus - Solve Differential Equations Using Laplace Transforms

17Calculus
Single Variable Calculus
Derivatives
Integrals
Multi-Variable Calculus
Precalculus
Functions

If you haven't seen Laplace Transforms before, the Laplace Transform section of 17calculus has a complete discussion of what they are and how to work with them.

Laplace Transforms Table

Laplace Transforms

\( f(t) \)

\(\displaystyle{ F(s) }\)

Basic Functions

\( t^n, ~ n = 1, 2, 3, \ldots \)

\(\displaystyle{ \frac{n!}{s^{n+1}} }\)

\( e^{at} \)

\(\displaystyle{ \frac{1}{s-a} }\)

\( \sin(\alpha t) \)

\(\displaystyle{ \frac{\alpha}{s^2 + \alpha^2} }\)

\( \cos(at) \)

\(\displaystyle{ \frac{s}{s^2 + a^2} }\)

\( \sinh(at) \)

\(\displaystyle{ \frac{a}{s^2 - a^2} }\)

\( \cosh(at) \)

\(\displaystyle{ \frac{s}{s^2 - a^2} }\)

Special Functions

\( \delta(t) \) unit impulse

\( 1 \)

\( \delta(t-\tau) \) shifted unit impulse

\( e^{-\tau s} \)

\( u(t) \) unit step

\(\displaystyle{ \frac{1}{s} }\)

\( u(t-\tau) \) shifted unit step

\(\displaystyle{ \frac{1}{s} e^{-\tau s} }\)

Combined Functions

\( e^{at}\sin(\alpha t) \)

\(\displaystyle{ \frac{\alpha}{(s-a)^2 + \alpha^2} }\)

\( f(t)u(t-a) \)

\(\displaystyle{ e^{-sa} \mathcal{L}\{ f(t+a) \} }\)

\( t^n e^{at}, ~ n = 1, 2, 3, \ldots \)

\(\displaystyle{ \frac{n!}{(s-a)^{n+1}} \} }\)

Derivatives and Integrals

\( f'(t) \)

\( sF(s) - f(0) \)

\( f''(t) \)

\( s^2F(s) - sf(0) - f'(0) \)

\(\displaystyle{ f^{(n)}(t) }\)

\(\displaystyle{ s^nF(s) - s^{n-1}f(0) - }\) \(\displaystyle{ s^{n-2}f'(0) - . . . - f^{(n-1)}(0) }\)

\( \int_0^t{ f(v)~dv }\)

\(\displaystyle{ \frac{F(s)}{s} }\)

Using Laplace Transforms to solve differential equation initial value problems is a great way to streamline solutions and, for forcing functions that are discontinuous, they are about the only way to solve them. The equations we use are

Laplace Transforms of Derivatives

\(\displaystyle{ \mathcal{L}\{y'(t)\} = sY(s) - y(0) }\)

\(\displaystyle{ \mathcal{L}\{y''(t)\} = s^2Y(s) - sy(0) - y'(0) }\)

where \(\mathcal{L}\{y(t)\} = Y(s)\)

\(\displaystyle{ \mathcal{L}\{ f^{(n)}(t) \} = }\) \(\displaystyle{ s^nF(s) - s^{n-1}f(0) - }\) \(\displaystyle{ s^{n-2}f'(0) - . . . - f^{(n-1)}(0) }\)

See the Laplace Transforms Involving Derivatives and Integrals page for more discussion on these equations.

The idea is that we are given a differential equation which we first transform using the Laplace Transform, we solve it in the s-domain and then perform an inverse Laplace transform back to the t-domain to get our answer. Here is a very quick video giving these steps in a flow-chart.

PatrickJMT - Laplace Transform flow-chart [1min-33secs]

video by PatrickJMT

This video clip explains in detail how to solve differential equations using Laplace Transforms including a quick calculation of the Laplace transform of the first derivative function above.

Dr Chris Tisdell - solve differential equations using Laplace Transforms [28min-50secs]

video by Dr Chris Tisdell

This video expands on the one above taking you step-by-step through this process and, at the end of the video, gives you a big picture of how to do this. This is a great video.

Dr Chris Tisdell - big picture [12min-45secs]

video by Dr Chris Tisdell

Elementary Differential Equations

Practice

Unless otherwise instructed, solve these initial value problems using Laplace transforms. Give your answers in exact, completely factored form.
Notes
1. In these problems, \(u(t)\) is the unit step function (Heaviside function).
2. Some of these problems may require convolution to solve.

Basic

Solve \( y'+2y = \sin(3t), y(0) = 0 \)

Problem Statement

Solve \( y'+2y = \sin(3t), y(0) = 0 \)

Solution

blackpenredpen - 3641 video solution

video by blackpenredpen

Log in to rate this practice problem and to see it's current rating.

