15.12 problem Problem 12

Internal problem ID [2895]
Internal file name [OUTPUT/2387_Sunday_June_05_2022_03_03_59_AM_82386455/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 12.
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]]

\[ \boxed {y^{\prime \prime }+16 y=4 \cos \left (3 t \right )+\delta \left (t -\frac {\pi }{3}\right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 0] \end {align*}

15.12.1 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) &=0\\ q(t) &=16\\ F &=4 \cos \left (3 t \right )+\delta \left (t -\frac {\pi }{3}\right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+16 y = 4 \cos \left (3 t \right )+\delta \left (t -\frac {\pi }{3}\right ) \end {align*}

The domain of \(p(t)=0\) 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 )+16 Y \left (s \right ) = \frac {4 s}{s^{2}+9}+{\mathrm e}^{-\frac {s \pi }{3}}\tag {1} \end {align*}

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

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )+16 Y \left (s \right ) = \frac {4 s}{s^{2}+9}+{\mathrm e}^{-\frac {s \pi }{3}} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {{\mathrm e}^{-\frac {s \pi }{3}} s^{2}+9 \,{\mathrm e}^{-\frac {s \pi }{3}}+4 s}{\left (s^{2}+9\right ) \left (s^{2}+16\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 }{3}} s^{2}+9 \,{\mathrm e}^{-\frac {s \pi }{3}}+4 s}{\left (s^{2}+9\right ) \left (s^{2}+16\right )}\right )\\ &= \frac {\operatorname {Heaviside}\left (t -\frac {\pi }{3}\right ) \cos \left (4 t +\frac {\pi }{6}\right )}{4}-\frac {4 \cos \left (4 t \right )}{7}+\frac {4 \cos \left (3 t \right )}{7} \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} -\frac {4 \cos \left (4 t \right )}{7}+\frac {4 \cos \left (3 t \right )}{7} & t <\frac {\pi }{3} \\ -\frac {4 \cos \left (4 t \right )}{7}+\frac {4 \cos \left (3 t \right )}{7}+\frac {\cos \left (4 t +\frac {\pi }{6}\right )}{4} & \frac {\pi }{3}\le t \end {array}\right . \] Simplifying the solution gives \[ y = -\frac {4 \cos \left (4 t \right )}{7}+\frac {4 \cos \left (3 t \right )}{7}+\left (\left \{\begin {array}{cc} 0 & t <\frac {\pi }{3} \\ \frac {\cos \left (4 t \right ) \sqrt {3}}{8}-\frac {\sin \left (4 t \right )}{8} & \frac {\pi }{3}\le t \end {array}\right .\right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {4 \cos \left (4 t \right )}{7}+\frac {4 \cos \left (3 t \right )}{7}+\left (\left \{\begin {array}{cc} 0 & t <\frac {\pi }{3} \\ \frac {\cos \left (4 t \right ) \sqrt {3}}{8}-\frac {\sin \left (4 t \right )}{8} & \frac {\pi }{3}\le t \end {array}\right .\right ) \\ \end{align*}

Verification of solutions

\[ y = -\frac {4 \cos \left (4 t \right )}{7}+\frac {4 \cos \left (3 t \right )}{7}+\left (\left \{\begin {array}{cc} 0 & t <\frac {\pi }{3} \\ \frac {\cos \left (4 t \right ) \sqrt {3}}{8}-\frac {\sin \left (4 t \right )}{8} & \frac {\pi }{3}\le t \end {array}\right .\right ) \] Verified OK.

