4.15 problem Problem 3(a)

Internal problem ID [12322]
Internal file name [OUTPUT/10975_Monday_October_02_2023_02_47_40_AM_58456927/index.tex]

Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A. Dobrushkin. CRC Press 2015
Section: Chapter 5.6 Laplace transform. Nonhomogeneous equations. Problems page 368
Problem number: Problem 3(a).
ODE order: 1.
ODE degree: 1.

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

Maple gives the following as the ode type

[[_linear, `class A`]]

\[ \boxed {y^{\prime }+y=\operatorname {Heaviside}\left (t \right )-\operatorname {Heaviside}\left (-2+t \right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 1] \end {align*}

4.15.1 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) &=\operatorname {Heaviside}\left (t \right )-\operatorname {Heaviside}\left (-2+t \right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime }+y = \operatorname {Heaviside}\left (t \right )-\operatorname {Heaviside}\left (-2+t \right ) \end {align*}

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

4.15.2 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 {1-{\mathrm e}^{-2 s}}{s}\tag {1} \end {align*}

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

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

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

Summary

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

Verification of solutions

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

4.15.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }+y=\mathit {Heaviside}\left (t \right )-\mathit {Heaviside}\left (-2+t \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+\mathit {Heaviside}\left (t \right )-\mathit {Heaviside}\left (-2+t \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=\mathit {Heaviside}\left (t \right )-\mathit {Heaviside}\left (-2+t \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 (\mathit {Heaviside}\left (t \right )-\mathit {Heaviside}\left (-2+t \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 (\mathit {Heaviside}\left (t \right )-\mathit {Heaviside}\left (-2+t \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 (\mathit {Heaviside}\left (t \right )-\mathit {Heaviside}\left (-2+t \right )\right )d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int \mu \left (t \right ) \left (\mathit {Heaviside}\left (t \right )-\mathit {Heaviside}\left (-2+t \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 (\mathit {Heaviside}\left (t \right )-\mathit {Heaviside}\left (-2+t \right )\right )d t +c_{1}}{{\mathrm e}^{t}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {{\mathrm e}^{t} \mathit {Heaviside}\left (t \right )-\mathit {Heaviside}\left (t \right )-{\mathrm e}^{t} \mathit {Heaviside}\left (-2+t \right )+\mathit {Heaviside}\left (-2+t \right ) {\mathrm e}^{2}+c_{1}}{{\mathrm e}^{t}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=\mathit {Heaviside}\left (-2+t \right ) {\mathrm e}^{2-t}-\mathit {Heaviside}\left (-2+t \right )+\left (c_{1} -\mathit {Heaviside}\left (t \right )\right ) {\mathrm e}^{-t}+\mathit {Heaviside}\left (t \right ) \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =1 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =1\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\mathit {Heaviside}\left (-2+t \right ) {\mathrm e}^{2-t}-\mathit {Heaviside}\left (t \right ) {\mathrm e}^{-t}+\mathit {Heaviside}\left (t \right )-\mathit {Heaviside}\left (-2+t \right )+{\mathrm e}^{-t} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\mathit {Heaviside}\left (-2+t \right ) {\mathrm e}^{2-t}-\mathit {Heaviside}\left (t \right ) {\mathrm e}^{-t}+\mathit {Heaviside}\left (t \right )-\mathit {Heaviside}\left (-2+t \right )+{\mathrm e}^{-t} \end {array} \]

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: 5.203 (sec). Leaf size: 22

dsolve([diff(y(t),t)+y(t)=Heaviside(t)-Heaviside(t-2),y(0) = 1],y(t), singsol=all)
 

\[ y \left (t \right ) = 1-\operatorname {Heaviside}\left (t -2\right )+\operatorname {Heaviside}\left (t -2\right ) {\mathrm e}^{2-t} \]

✓ Solution by Mathematica

Time used: 0.108 (sec). Leaf size: 31

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

\[ y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} 1 & 0\leq t\leq 2 \\ e^{2-t} & t>2 \\ e^{-t} & \text {True} \\ \end {array} \\ \end {array} \]