14.16 problem Problem 46 part a

Internal problem ID [2882]
Internal file name [OUTPUT/2374_Sunday_June_05_2022_03_02_20_AM_82997815/index.tex]

Book: Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth edition, 2015
Section: Chapter 10, The Laplace Transform and Some Elementary Applications. Exercises for 10.7. page 704
Problem number: Problem 46 part a.
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "linear", "first_order_ode_lie_symmetry_lookup"

Maple gives the following as the ode type

[[_linear, `class A`]]

\[ \boxed {y^{\prime }-y=\left \{\begin {array}{cc} 2 & 0\le t <1 \\ -1 & 1\le t \end {array}\right .} \] With initial conditions \begin {align*} [y \left (0\right ) = 1] \end {align*}

14.16.1 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) &=-1\\ q(t) &=\left \{\begin {array}{cc} 2 & 0\le t <1 \\ -1 & 1\le t \end {array}\right . \end {align*}

Hence the ode is \begin {align*} y^{\prime }-y = \left \{\begin {array}{cc} 2 & 0\le t <1 \\ -1 & 1\le t \end {array}\right . \end {align*}

The domain of \(p(t)=-1\) is \[ \{-\infty

14.16.2 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 )-Y \left (s \right ) = \frac {2-3 \,{\mathrm e}^{-s}}{s}\tag {1} \end {align*}

Replacing initial condition gives \begin {align*} s Y \left (s \right )-1-Y \left (s \right ) = \frac {2-3 \,{\mathrm e}^{-s}}{s} \end {align*}

Solving for \(Y(s)\) gives \begin {align*} Y(s) = -\frac {-2+3 \,{\mathrm e}^{-s}-s}{s \left (s -1\right )} \end {align*}

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

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} -2+3 \,{\mathrm e}^{t} & t <1 \\ 1+3 \,{\mathrm e} & t =1 \\ 1+3 \,{\mathrm e}^{t}-3 \,{\mathrm e}^{t -1} & 1

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left \{\begin {array}{cc} -2+3 \,{\mathrm e}^{t} & t <1 \\ 1+3 \,{\mathrm e} & t &=1 \\ 1+3 \,{\mathrm e}^{t}-3 \,{\mathrm e}^{t -1} & 1

Verification of solutions

\[ y = \left \{\begin {array}{cc} -2+3 \,{\mathrm e}^{t} & t <1 \\ 1+3 \,{\mathrm e} & t =1 \\ 1+3 \,{\mathrm e}^{t}-3 \,{\mathrm e}^{t -1} & 1

14.16.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-y=\left \{\begin {array}{cc} 2 & 0\le t <1 \\ -1 & 1\le t \end {array}\right ., y \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y+\left \{\begin {array}{cc} 2 & 0\le t <1 \\ -1 & 1\le t \end {array}\right . \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & y^{\prime }-y=\left \{\begin {array}{cc} 2 & 0\le t <1 \\ -1 & 1\le t \end {array}\right . \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (t \right ) \\ {} & {} & \mu \left (t \right ) \left (y^{\prime }-y\right )=\mu \left (t \right ) \left (\left \{\begin {array}{cc} 2 & 0\le t <1 \\ -1 & 1\le t \end {array}\right .\right ) \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d t}\left (y \mu \left (t \right )\right ) \\ {} & {} & \mu \left (t \right ) \left (y^{\prime }-y\right )=y^{\prime } \mu \left (t \right )+y \mu ^{\prime }\left (t \right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \mu ^{\prime }\left (t \right ) \\ {} & {} & \mu ^{\prime }\left (t \right )=-\mu \left (t \right ) \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (t \right )={\mathrm e}^{-t} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \left (\frac {d}{d t}\left (y \mu \left (t \right )\right )\right )d t =\int \mu \left (t \right ) \left (\left \{\begin {array}{cc} 2 & 0\le t <1 \\ -1 & 1\le t \end {array}\right .\right )d t +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (t \right )=\int \mu \left (t \right ) \left (\left \{\begin {array}{cc} 2 & 0\le t <1 \\ -1 & 1\le t \end {array}\right .\right )d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int \mu \left (t \right ) \left (\left \{\begin {array}{cc} 2 & 0\le t <1 \\ -1 & 1\le t \end {array}\right .\right )d t +c_{1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )={\mathrm e}^{-t} \\ {} & {} & y=\frac {\int {\mathrm e}^{-t} \left (\left \{\begin {array}{cc} 2 & 0\le t <1 \\ -1 & 1\le t \end {array}\right .\right )d t +c_{1}}{{\mathrm e}^{-t}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {\left \{\begin {array}{cc} 0 & t \le 0 \\ -2 \,{\mathrm e}^{-t}+2 & t \le 1 \\ {\mathrm e}^{-t}-3 \,{\mathrm e}^{-1}+2 & 1

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
<- 1st order linear successful`
 

✓ Solution by Maple

Time used: 2.938 (sec). Leaf size: 38

dsolve([diff(y(t),t)-y(t)=piecewise(0<=t and t<1,2,t>=1,-1),y(0) = 1],y(t), singsol=all)
 

\[ y \left (t \right ) = \left \{\begin {array}{cc} -2+3 \,{\mathrm e}^{t} & t <1 \\ 1+3 \,{\mathrm e} & t =1 \\ 1+3 \,{\mathrm e}^{t}-3 \,{\mathrm e}^{t -1} & 1

✓ Solution by Mathematica

Time used: 0.072 (sec). Leaf size: 42

DSolve[{y'[t]-y[t]==Piecewise[{{2,0<=t<1},{-1,t>=1}}],{y[0]==1}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} e^t & t\leq 0 \\ -2+3 e^t & 0