14.14 problem Problem 40

Internal problem ID [2880]
Internal file name [OUTPUT/2372_Sunday_June_05_2022_03_02_03_AM_30140179/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 40.
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_laplace", "second_order_linear_constant_coeff"

Maple gives the following as the ode type

[[_2nd_order, _linear, _nonhomogeneous]]

\[ \boxed {y^{\prime \prime }+4 y^{\prime }+5 y=5 \operatorname {Heaviside}\left (t -3\right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 2, y^{\prime }\left (0\right ) = 1] \end {align*}

14.14.1 Existence and uniqueness analysis

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

Where here \begin {align*} p(t) &=4\\ q(t) &=5\\ F &=5 \operatorname {Heaviside}\left (t -3\right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+4 y^{\prime }+5 y = 5 \operatorname {Heaviside}\left (t -3\right ) \end {align*}

The domain of \(p(t)=4\) is \[ \{-\infty

Solving using the Laplace transform method. Let \begin {align*} \mathcal {L}\left (y\right ) =Y(s) \end {align*}

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 )\\ \mathcal {L}\left (y^{\prime \prime }\right ) &= s^2 Y(s) - y'(0) - s y \left (0\right ) \end {align*}

The given ode now becomes an algebraic equation in the Laplace domain \begin {align*} s^{2} Y \left (s \right )-y^{\prime }\left (0\right )-s y \left (0\right )+4 s Y \left (s \right )-4 y \left (0\right )+5 Y \left (s \right ) = \frac {5 \,{\mathrm e}^{-3 s}}{s}\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&=2\\ y'(0) &=1 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-9-2 s +4 s Y \left (s \right )+5 Y \left (s \right ) = \frac {5 \,{\mathrm e}^{-3 s}}{s} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {2 s^{2}+5 \,{\mathrm e}^{-3 s}+9 s}{s \left (s^{2}+4 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 {2 s^{2}+5 \,{\mathrm e}^{-3 s}+9 s}{s \left (s^{2}+4 s +5\right )}\right )\\ &= {\mathrm e}^{-2 t} \left (2 \cos \left (t \right )+5 \sin \left (t \right )\right )+\left (\frac {1}{10}+\frac {i}{5}\right ) \left (2-4 i-5 \,{\mathrm e}^{\left (-2-i\right ) \left (t -3\right )}+\left (3+4 i\right ) {\mathrm e}^{\left (-2+i\right ) \left (t -3\right )}\right ) \operatorname {Heaviside}\left (t -3\right ) \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} {\mathrm e}^{-2 t} \left (2 \cos \left (t \right )+5 \sin \left (t \right )\right ) & t <3 \\ {\mathrm e}^{-2 t} \left (2 \cos \left (t \right )+5 \sin \left (t \right )\right )+\left (\frac {1}{10}+\frac {i}{5}\right ) \left (2-4 i-5 \,{\mathrm e}^{\left (-2-i\right ) \left (t -3\right )}+\left (3+4 i\right ) {\mathrm e}^{\left (-2+i\right ) \left (t -3\right )}\right ) & 3\le t \end {array}\right . \] Simplifying the solution gives \[ y = {\mathrm e}^{-2 t} \left (2 \cos \left (t \right )+5 \sin \left (t \right )\right )-\left (\left \{\begin {array}{cc} 0 & t <3 \\ -1+\left (\frac {1}{2}+i\right ) {\mathrm e}^{\left (-2-i\right ) \left (t -3\right )}+\left (\frac {1}{2}-i\right ) {\mathrm e}^{\left (-2+i\right ) \left (t -3\right )} & 3\le t \end {array}\right .\right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{-2 t} \left (2 \cos \left (t \right )+5 \sin \left (t \right )\right )-\left (\left \{\begin {array}{cc} 0 & t <3 \\ -1+\left (\frac {1}{2}+i\right ) {\mathrm e}^{\left (-2-i\right ) \left (t -3\right )}+\left (\frac {1}{2}-i\right ) {\mathrm e}^{\left (-2+i\right ) \left (t -3\right )} & 3\le t \end {array}\right .\right ) \\ \end{align*}

Verification of solutions

\[ y = {\mathrm e}^{-2 t} \left (2 \cos \left (t \right )+5 \sin \left (t \right )\right )-\left (\left \{\begin {array}{cc} 0 & t <3 \\ -1+\left (\frac {1}{2}+i\right ) {\mathrm e}^{\left (-2-i\right ) \left (t -3\right )}+\left (\frac {1}{2}-i\right ) {\mathrm e}^{\left (-2+i\right ) \left (t -3\right )} & 3\le t \end {array}\right .\right ) \] Verified OK.

