15.10 problem Problem 10

Internal problem ID [2893]
Internal file name [OUTPUT/2385_Sunday_June_05_2022_03_03_45_AM_83974142/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 10.
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]]

\[ \boxed {y^{\prime \prime }+6 y^{\prime }+13 y=\delta \left (t -\frac {\pi }{4}\right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 5, y^{\prime }\left (0\right ) = 5] \end {align*}

15.10.1 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) &=6\\ q(t) &=13\\ F &=\delta \left (t -\frac {\pi }{4}\right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+6 y^{\prime }+13 y = \delta \left (t -\frac {\pi }{4}\right ) \end {align*}

The domain of \(p(t)=6\) 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 )+6 s Y \left (s \right )-6 y \left (0\right )+13 Y \left (s \right ) = {\mathrm e}^{-\frac {s \pi }{4}}\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&=5\\ y'(0) &=5 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-35-5 s +6 s Y \left (s \right )+13 Y \left (s \right ) = {\mathrm e}^{-\frac {s \pi }{4}} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {{\mathrm e}^{-\frac {s \pi }{4}}+5 s +35}{s^{2}+6 s +13} \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}^{-\frac {s \pi }{4}}+5 s +35}{s^{2}+6 s +13}\right )\\ &= -\frac {\operatorname {Heaviside}\left (t -\frac {\pi }{4}\right ) {\mathrm e}^{-3 t +\frac {3 \pi }{4}} \cos \left (2 t \right )}{2}+5 \,{\mathrm e}^{-3 t} \left (\cos \left (2 t \right )+2 \sin \left (2 t \right )\right ) \end {align*}

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

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= 5 \,{\mathrm e}^{-3 t} \left (\cos \left (2 t \right )+2 \sin \left (2 t \right )\right )-\left (\left \{\begin {array}{cc} 0 & t <\frac {\pi }{4} \\ \frac {{\mathrm e}^{-3 t +\frac {3 \pi }{4}} \cos \left (2 t \right )}{2} & \frac {\pi }{4}\le t \end {array}\right .\right ) \\ \end{align*}

Verification of solutions

\[ y = 5 \,{\mathrm e}^{-3 t} \left (\cos \left (2 t \right )+2 \sin \left (2 t \right )\right )-\left (\left \{\begin {array}{cc} 0 & t <\frac {\pi }{4} \\ \frac {{\mathrm e}^{-3 t +\frac {3 \pi }{4}} \cos \left (2 t \right )}{2} & \frac {\pi }{4}\le t \end {array}\right .\right ) \] Verified OK.

15.10.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+6 y^{\prime }+13 y=\mathit {Dirac}\left (t -\frac {\pi }{4}\right ), y \left (0\right )=5, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=5\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d t}y^{\prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+6 r +13=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-6\right )\pm \left (\sqrt {-16}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-3-2 \,\mathrm {I}, -3+2 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-3 t} \cos \left (2 t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-3 t} \sin \left (2 t \right ) \\ \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} {\mathrm e}^{-3 t} \cos \left (2 t \right )+c_{2} {\mathrm e}^{-3 t} \sin \left (2 t \right )+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 )=\mathit {Dirac}\left (t -\frac {\pi }{4}\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} {\mathrm e}^{-3 t} \cos \left (2 t \right ) & {\mathrm e}^{-3 t} \sin \left (2 t \right ) \\ -3 \,{\mathrm e}^{-3 t} \cos \left (2 t \right )-2 \,{\mathrm e}^{-3 t} \sin \left (2 t \right ) & -3 \,{\mathrm e}^{-3 t} \sin \left (2 t \right )+2 \,{\mathrm e}^{-3 t} \cos \left (2 t \right ) \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=2 \,{\mathrm e}^{-6 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\frac {\left (\int \mathit {Dirac}\left (t -\frac {\pi }{4}\right )d t \right ) \cos \left (2 t \right ) {\mathrm e}^{-3 t +\frac {3 \pi }{4}}}{2} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\frac {\mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) {\mathrm e}^{-3 t +\frac {3 \pi }{4}} \cos \left (2 t \right )}{2} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-3 t} \cos \left (2 t \right )+c_{2} {\mathrm e}^{-3 t} \sin \left (2 t \right )-\frac {\mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) {\mathrm e}^{-3 t +\frac {3 \pi }{4}} \cos \left (2 t \right )}{2} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-3 t} \cos \left (2 t \right )+c_{2} {\mathrm e}^{-3 t} \sin \left (2 t \right )-\frac {\mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) {\mathrm e}^{-3 t +\frac {3 \pi }{4}} \cos \left (2 t \right )}{2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=5 \\ {} & {} & 5=c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-3 c_{1} {\mathrm e}^{-3 t} \cos \left (2 t \right )-2 c_{1} {\mathrm e}^{-3 t} \sin \left (2 t \right )-3 c_{2} {\mathrm e}^{-3 t} \sin \left (2 t \right )+2 c_{2} {\mathrm e}^{-3 t} \cos \left (2 t \right )-\frac {\mathit {Dirac}\left (t -\frac {\pi }{4}\right ) {\mathrm e}^{-3 t +\frac {3 \pi }{4}} \cos \left (2 t \right )}{2}+\frac {3 \mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) {\mathrm e}^{-3 t +\frac {3 \pi }{4}} \cos \left (2 t \right )}{2}+\mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) {\mathrm e}^{-3 t +\frac {3 \pi }{4}} \sin \left (2 t \right ) \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=5 \\ {} & {} & 5=-3 c_{1} +2 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =5, c_{2} =10\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {\mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) {\mathrm e}^{-3 t +\frac {3 \pi }{4}} \cos \left (2 t \right )}{2}+5 \,{\mathrm e}^{-3 t} \left (\cos \left (2 t \right )+2 \sin \left (2 t \right )\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {\mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) {\mathrm e}^{-3 t +\frac {3 \pi }{4}} \cos \left (2 t \right )}{2}+5 \,{\mathrm e}^{-3 t} \left (\cos \left (2 t \right )+2 \sin \left (2 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: 2.531 (sec). Leaf size: 42

dsolve([diff(y(t),t$2)+6*diff(y(t),t)+13*y(t)=Dirac(t-Pi/4),y(0) = 5, D(y)(0) = 5],y(t), singsol=all)
 

\[ y \left (t \right ) = -\frac {\operatorname {Heaviside}\left (t -\frac {\pi }{4}\right ) \cos \left (2 t \right ) {\mathrm e}^{\frac {3 \pi }{4}-3 t}}{2}+5 \,{\mathrm e}^{-3 t} \left (\cos \left (2 t \right )+2 \sin \left (2 t \right )\right ) \]

✓ Solution by Mathematica

Time used: 0.287 (sec). Leaf size: 121

DSolve[{y''[t]+46*y'[t]+13*y[t]==DiracDelta[t-Pi/4],{y[0]==1,y'[0]==-1}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {1}{516} e^{-2 \sqrt {129} t-23 t-\frac {\sqrt {129} \pi }{2}} \left (2 e^{\frac {\sqrt {129} \pi }{2}} \left (\left (129+11 \sqrt {129}\right ) e^{4 \sqrt {129} t}+129-11 \sqrt {129}\right )-\sqrt {129} e^{23 \pi /4} \left (e^{\sqrt {129} \pi }-e^{4 \sqrt {129} t}\right ) \theta (4 t-\pi )\right ) \]