Internal problem ID [2891]
Internal file name [OUTPUT/2383_Sunday_June_05_2022_03_03_31_AM_69290317/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 8.
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]]
\[ \boxed {y^{\prime \prime }-4 y^{\prime }+13 y=\delta \left (t -\frac {\pi }{4}\right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 3, y^{\prime }\left (0\right ) = 0] \end {align*}
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) &=13\\ F &=\delta \left (t -\frac {\pi }{4}\right ) \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }-4 y^{\prime }+13 y = \delta \left (t -\frac {\pi }{4}\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 )+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 )&=3\\ y'(0) &=0 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )+12-3 s -4 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}}+3 s -12}{s^{2}-4 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}}+3 s -12}{s^{2}-4 s +13}\right )\\ &= \frac {{\mathrm e}^{2 t -\frac {\pi }{2}} \sin \left (3 t +\frac {\pi }{4}\right ) \left (-1+\operatorname {Heaviside}\left (-t +\frac {\pi }{4}\right )\right )}{3}+{\mathrm e}^{2 t} \left (3 \cos \left (3 t \right )-2 \sin \left (3 t \right )\right ) \end {align*}
Converting the above solution to piecewise it becomes \[
y = \left \{\begin {array}{cc} {\mathrm e}^{2 t} \left (3 \cos \left (3 t \right )-2 \sin \left (3 t \right )\right ) & t \le \frac {\pi }{4} \\ {\mathrm e}^{2 t} \left (3 \cos \left (3 t \right )-2 \sin \left (3 t \right )\right )-\frac {{\mathrm e}^{2 t -\frac {\pi }{2}} \sin \left (3 t +\frac {\pi }{4}\right )}{3} & \frac {\pi }{4} The solution(s) found are the following \begin{align*}
\tag{1} y &= \left \{\begin {array}{cc} {\mathrm e}^{2 t} \left (3 \cos \left (3 t \right )-2 \sin \left (3 t \right )\right ) & t \le \frac {\pi }{4} \\ -\frac {{\mathrm e}^{2 t -\frac {\pi }{2}} \sqrt {2}\, \left (\sin \left (3 t \right )+\cos \left (3 t \right )\right )}{6}+3 \,{\mathrm e}^{2 t} \left (\cos \left (3 t \right )-\frac {2 \sin \left (3 t \right )}{3}\right ) & \frac {\pi }{4} Verification of solutions
\[
y = \left \{\begin {array}{cc} {\mathrm e}^{2 t} \left (3 \cos \left (3 t \right )-2 \sin \left (3 t \right )\right ) & t \le \frac {\pi }{4} \\ -\frac {{\mathrm e}^{2 t -\frac {\pi }{2}} \sqrt {2}\, \left (\sin \left (3 t \right )+\cos \left (3 t \right )\right )}{6}+3 \,{\mathrm e}^{2 t} \left (\cos \left (3 t \right )-\frac {2 \sin \left (3 t \right )}{3}\right ) & \frac {\pi }{4}
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }-4 y^{\prime }+13 y=\mathit {Dirac}\left (t -\frac {\pi }{4}\right ), y \left (0\right )=3, 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}-4 r +13=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {4\pm \left (\sqrt {-36}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (2-3 \,\mathrm {I}, 2+3 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{2 t} \cos \left (3 t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{2 t} \sin \left (3 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}^{2 t} \cos \left (3 t \right )+c_{2} {\mathrm e}^{2 t} \sin \left (3 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}^{2 t} \cos \left (3 t \right ) & {\mathrm e}^{2 t} \sin \left (3 t \right ) \\ 2 \,{\mathrm e}^{2 t} \cos \left (3 t \right )-3 \,{\mathrm e}^{2 t} \sin \left (3 t \right ) & 2 \,{\mathrm e}^{2 t} \sin \left (3 t \right )+3 \,{\mathrm e}^{2 t} \cos \left (3 t \right ) \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=3 \,{\mathrm e}^{4 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\frac {\sqrt {2}\, {\mathrm e}^{2 t -\frac {\pi }{2}} \left (\sin \left (3 t \right )+\cos \left (3 t \right )\right ) \left (\int \mathit {Dirac}\left (t -\frac {\pi }{4}\right )d t \right )}{6} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\frac {\sqrt {2}\, {\mathrm e}^{2 t -\frac {\pi }{2}} \left (\sin \left (3 t \right )+\cos \left (3 t \right )\right ) \mathit {Heaviside}\left (t -\frac {\pi }{4}\right )}{6} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{2 t} \cos \left (3 t \right )+c_{2} {\mathrm e}^{2 t} \sin \left (3 t \right )-\frac {\sqrt {2}\, {\mathrm e}^{2 t -\frac {\pi }{2}} \left (\sin \left (3 t \right )+\cos \left (3 t \right )\right ) \mathit {Heaviside}\left (t -\frac {\pi }{4}\right )}{6} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{2 t} \cos \left (3 t \right )+c_{2} {\mathrm e}^{2 t} \sin \left (3 t \right )-\frac {\sqrt {2}\, {\mathrm e}^{2 t -\frac {\pi }{2}} \left (\sin \left (3 t \right )+\cos \left (3 t \right )\right ) \mathit {Heaviside}\left (t -\frac {\pi }{4}\right )}{6} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=3 \\ {} & {} & 3=c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=2 c_{1} {\mathrm e}^{2 t} \cos \left (3 t \right )-3 c_{1} {\mathrm e}^{2 t} \sin \left (3 t \right )+2 c_{2} {\mathrm e}^{2 t} \sin \left (3 t \right )+3 c_{2} {\mathrm e}^{2 t} \cos \left (3 t \right )-\frac {\sqrt {2}\, {\mathrm e}^{2 t -\frac {\pi }{2}} \left (\sin \left (3 t \right )+\cos \left (3 t \right )\right ) \mathit {Heaviside}\left (t -\frac {\pi }{4}\right )}{3}-\frac {\sqrt {2}\, {\mathrm e}^{2 t -\frac {\pi }{2}} \left (3 \cos \left (3 t \right )-3 \sin \left (3 t \right )\right ) \mathit {Heaviside}\left (t -\frac {\pi }{4}\right )}{6}-\frac {\sqrt {2}\, {\mathrm e}^{2 t -\frac {\pi }{2}} \left (\sin \left (3 t \right )+\cos \left (3 t \right )\right ) \mathit {Dirac}\left (t -\frac {\pi }{4}\right )}{6} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=2 c_{1} +3 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =3, c_{2} =-2\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {\sqrt {2}\, {\mathrm e}^{2 t -\frac {\pi }{2}} \left (\sin \left (3 t \right )+\cos \left (3 t \right )\right ) \mathit {Heaviside}\left (t -\frac {\pi }{4}\right )}{6}+3 \,{\mathrm e}^{2 t} \left (\cos \left (3 t \right )-\frac {2 \sin \left (3 t \right )}{3}\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {\sqrt {2}\, {\mathrm e}^{2 t -\frac {\pi }{2}} \left (\sin \left (3 t \right )+\cos \left (3 t \right )\right ) \mathit {Heaviside}\left (t -\frac {\pi }{4}\right )}{6}+3 \,{\mathrm e}^{2 t} \left (\cos \left (3 t \right )-\frac {2 \sin \left (3 t \right )}{3}\right ) \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 3.344 (sec). Leaf size: 51
\[
y \left (t \right ) = -\frac {\sqrt {2}\, {\mathrm e}^{-\frac {\pi }{2}+2 t} \operatorname {Heaviside}\left (t -\frac {\pi }{4}\right ) \left (\sin \left (3 t \right )+\cos \left (3 t \right )\right )}{6}+3 \,{\mathrm e}^{2 t} \left (\cos \left (3 t \right )-\frac {2 \sin \left (3 t \right )}{3}\right )
\]
✓ Solution by Mathematica
Time used: 0.211 (sec). Leaf size: 61
\[
y(t)\to \frac {1}{6} e^{2 t} \left (6 (3 \cos (3 t)-2 \sin (3 t))-\sqrt {2} e^{-\pi /2} \theta (12 t-3 \pi ) (\sin (3 t)+\cos (3 t))\right )
\]
15.8.2 Maple step by step solution
`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`
dsolve([diff(y(t),t$2)-4*diff(y(t),t)+13*y(t)=Dirac(t-Pi/4),y(0) = 3, D(y)(0) = 0],y(t), singsol=all)
DSolve[{y''[t]-4*y'[t]+13*y[t]==DiracDelta[t-Pi/4],{y[0]==3,y'[0]==0}},y[t],t,IncludeSingularSolutions -> True]