15.13 problem Problem 13

Internal problem ID [2896]
Internal file name [OUTPUT/2388_Sunday_June_05_2022_03_04_08_AM_19570783/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 13.
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]]

\[ \boxed {y^{\prime \prime }+2 y^{\prime }+5 y=4 \sin \left (t \right )+\delta \left (t -\frac {\pi }{6}\right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 1] \end {align*}

15.13.1 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) &=2\\ q(t) &=5\\ F &=4 \sin \left (t \right )+\delta \left (t -\frac {\pi }{6}\right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+2 y^{\prime }+5 y = 4 \sin \left (t \right )+\delta \left (t -\frac {\pi }{6}\right ) \end {align*}

The domain of \(p(t)=2\) 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 )+2 s Y \left (s \right )-2 y \left (0\right )+5 Y \left (s \right ) = \frac {4}{s^{2}+1}+{\mathrm e}^{-\frac {s \pi }{6}}\tag {1} \end {align*}

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

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

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {{\mathrm e}^{-\frac {s \pi }{6}} s^{2}+s^{2}+{\mathrm e}^{-\frac {s \pi }{6}}+5}{\left (s^{2}+1\right ) \left (s^{2}+2 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}^{-\frac {s \pi }{6}} s^{2}+s^{2}+{\mathrm e}^{-\frac {s \pi }{6}}+5}{\left (s^{2}+1\right ) \left (s^{2}+2 s +5\right )}\right )\\ &= -\frac {\operatorname {Heaviside}\left (t -\frac {\pi }{6}\right ) {\mathrm e}^{-t +\frac {\pi }{6}} \cos \left (2 t +\frac {\pi }{6}\right )}{2}-\frac {2 \cos \left (t \right )}{5}+\frac {4 \sin \left (t \right )}{5}+\frac {{\mathrm e}^{-t} \left (4 \cos \left (2 t \right )+3 \sin \left (2 t \right )\right )}{10} \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} -\frac {2 \cos \left (t \right )}{5}+\frac {4 \sin \left (t \right )}{5}+\frac {{\mathrm e}^{-t} \left (4 \cos \left (2 t \right )+3 \sin \left (2 t \right )\right )}{10} & t <\frac {\pi }{6} \\ -\frac {2 \cos \left (t \right )}{5}+\frac {4 \sin \left (t \right )}{5}+\frac {{\mathrm e}^{-t} \left (4 \cos \left (2 t \right )+3 \sin \left (2 t \right )\right )}{10}-\frac {{\mathrm e}^{-t +\frac {\pi }{6}} \cos \left (2 t +\frac {\pi }{6}\right )}{2} & \frac {\pi }{6}\le t \end {array}\right . \] Simplifying the solution gives \[ y = -\left (\left \{\begin {array}{cc} 0 & t <\frac {\pi }{6} \\ \frac {{\mathrm e}^{-t +\frac {\pi }{6}} \left (\cos \left (2 t \right ) \sqrt {3}-\sin \left (2 t \right )\right )}{4} & \frac {\pi }{6}\le t \end {array}\right .\right )+\frac {{\mathrm e}^{-t} \left (4 \cos \left (2 t \right )+3 \sin \left (2 t \right )\right )}{10}-\frac {2 \cos \left (t \right )}{5}+\frac {4 \sin \left (t \right )}{5} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\left (\left \{\begin {array}{cc} 0 & t <\frac {\pi }{6} \\ \frac {{\mathrm e}^{-t +\frac {\pi }{6}} \left (\cos \left (2 t \right ) \sqrt {3}-\sin \left (2 t \right )\right )}{4} & \frac {\pi }{6}\le t \end {array}\right .\right )+\frac {{\mathrm e}^{-t} \left (4 \cos \left (2 t \right )+3 \sin \left (2 t \right )\right )}{10}-\frac {2 \cos \left (t \right )}{5}+\frac {4 \sin \left (t \right )}{5} \\ \end{align*}

Verification of solutions

\[ y = -\left (\left \{\begin {array}{cc} 0 & t <\frac {\pi }{6} \\ \frac {{\mathrm e}^{-t +\frac {\pi }{6}} \left (\cos \left (2 t \right ) \sqrt {3}-\sin \left (2 t \right )\right )}{4} & \frac {\pi }{6}\le t \end {array}\right .\right )+\frac {{\mathrm e}^{-t} \left (4 \cos \left (2 t \right )+3 \sin \left (2 t \right )\right )}{10}-\frac {2 \cos \left (t \right )}{5}+\frac {4 \sin \left (t \right )}{5} \] Verified OK.