14.14.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+4 y^{\prime }+5 y=5 \mathit {Heaviside}\left (t -3\right ), y \left (0\right )=2, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+4 r +5=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-4\right )\pm \left (\sqrt {-4}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-2-\mathrm {I}, -2+\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )=\cos \left (t \right ) {\mathrm e}^{-2 t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=\sin \left (t \right ) {\mathrm e}^{-2 t} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y=c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right )+y_{p}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y=c_{1} \cos \left (t \right ) {\mathrm e}^{-2 t}+c_{2} \sin \left (t \right ) {\mathrm e}^{-2 t}+y_{p}\left (t \right ) \\ \square & {} & \textrm {Find a particular solution}\hspace {3pt} y_{p}\left (t \right )\hspace {3pt}\textrm {of the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Use variation of parameters to find}\hspace {3pt} y_{p}\hspace {3pt}\textrm {here}\hspace {3pt} f \left (t \right )\hspace {3pt}\textrm {is the forcing function}\hspace {3pt} \\ {} & {} & \left [y_{p}\left (t \right )=-y_{1}\left (t \right ) \left (\int \frac {y_{2}\left (t \right ) f \left (t \right )}{W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )}d t \right )+y_{2}\left (t \right ) \left (\int \frac {y_{1}\left (t \right ) f \left (t \right )}{W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )}d t \right ), f \left (t \right )=5 \mathit {Heaviside}\left (t -3\right )\right ] \\ {} & \circ & \textrm {Wronskian of solutions of the homogeneous equation}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=\left [\begin {array}{cc} \cos \left (t \right ) {\mathrm e}^{-2 t} & \sin \left (t \right ) {\mathrm e}^{-2 t} \\ -\sin \left (t \right ) {\mathrm e}^{-2 t}-2 \cos \left (t \right ) {\mathrm e}^{-2 t} & \cos \left (t \right ) {\mathrm e}^{-2 t}-2 \sin \left (t \right ) {\mathrm e}^{-2 t} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )={\mathrm e}^{-4 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-5 \,{\mathrm e}^{-2 t} \left (\cos \left (t \right ) \left (\int \mathit {Heaviside}\left (t -3\right ) \sin \left (t \right ) {\mathrm e}^{2 t}d t \right )-\sin \left (t \right ) \left (\int \mathit {Heaviside}\left (t -3\right ) \cos \left (t \right ) {\mathrm e}^{2 t}d t \right )\right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\mathit {Heaviside}\left (t -3\right ) \left (-1+\left (\left (\cos \left (t \right )+2 \sin \left (t \right )\right ) \cos \left (3\right )-2 \left (\cos \left (t \right )-\frac {\sin \left (t \right )}{2}\right ) \sin \left (3\right )\right ) {\mathrm e}^{-2 t +6}\right ) \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} \cos \left (t \right ) {\mathrm e}^{-2 t}+c_{2} \sin \left (t \right ) {\mathrm e}^{-2 t}-\mathit {Heaviside}\left (t -3\right ) \left (-1+\left (\left (\cos \left (t \right )+2 \sin \left (t \right )\right ) \cos \left (3\right )-2 \left (\cos \left (t \right )-\frac {\sin \left (t \right )}{2}\right ) \sin \left (3\right )\right ) {\mathrm e}^{-2 t +6}\right ) \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} \cos \left (t \right ) {\mathrm e}^{-2 t}+c_{2} \sin \left (t \right ) {\mathrm e}^{-2 t}-\mathit {Heaviside}\left (t -3\right ) \left (-1+\left (\left (\cos \left (t \right )+2 \sin \left (t \right )\right ) \cos \left (3\right )-2 \left (\cos \left (t \right )-\frac {\sin \left (t \right )}{2}\right ) \sin \left (3\right )\right ) {\mathrm e}^{-2 t +6}\right ) \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=2 \\ {} & {} & 2=c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-c_{1} \sin \left (t \right ) {\mathrm e}^{-2 t}-2 c_{1} \cos \left (t \right ) {\mathrm e}^{-2 t}+c_{2} \cos \left (t \right ) {\mathrm e}^{-2 t}-2 c_{2} \sin \left (t \right ) {\mathrm e}^{-2 t}-\mathit {Dirac}\left (t -3\right ) \left (-1+\left (\left (\cos \left (t \right )+2 \sin \left (t \right )\right ) \cos \left (3\right )-2 \left (\cos \left (t \right )-\frac {\sin \left (t \right )}{2}\right ) \sin \left (3\right )\right ) {\mathrm e}^{-2 t +6}\right )-\mathit {Heaviside}\left (t -3\right ) \left (\left (\left (2 \cos \left (t \right )-\sin \left (t \right )\right ) \cos \left (3\right )-2 \left (-\sin \left (t \right )-\frac {\cos \left (t \right )}{2}\right ) \sin \left (3\right )\right ) {\mathrm e}^{-2 t +6}-2 \left (\left (\cos \left (t \right )+2 \sin \left (t \right )\right ) \cos \left (3\right )-2 \left (\cos \left (t \right )-\frac {\sin \left (t \right )}{2}\right ) \sin \left (3\right )\right ) {\mathrm e}^{-2 t +6}\right ) \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=-2 c_{1} +c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =2, c_{2} =5\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\left (\left (\cos \left (3\right )-2 \sin \left (3\right )\right ) \cos \left (t \right )+2 \left (\cos \left (3\right )+\frac {\sin \left (3\right )}{2}\right ) \sin \left (t \right )\right ) \mathit {Heaviside}\left (t -3\right ) {\mathrm e}^{-2 t +6}+\mathit {Heaviside}\left (t -3\right )+{\mathrm e}^{-2 t} \left (2 \cos \left (t \right )+5 \sin \left (t \right )\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\left (\left (\cos \left (3\right )-2 \sin \left (3\right )\right ) \cos \left (t \right )+2 \left (\cos \left (3\right )+\frac {\sin \left (3\right )}{2}\right ) \sin \left (t \right )\right ) \mathit {Heaviside}\left (t -3\right ) {\mathrm e}^{-2 t +6}+\mathit {Heaviside}\left (t -3\right )+{\mathrm e}^{-2 t} \left (2 \cos \left (t \right )+5 \sin \left (t \right )\right ) \end {array} \]