Solve \( y''-4y = 3t^2, y(0)=2, y'(0) = 1 \)

Problem Statement

Solve \( y''-4y = 3t^2, y(0)=2, y'(0) = 1 \)

Solution

blackpenredpen - 3642 video solution

video by blackpenredpen

Log in to rate this practice problem and to see it's current rating.

Solve \( y''-2y'-15y=0, y(0) = -2, y'(0)=1 \)

Problem Statement

Solve \( y''-2y'-15y=0, y(0) = -2, y'(0)=1 \)

Solution

blackpenredpen - 3643 video solution

video by blackpenredpen

Log in to rate this practice problem and to see it's current rating.

Solve \( y''+16y = \cos(4t), y(0)=0, y'(0)=1 \)

Problem Statement

Solve \( y''+16y = \cos(4t), y(0)=0, y'(0)=1 \)

Solution

blackpenredpen - 3644 video solution

video by blackpenredpen

Log in to rate this practice problem and to see it's current rating.

Solve \( y''+y = u(t-3), y(0)=0, y'(0) = 1 \)

Problem Statement

Solve \( y''+y = u(t-3), y(0)=0, y'(0) = 1 \)

Solution

blackpenredpen - 3645 video solution

video by blackpenredpen

Log in to rate this practice problem and to see it's current rating.

Solve \(\displaystyle{ y''-4y+5y=0, y(0)=1, y'(0)=2 }\)

Problem Statement

Solve \(\displaystyle{ y''-4y+5y=0, y(0)=1, y'(0)=2 }\)

Solution

blackpenredpen - 3646 video solution

video by blackpenredpen

Log in to rate this practice problem and to see it's current rating.

Solve \( y' + y = u(t-1) \); \( y(0) = 0 \)

Problem Statement

Solve \( y' + y = u(t-1) \); \( y(0) = 0 \)

Solution

Dr Chris Tisdell - 646 video solution

video by Dr Chris Tisdell

Log in to rate this practice problem and to see it's current rating.

Solve \( y^{(4)}-y=0 \); \( y(0) = 0 \), \( y'(0) = 1 \), \( y''(0) = 0 \), \( y'''(0) = 0 \)

Problem Statement

Solve \( y^{(4)}-y=0 \); \( y(0) = 0 \), \( y'(0) = 1 \), \( y''(0) = 0 \), \( y'''(0) = 0 \)

Final Answer

\(\displaystyle{ y(t) = \frac{1}{2}\left[ \sin(t) + \sinh(t) \right] }\)

Problem Statement

Solve \( y^{(4)}-y=0 \); \( y(0) = 0 \), \( y'(0) = 1 \), \( y''(0) = 0 \), \( y'''(0) = 0 \)

Solution

Dr Chris Tisdell - 648 video solution

video by Dr Chris Tisdell

Final Answer

\(\displaystyle{ y(t) = \frac{1}{2}\left[ \sin(t) + \sinh(t) \right] }\)

Log in to rate this practice problem and to see it's current rating.

Intermediate

Solve \(y''+4y=q(t), t > 0 \); \(y(0)=y'(0)=0\) where \(\displaystyle{ q(t) = \left\{\begin{array}{crl} t & & 0 \leq t \leq \pi/2 \\ \pi/2 & & t > \pi/2 \end{array} \right. }\)

Problem Statement

Solve \(y''+4y=q(t), t > 0 \); \(y(0)=y'(0)=0\) where \(\displaystyle{ q(t) = \left\{\begin{array}{crl} t & & 0 \leq t \leq \pi/2 \\ \pi/2 & & t > \pi/2 \end{array} \right. }\)

Final Answer

\(\displaystyle{ y(t) = \frac{1}{4} \left[ t- \frac{1}{2} \sin(2t) \right] - \frac{u(t-\pi/2)}{4}\left[(t-\pi/2) - \frac{1}{2} \sin[2(t-\pi/2)] \right] }\)

Problem Statement

Solve \(y''+4y=q(t), t > 0 \); \(y(0)=y'(0)=0\) where \(\displaystyle{ q(t) = \left\{\begin{array}{crl} t & & 0 \leq t \leq \pi/2 \\ \pi/2 & & t > \pi/2 \end{array} \right. }\)

Solution

Dr Chris Tisdell - 647 video solution

video by Dr Chris Tisdell

Final Answer

\(\displaystyle{ y(t) = \frac{1}{4} \left[ t- \frac{1}{2} \sin(2t) \right] - \frac{u(t-\pi/2)}{4}\left[(t-\pi/2) - \frac{1}{2} \sin[2(t-\pi/2)] \right] }\)

Log in to rate this practice problem and to see it's current rating.