15.13.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+2 y^{\prime }+5 y=4 \sin \left (t \right )+\mathit {Dirac}\left (t -\frac {\pi }{6}\right ), y \left (0\right )=0, 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}+2 r +5=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-2\right )\pm \left (\sqrt {-16}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-1-2 \,\mathrm {I}, -1+2 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-t} \cos \left (2 t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-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}^{-t} \cos \left (2 t \right )+c_{2} {\mathrm e}^{-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 )=4 \sin \left (t \right )+\mathit {Dirac}\left (t -\frac {\pi }{6}\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}^{-t} \cos \left (2 t \right ) & {\mathrm e}^{-t} \sin \left (2 t \right ) \\ -{\mathrm e}^{-t} \cos \left (2 t \right )-2 \,{\mathrm e}^{-t} \sin \left (2 t \right ) & -{\mathrm e}^{-t} \sin \left (2 t \right )+2 \,{\mathrm e}^{-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}^{-2 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\frac {{\mathrm e}^{-t} \left (\cos \left (2 t \right ) \left (\int \left (16 \sin \left (t \right )^{2} \cos \left (t \right ) {\mathrm e}^{t}+{\mathrm e}^{\frac {\pi }{6}} \sqrt {3}\, \mathit {Dirac}\left (t -\frac {\pi }{6}\right )\right )d t \right )-\sin \left (2 t \right ) \left (\int \left (8 \cos \left (2 t \right ) \sin \left (t \right ) {\mathrm e}^{t}+{\mathrm e}^{\frac {\pi }{6}} \mathit {Dirac}\left (t -\frac {\pi }{6}\right )\right )d t \right )\right )}{4} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\frac {\mathit {Heaviside}\left (t -\frac {\pi }{6}\right ) \left (\cos \left (t \right )^{2} \sqrt {3}-\frac {\sqrt {3}}{2}-\sin \left (t \right ) \cos \left (t \right )\right ) {\mathrm e}^{-t +\frac {\pi }{6}}}{2}-\frac {2 \cos \left (t \right )}{5}+\frac {4 \sin \left (t \right )}{5} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-t} \cos \left (2 t \right )+c_{2} {\mathrm e}^{-t} \sin \left (2 t \right )-\frac {\mathit {Heaviside}\left (t -\frac {\pi }{6}\right ) \left (\cos \left (t \right )^{2} \sqrt {3}-\frac {\sqrt {3}}{2}-\sin \left (t \right ) \cos \left (t \right )\right ) {\mathrm e}^{-t +\frac {\pi }{6}}}{2}-\frac {2 \cos \left (t \right )}{5}+\frac {4 \sin \left (t \right )}{5} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-t} \cos \left (2 t \right )+c_{2} {\mathrm e}^{-t} \sin \left (2 t \right )-\frac {\mathit {Heaviside}\left (t -\frac {\pi }{6}\right ) \left (\cos \left (t \right )^{2} \sqrt {3}-\frac {\sqrt {3}}{2}-\sin \left (t \right ) \cos \left (t \right )\right ) {\mathrm e}^{-t +\frac {\pi }{6}}}{2}-\frac {2 \cos \left (t \right )}{5}+\frac {4 \sin \left (t \right )}{5} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} -\frac {2}{5} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-c_{1} {\mathrm e}^{-t} \cos \left (2 t \right )-2 c_{1} {\mathrm e}^{-t} \sin \left (2 t \right )-c_{2} {\mathrm e}^{-t} \sin \left (2 t \right )+2 c_{2} {\mathrm e}^{-t} \cos \left (2 t \right )-\frac {\mathit {Dirac}\left (t -\frac {\pi }{6}\right ) \left (\cos \left (t \right )^{2} \sqrt {3}-\frac {\sqrt {3}}{2}-\sin \left (t \right ) \cos \left (t \right )\right ) {\mathrm e}^{-t +\frac {\pi }{6}}}{2}-\frac {\mathit {Heaviside}\left (t -\frac {\pi }{6}\right ) \left (-2 \sqrt {3}\, \sin \left (t \right ) \cos \left (t \right )-\cos \left (t \right )^{2}+\sin \left (t \right )^{2}\right ) {\mathrm e}^{-t +\frac {\pi }{6}}}{2}+\frac {\mathit {Heaviside}\left (t -\frac {\pi }{6}\right ) \left (\cos \left (t \right )^{2} \sqrt {3}-\frac {\sqrt {3}}{2}-\sin \left (t \right ) \cos \left (t \right )\right ) {\mathrm e}^{-t +\frac {\pi }{6}}}{2}+\frac {2 \sin \left (t \right )}{5}+\frac {4 \cos \left (t \right )}{5} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=-c_{1} +\frac {4}{5}+2 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =\frac {2}{5}, c_{2} =\frac {3}{10}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {\mathit {Heaviside}\left (t -\frac {\pi }{6}\right ) \left (\cos \left (t \right )^{2} \sqrt {3}-\frac {\sqrt {3}}{2}-\sin \left (t \right ) \cos \left (t \right )\right ) {\mathrm e}^{-t +\frac {\pi }{6}}}{2}+\frac {\left (4 \cos \left (t \right )^{2}+3 \sin \left (t \right ) \cos \left (t \right )-2\right ) {\mathrm e}^{-t}}{5}-\frac {2 \cos \left (t \right )}{5}+\frac {4 \sin \left (t \right )}{5} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {\mathit {Heaviside}\left (t -\frac {\pi }{6}\right ) \left (\cos \left (t \right )^{2} \sqrt {3}-\frac {\sqrt {3}}{2}-\sin \left (t \right ) \cos \left (t \right )\right ) {\mathrm e}^{-t +\frac {\pi }{6}}}{2}+\frac {\left (4 \cos \left (t \right )^{2}+3 \sin \left (t \right ) \cos \left (t \right )-2\right ) {\mathrm e}^{-t}}{5}-\frac {2 \cos \left (t \right )}{5}+\frac {4 \sin \left (t \right )}{5} \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: 3.406 (sec). Leaf size: 56

