15.4 problem Problem 4

Internal problem ID [2887]
Internal file name [OUTPUT/2379_Sunday_June_05_2022_03_02_55_AM_59375699/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 10.8. page 710
Problem number: Problem 4.
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`]]

\[ \boxed {y^{\prime }-5 y=2 \,{\mathrm e}^{-t}+\delta \left (t -3\right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 0] \end {align*}

15.4.1 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) &=-5\\ q(t) &=2 \,{\mathrm e}^{-t}+\delta \left (t -3\right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime }-5 y = 2 \,{\mathrm e}^{-t}+\delta \left (t -3\right ) \end {align*}

The domain of \(p(t)=-5\) is \[ \{-\infty

15.4.2 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 )-5 Y \left (s \right ) = \frac {2}{s +1}+{\mathrm e}^{-3 s}\tag {1} \end {align*}

Replacing initial condition gives \begin {align*} s Y \left (s \right )-5 Y \left (s \right ) = \frac {2}{s +1}+{\mathrm e}^{-3 s} \end {align*}

Solving for \(Y(s)\) gives \begin {align*} Y(s) = \frac {{\mathrm e}^{-3 s} s +{\mathrm e}^{-3 s}+2}{\left (s +1\right ) \left (s -5\right )} \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}^{-3 s} s +{\mathrm e}^{-3 s}+2}{\left (s +1\right ) \left (s -5\right )}\right )\\ &= \frac {2 \,{\mathrm e}^{2 t} \sinh \left (3 t \right )}{3}+{\mathrm e}^{-15+5 t} \left (-\operatorname {Heaviside}\left (-t +3\right )+1\right ) \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} \frac {2 \,{\mathrm e}^{2 t} \sinh \left (3 t \right )}{3} & t \le 3 \\ \frac {2 \,{\mathrm e}^{2 t} \sinh \left (3 t \right )}{3}+{\mathrm e}^{-15+5 t} & 3

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left \{\begin {array}{cc} \frac {2 \,{\mathrm e}^{2 t} \sinh \left (3 t \right )}{3} & t \le 3 \\ \frac {2 \,{\mathrm e}^{2 t} \sinh \left (3 t \right )}{3}+{\mathrm e}^{-15+5 t} & 3

Verification of solutions

\[ y = \left \{\begin {array}{cc} \frac {2 \,{\mathrm e}^{2 t} \sinh \left (3 t \right )}{3} & t \le 3 \\ \frac {2 \,{\mathrm e}^{2 t} \sinh \left (3 t \right )}{3}+{\mathrm e}^{-15+5 t} & 3

15.4.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-5 y=2 \,{\mathrm e}^{-t}+\mathit {Dirac}\left (t -3\right ), y \left (0\right )=0\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 }=5 y+2 \,{\mathrm e}^{-t}+\mathit {Dirac}\left (t -3\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 }-5 y=2 \,{\mathrm e}^{-t}+\mathit {Dirac}\left (t -3\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 }-5 y\right )=\mu \left (t \right ) \left (2 \,{\mathrm e}^{-t}+\mathit {Dirac}\left (t -3\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 }-5 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 )=-5 \mu \left (t \right ) \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (t \right )={\mathrm e}^{-5 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 (2 \,{\mathrm e}^{-t}+\mathit {Dirac}\left (t -3\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 (2 \,{\mathrm e}^{-t}+\mathit {Dirac}\left (t -3\right )\right )d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int \mu \left (t \right ) \left (2 \,{\mathrm e}^{-t}+\mathit {Dirac}\left (t -3\right )\right )d t +c_{1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )={\mathrm e}^{-5 t} \\ {} & {} & y=\frac {\int {\mathrm e}^{-5 t} \left (2 \,{\mathrm e}^{-t}+\mathit {Dirac}\left (t -3\right )\right )d t +c_{1}}{{\mathrm e}^{-5 t}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {\mathit {Heaviside}\left (t -3\right ) {\mathrm e}^{-15}-\frac {{\mathrm e}^{-6 t}}{3}+c_{1}}{{\mathrm e}^{-5 t}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{5 t} c_{1} +{\mathrm e}^{-15+5 t} \mathit {Heaviside}\left (t -3\right )-\frac {{\mathrm e}^{-t}}{3} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} -\frac {1}{3} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {1}{3} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {1}{3}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {{\mathrm e}^{5 t}}{3}+{\mathrm e}^{-15+5 t} \mathit {Heaviside}\left (t -3\right )-\frac {{\mathrm e}^{-t}}{3} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {{\mathrm e}^{5 t}}{3}+{\mathrm e}^{-15+5 t} \mathit {Heaviside}\left (t -3\right )-\frac {{\mathrm e}^{-t}}{3} \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: 2.578 (sec). Leaf size: 32

dsolve([diff(y(t),t)-5*y(t)=2*exp(-t)+Dirac(t-3),y(0) = 0],y(t), singsol=all)
 

\[ y \left (t \right ) = \frac {2 \,{\mathrm e}^{2 t} \sinh \left (3 t \right )}{3}+\operatorname {Heaviside}\left (-3+t \right ) {\mathrm e}^{5 t -15} \]

✓ Solution by Mathematica

Time used: 0.093 (sec). Leaf size: 34

DSolve[{y'[t]-5*y[t]==2*Exp[-t]+DiracDelta[t-3],{y[0]==0}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {1}{3} e^{-t} \left (3 e^{6 t-15} \theta (t-3)+e^{6 t}-1\right ) \]