15.12.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+16 y=4 \cos \left (3 t \right )+\mathit {Dirac}\left (t -\frac {\pi }{3}\right ), y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0\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}+16=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {-64}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-4 \,\mathrm {I}, 4 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )=\cos \left (4 t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=\sin \left (4 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} \cos \left (4 t \right )+c_{2} \sin \left (4 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 \cos \left (3 t \right )+\mathit {Dirac}\left (t -\frac {\pi }{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 (4 t \right ) & \sin \left (4 t \right ) \\ -4 \sin \left (4 t \right ) & 4 \cos \left (4 t \right ) \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=4 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\frac {\cos \left (4 t \right ) \left (\int \left (-\sqrt {3}\, \mathit {Dirac}\left (t -\frac {\pi }{3}\right )+8 \sin \left (4 t \right ) \cos \left (3 t \right )\right )d t \right )}{8}+\frac {\sin \left (4 t \right ) \left (\int \left (8 \cos \left (4 t \right ) \cos \left (3 t \right )-\mathit {Dirac}\left (t -\frac {\pi }{3}\right )\right )d t \right )}{8} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\frac {\left (\left (8 \cos \left (t \right )^{4}-8 \cos \left (t \right )^{2}+1\right ) \sqrt {3}-8 \sin \left (t \right ) \cos \left (t \right )^{3}+4 \cos \left (t \right ) \sin \left (t \right )\right ) \mathit {Heaviside}\left (t -\frac {\pi }{3}\right )}{8}+\frac {16 \cos \left (t \right )^{3}}{7}-\frac {12 \cos \left (t \right )}{7} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} \cos \left (4 t \right )+c_{2} \sin \left (4 t \right )+\frac {\left (\left (8 \cos \left (t \right )^{4}-8 \cos \left (t \right )^{2}+1\right ) \sqrt {3}-8 \sin \left (t \right ) \cos \left (t \right )^{3}+4 \cos \left (t \right ) \sin \left (t \right )\right ) \mathit {Heaviside}\left (t -\frac {\pi }{3}\right )}{8}+\frac {16 \cos \left (t \right )^{3}}{7}-\frac {12 \cos \left (t \right )}{7} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} \cos \left (4 t \right )+c_{2} \sin \left (4 t \right )+\frac {\left (\left (8 \cos \left (t \right )^{4}-8 \cos \left (t \right )^{2}+1\right ) \sqrt {3}-8 \sin \left (t \right ) \cos \left (t \right )^{3}+4 \cos \left (t \right ) \sin \left (t \right )\right ) \mathit {Heaviside}\left (t -\frac {\pi }{3}\right )}{8}+\frac {16 \cos \left (t \right )^{3}}{7}-\frac {12 \cos \left (t \right )}{7} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} +\frac {4}{7} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-4 c_{1} \sin \left (4 t \right )+4 c_{2} \cos \left (4 t \right )+\frac {\left (\left (-32 \sin \left (t \right ) \cos \left (t \right )^{3}+16 \cos \left (t \right ) \sin \left (t \right )\right ) \sqrt {3}-8 \cos \left (t \right )^{4}+24 \sin \left (t \right )^{2} \cos \left (t \right )^{2}-4 \sin \left (t \right )^{2}+4 \cos \left (t \right )^{2}\right ) \mathit {Heaviside}\left (t -\frac {\pi }{3}\right )}{8}+\frac {\left (\left (8 \cos \left (t \right )^{4}-8 \cos \left (t \right )^{2}+1\right ) \sqrt {3}-8 \sin \left (t \right ) \cos \left (t \right )^{3}+4 \cos \left (t \right ) \sin \left (t \right )\right ) \mathit {Dirac}\left (t -\frac {\pi }{3}\right )}{8}-\frac {48 \cos \left (t \right )^{2} \sin \left (t \right )}{7}+\frac {12 \sin \left (t \right )}{7} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=4 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\frac {4}{7}, c_{2} =0\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {4}{7}+\frac {\left (\left (8 \cos \left (t \right )^{4}-8 \cos \left (t \right )^{2}+1\right ) \sqrt {3}-8 \sin \left (t \right ) \cos \left (t \right )^{3}+4 \cos \left (t \right ) \sin \left (t \right )\right ) \mathit {Heaviside}\left (t -\frac {\pi }{3}\right )}{8}-\frac {32 \cos \left (t \right )^{4}}{7}+\frac {16 \cos \left (t \right )^{3}}{7}+\frac {32 \cos \left (t \right )^{2}}{7}-\frac {12 \cos \left (t \right )}{7} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {4}{7}+\frac {\left (\left (8 \cos \left (t \right )^{4}-8 \cos \left (t \right )^{2}+1\right ) \sqrt {3}-8 \sin \left (t \right ) \cos \left (t \right )^{3}+4 \cos \left (t \right ) \sin \left (t \right )\right ) \mathit {Heaviside}\left (t -\frac {\pi }{3}\right )}{8}-\frac {32 \cos \left (t \right )^{4}}{7}+\frac {16 \cos \left (t \right )^{3}}{7}+\frac {32 \cos \left (t \right )^{2}}{7}-\frac {12 \cos \left (t \right )}{7} \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: 3.609 (sec). Leaf size: 33

dsolve([diff(y(t),t$2)+16*y(t)=4*cos(3*t)+Dirac(t-Pi/3),y(0) = 0, D(y)(0) = 0],y(t), singsol=all)
 

\[ y \left (t \right ) = \frac {\left (\cos \left (4 t \right ) \sqrt {3}-\sin \left (4 t \right )\right ) \operatorname {Heaviside}\left (t -\frac {\pi }{3}\right )}{8}-\frac {4 \cos \left (4 t \right )}{7}+\frac {4 \cos \left (3 t \right )}{7} \]

✓ Solution by Mathematica

Time used: 0.159 (sec). Leaf size: 50

DSolve[{y''[t]+16*y[t]==4*Cos[3*t]+DiracDelta[t-Pi/3],{y[0]==0,y'[0]==0}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {1}{8} \theta (3 t-\pi ) \left (\sqrt {3} \cos (4 t)-\sin (4 t)\right )+\frac {4}{7} (\cos (3 t)-\cos (4 t)) \]