dsolve([diff(y(t),t$2)+2*diff(y(t),t)+5*y(t)=4*sin(t)+Dirac(t-Pi/6),y(0) = 0, D(y)(0) = 1],y(t), singsol=all)
 

\[ y \left (t \right ) = -\frac {\operatorname {Heaviside}\left (t -\frac {\pi }{6}\right ) \left (\sqrt {3}\, \cos \left (t \right )^{2}-\cos \left (t \right ) \sin \left (t \right )-\frac {\sqrt {3}}{2}\right ) {\mathrm e}^{-t +\frac {\pi }{6}}}{2}+\frac {\left (4 \cos \left (t \right )^{2}+3 \cos \left (t \right ) \sin \left (t \right )-2\right ) {\mathrm e}^{-t}}{5}-\frac {2 \cos \left (t \right )}{5}+\frac {4 \sin \left (t \right )}{5} \]

✓ Solution by Mathematica

Time used: 0.644 (sec). Leaf size: 75

DSolve[{y''[t]+2*y'[t]+5*y[t]==4*Sin[t]+DiracDelta[t-Pi/6],{y[0]==0,y'[0]==1}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {1}{20} e^{-t} \left (-5 e^{\pi /6} \theta (6 t-\pi ) \left (\sqrt {3} \cos (2 t)-\sin (2 t)\right )+16 e^t \sin (t)+6 \sin (2 t)-8 e^t \cos (t)+8 \cos (2 t)\right ) \]