Solve \( y'' - 3y' + 2y = r(t) \); \(y(0)=1, y'(0)=3\) where \( r(t) = u(t-1) - u(t-2) \)

Problem Statement

Solve \( y'' - 3y' + 2y = r(t) \); \(y(0)=1, y'(0)=3\) where \( r(t) = u(t-1) - u(t-2) \)

Final Answer

\(\displaystyle{ y(t) = -e^t + 2e^{2t} + u(t-1)\left[ \frac{1}{2} - e^{t-1} + \frac{1}{2}e^{2(t-1)} \right] - u(t-2)\left[ \frac{1}{2} - e^{t-2} + \frac{1}{2}e^{2(t-2)} \right] }\)

Problem Statement

Solve \( y'' - 3y' + 2y = r(t) \); \(y(0)=1, y'(0)=3\) where \( r(t) = u(t-1) - u(t-2) \)

Solution

Dr Chris Tisdell - 649 video solution

video by Dr Chris Tisdell

Final Answer

\(\displaystyle{ y(t) = -e^t + 2e^{2t} + u(t-1)\left[ \frac{1}{2} - e^{t-1} + \frac{1}{2}e^{2(t-1)} \right] - u(t-2)\left[ \frac{1}{2} - e^{t-2} + \frac{1}{2}e^{2(t-2)} \right] }\)

Log in to rate this practice problem and to see it's current rating.

Solve \( 4y'' + y = g(t); \) \( y(0)=3, y'(0)=-7 \)

Problem Statement

Solve \( 4y'' + y = g(t); \) \( y(0)=3, y'(0)=-7 \)

Final Answer

\( y(t) = 3\cos(t/2) - 14\sin(t/2) + (1/2)\int_0^t{\sin(t/2)g(t-\tau)~d\tau} \)

Problem Statement

Solve \( 4y'' + y = g(t); \) \( y(0)=3, y'(0)=-7 \)

Solution

Krista King Math - 2242 video solution

video by Krista King Math

Final Answer

\( y(t) = 3\cos(t/2) - 14\sin(t/2) + (1/2)\int_0^t{\sin(t/2)g(t-\tau)~d\tau} \)

Log in to rate this practice problem and to see it's current rating.

Solve \( y'' + \omega_0^2 y = F_0 \sin(\omega t); y’(0)=0, y(0)=0 \)

Problem Statement

Solve \( y'' + \omega_0^2 y = F_0 \sin(\omega t); y’(0)=0, y(0)=0 \)

Final Answer

\(\displaystyle{ y(t) = \frac{F_0}{\omega_0^2-\omega^2} \left[ \sin(\omega t) - \frac{\omega}{\omega_0}\sin(\omega_0 t) \right] }\)

Problem Statement

Solve \( y'' + \omega_0^2 y = F_0 \sin(\omega t); y’(0)=0, y(0)=0 \)

Solution

Michel vanBiezen - 2259 video solution

video by Michel vanBiezen

Final Answer

\(\displaystyle{ y(t) = \frac{F_0}{\omega_0^2-\omega^2} \left[ \sin(\omega t) - \frac{\omega}{\omega_0}\sin(\omega_0 t) \right] }\)

Log in to rate this practice problem and to see it's current rating.

Solve \( y’ + 3y + 2\int_0^ty(\tau)~d\tau = 2e^{-3t}; y(0)=0 \)

Problem Statement

Solve \( y’ + 3y + 2\int_0^ty(\tau)~d\tau = 2e^{-3t}; y(0)=0 \)

Final Answer

\( y(t) = e^{-3t}(-3+4e^t-e^{2t}) \)

Problem Statement

Solve \( y’ + 3y + 2\int_0^ty(\tau)~d\tau = 2e^{-3t}; y(0)=0 \)

Solution

Michel vanBiezen - 2261 video solution

video by Michel vanBiezen

Final Answer

\( y(t) = e^{-3t}(-3+4e^t-e^{2t}) \)

Log in to rate this practice problem and to see it's current rating.

Really UNDERSTAND Differential Equations

Log in to rate this page and to see it's current rating.

Topics You Need To Understand For This Page

Related Topics and Links

external links you may find helpful

laplace transforms youtube playlist

To bookmark this page and practice problems, log in to your account or set up a free account.

Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

free ideas to save on books

Shop Amazon - Wearable Technology: Electronics

As an Amazon Associate I earn from qualifying purchases.

I recently started a Patreon account to help defray the expenses associated with this site. To keep this site free, please consider supporting me.

Support 17Calculus on Patreon

Practice Search

Practice Instructions

Unless otherwise instructed, solve these initial value problems using Laplace transforms. Give your answers in exact, completely factored form.
Notes
1. In these problems, \(u(t)\) is the unit step function (Heaviside function).
2. Some of these problems may require convolution to solve.

Do NOT follow this link or you will be banned from the site!

When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.

Links and banners on this page are affiliate links. We carefully choose only the affiliates that we think will help you learn. Clicking on them and making purchases help you support 17Calculus at no extra charge to you. However, only you can decide what will actually help you learn. So think carefully about what you need and purchase only what you think will help you.

We use cookies on this site to enhance your learning experience.

17calculus

Copyright © 2010-2022 17Calculus, All Rights Reserved     [Privacy Policy]     [Support]     [About]

mathjax.org
Real Time Web Analytics