14.4 problem Problem 30

Internal problem ID [2870]
Internal file name [OUTPUT/2362_Sunday_June_05_2022_03_00_46_AM_64087347/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 30.
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`]]

\[ \boxed {y^{\prime }+2 y=\operatorname {Heaviside}\left (t -\pi \right ) \sin \left (2 t \right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 3] \end {align*}

14.4.1 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) &=2\\ q(t) &=\operatorname {Heaviside}\left (t -\pi \right ) \sin \left (2 t \right ) \end {align*}

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

The domain of \(p(t)=2\) is \[ \{-\infty

14.4.2 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 )+2 Y \left (s \right ) = \frac {2 \,{\mathrm e}^{-s \pi }}{s^{2}+4}\tag {1} \end {align*}

Replacing initial condition gives \begin {align*} s Y \left (s \right )-3+2 Y \left (s \right ) = \frac {2 \,{\mathrm e}^{-s \pi }}{s^{2}+4} \end {align*}

Solving for \(Y(s)\) gives \begin {align*} Y(s) = \frac {3 s^{2}+2 \,{\mathrm e}^{-s \pi }+12}{\left (s^{2}+4\right ) \left (s +2\right )} \end {align*}

Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (\frac {3 s^{2}+2 \,{\mathrm e}^{-s \pi }+12}{\left (s^{2}+4\right ) \left (s +2\right )}\right )\\ &= 3 \,{\mathrm e}^{-2 t}+\frac {\operatorname {Heaviside}\left (t -\pi \right ) \left ({\mathrm e}^{-t +\pi } \cosh \left (t -\pi \right )-\cos \left (t \right )^{2}+\sin \left (t \right ) \cos \left (t \right )\right )}{2} \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} 3 \,{\mathrm e}^{-2 t} & t <\pi \\ 3 \,{\mathrm e}^{-2 t}+\frac {{\mathrm e}^{-t +\pi } \cosh \left (t -\pi \right )}{2}-\frac {\cos \left (t \right )^{2}}{2}+\frac {\sin \left (t \right ) \cos \left (t \right )}{2} & \pi \le t \end {array}\right . \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left \{\begin {array}{cc} 3 \,{\mathrm e}^{-2 t} & t <\pi \\ 3 \,{\mathrm e}^{-2 t}+\frac {{\mathrm e}^{-t +\pi } \cosh \left (t -\pi \right )}{2}-\frac {\cos \left (t \right )^{2}}{2}+\frac {\sin \left (t \right ) \cos \left (t \right )}{2} & \pi \le t \end {array}\right . \\ \end{align*}

Verification of solutions

\[ y = \left \{\begin {array}{cc} 3 \,{\mathrm e}^{-2 t} & t <\pi \\ 3 \,{\mathrm e}^{-2 t}+\frac {{\mathrm e}^{-t +\pi } \cosh \left (t -\pi \right )}{2}-\frac {\cos \left (t \right )^{2}}{2}+\frac {\sin \left (t \right ) \cos \left (t \right )}{2} & \pi \le t \end {array}\right . \] Verified OK.

14.4.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }+2 y=\mathit {Heaviside}\left (t -\pi \right ) \sin \left (2 t \right ), y \left (0\right )=3\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 }=-2 y+\mathit {Heaviside}\left (t -\pi \right ) \sin \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 }+2 y=\mathit {Heaviside}\left (t -\pi \right ) \sin \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 }+2 y\right )=\mu \left (t \right ) \mathit {Heaviside}\left (t -\pi \right ) \sin \left (2 t \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 }+2 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 )=2 \mu \left (t \right ) \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (t \right )={\mathrm e}^{2 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 ) \mathit {Heaviside}\left (t -\pi \right ) \sin \left (2 t \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 ) \mathit {Heaviside}\left (t -\pi \right ) \sin \left (2 t \right )d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int \mu \left (t \right ) \mathit {Heaviside}\left (t -\pi \right ) \sin \left (2 t \right )d t +c_{1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )={\mathrm e}^{2 t} \\ {} & {} & y=\frac {\int {\mathrm e}^{2 t} \mathit {Heaviside}\left (t -\pi \right ) \sin \left (2 t \right )d t +c_{1}}{{\mathrm e}^{2 t}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {\left (-\frac {{\mathrm e}^{2 t} \cos \left (2 t \right )}{4}+\frac {{\mathrm e}^{2 t} \sin \left (2 t \right )}{4}\right ) \mathit {Heaviside}\left (t -\pi \right )+\frac {\mathit {Heaviside}\left (t -\pi \right ) {\mathrm e}^{2 \pi }}{4}+c_{1}}{{\mathrm e}^{2 t}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=\frac {{\mathrm e}^{-2 t +2 \pi } \mathit {Heaviside}\left (t -\pi \right )}{4}+\frac {\mathit {Heaviside}\left (t -\pi \right ) \left (-\cos \left (2 t \right )+\sin \left (2 t \right )\right )}{4}+c_{1} {\mathrm e}^{-2 t} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=3 \\ {} & {} & 3=c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =3 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =3\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {{\mathrm e}^{-2 t +2 \pi } \mathit {Heaviside}\left (t -\pi \right )}{4}+\frac {\mathit {Heaviside}\left (t -\pi \right ) \left (-\cos \left (2 t \right )+\sin \left (2 t \right )\right )}{4}+3 \,{\mathrm e}^{-2 t} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {{\mathrm e}^{-2 t +2 \pi } \mathit {Heaviside}\left (t -\pi \right )}{4}+\frac {\mathit {Heaviside}\left (t -\pi \right ) \left (-\cos \left (2 t \right )+\sin \left (2 t \right )\right )}{4}+3 \,{\mathrm e}^{-2 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: 2.312 (sec). Leaf size: 43

dsolve([diff(y(t),t)+2*y(t)=Heaviside(t-Pi)*sin(2*t),y(0) = 3],y(t), singsol=all)
 

\[ y \left (t \right ) = \frac {\operatorname {Heaviside}\left (t -\pi \right ) {\mathrm e}^{-2 t +2 \pi }}{4}+\frac {\operatorname {Heaviside}\left (t -\pi \right ) \left (-\cos \left (2 t \right )+\sin \left (2 t \right )\right )}{4}+3 \,{\mathrm e}^{-2 t} \]

✓ Solution by Mathematica

Time used: 0.117 (sec). Leaf size: 55

DSolve[{y'[t]+2*y[t]==UnitStep[t-Pi]*Sin[2*t],{y[0]==3}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} 3 e^{-2 t} & t\leq \pi \\ \frac {1}{4} e^{-2 t} \left (-e^{2 t} \cos (2 t)+e^{2 t} \sin (2 t)+e^{2 \pi }+12\right ) & \text {True} \\ \end {array} \\ \end {array} \]