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
trying a double symmetry of the form [xi=0, eta=F(x)] 
-> Try solving first the homogeneous part of the ODE 
   checking if the LODE has constant coefficients 
   <- constant coefficients successful 
<- solving first the homogeneous part of the ODE successful`
 

✓ Solution by Maple

Time used: 4.297 (sec). Leaf size: 53

dsolve([diff(y(t),t$2)+4*diff(y(t),t)+5*y(t)=5*Heaviside(t-3),y(0) = 2, D(y)(0) = 1],y(t), singsol=all)
 

\[ y \left (t \right ) = \left (-\frac {1}{2}-i\right ) \operatorname {Heaviside}\left (-3+t \right ) {\mathrm e}^{\left (-2-i\right ) \left (-3+t \right )}+\left (-\frac {1}{2}+i\right ) \operatorname {Heaviside}\left (-3+t \right ) {\mathrm e}^{\left (-2+i\right ) \left (-3+t \right )}+\operatorname {Heaviside}\left (-3+t \right )+{\mathrm e}^{-2 t} \left (2 \cos \left (t \right )+5 \sin \left (t \right )\right ) \]

✓ Solution by Mathematica

Time used: 0.037 (sec). Leaf size: 68

DSolve[{y''[t]+4*y'[t]+5*y[t]==5*UnitStep[t-3],{y[0]==2,y'[0]==1}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} e^{-2 t} (2 \cos (t)+5 \sin (t)) & t\leq 3 \\ e^{-2 t} \left (-e^6 \cos (3-t)+e^{2 t}+2 \cos (t)+2 e^6 \sin (3-t)+5 \sin (t)\right ) & \text {True} \\ \end {array} \\ \end {array} \]