21.6 problem 30.10 (a)

Internal problem ID [13888]
Internal file name [OUTPUT/13060_Friday_February_23_2024_06_54_26_AM_65771952/index.tex]

Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell. second edition. CRC Press. FL, USA. 2020
Section: Chapter 30. Piecewise-defined functions and periodic functions. Additional Exercises. page 553
Problem number: 30.10 (a).
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "quadrature"

Maple gives the following as the ode type

[_quadrature]

\[ \boxed {y^{\prime }=\left \{\begin {array}{cc} 0 & t <1 \\ 1 & 1

21.6.1 Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime } + p(t)y &= q(t) \end {align*}

Where here \begin {align*} p(t) &=0\\ q(t) &=\left \{\begin {array}{cc} 0 & t <1 \\ 1 & 1

Hence the ode is \begin {align*} y^{\prime } = \left \{\begin {array}{cc} 0 & t <1 \\ 1 & 1

The domain of \(p(t)=0\) is \[ \{-\infty

21.6.2 Solving as laplace ode

Solving using the Laplace transform method. Let \[ \mathcal {L}\left (y\right ) =Y(s) \] Taking the Laplace transform of the ode and using the relations that \begin {align*} \mathcal {L}\left (y^{\prime }\right )&= s Y(s) - y \left (0\right ) \end {align*}

The given ode now becomes an algebraic equation in the Laplace domain \begin {align*} s Y \left (s \right )-y \left (0\right ) = \frac {{\mathrm e}^{-s}-{\mathrm e}^{-3 s}}{s}\tag {1} \end {align*}

Replacing initial condition gives \begin {align*} s Y \left (s \right ) = \frac {{\mathrm e}^{-s}-{\mathrm e}^{-3 s}}{s} \end {align*}

Solving for \(Y(s)\) gives \begin {align*} Y(s) = \frac {{\mathrm e}^{-s}-{\mathrm e}^{-3 s}}{s^{2}} \end {align*}

Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (\frac {{\mathrm e}^{-s}-{\mathrm e}^{-3 s}}{s^{2}}\right )\\ &= \operatorname {Heaviside}\left (t -1\right ) \left (t -1\right )-\operatorname {Heaviside}\left (t -3\right ) \left (t -3\right ) \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} 0 & t <1 \\ t -1 & t <3 \\ 2 & 3\le t \end {array}\right . \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left \{\begin {array}{cc} 0 & t <1 \\ t -1 & t <3 \\ 2 & 3\le t \end {array}\right . \\ \end{align*}

Verification of solutions

\[ y = \left \{\begin {array}{cc} 0 & t <1 \\ t -1 & t <3 \\ 2 & 3\le t \end {array}\right . \] Verified OK.

21.6.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\left \{\begin {array}{cc} 0 & t <1 \\ 1 & 1

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
<- quadrature successful`
 

✓ Solution by Maple

Time used: 5.203 (sec). Leaf size: 19

dsolve([diff(y(t),t)=piecewise(t<1,0,1<t and t<3,1,t>3,0),y(0) = 0],y(t), singsol=all)
 

\[ y \left (t \right ) = \left \{\begin {array}{cc} 0 & t <1 \\ t -1 & t <3 \\ 2 & 3\le t \end {array}\right . \]

✓ Solution by Mathematica

Time used: 0.006 (sec). Leaf size: 23

DSolve[{y'[t]==Piecewise[{ {0,t<1},{1,1<t<3},{0,t>3}}],{y[0]==0}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} 0 & t\leq 1 \\ t-1 & 1