Internal problem ID [12339]
Internal file name [OUTPUT/10992_Monday_October_02_2023_02_47_45_AM_90181001/index.tex
]
Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A.
Dobrushkin. CRC Press 2015
Section: Chapter 5.6 Laplace transform. Nonhomogeneous equations. Problems page
368
Problem number: Problem 5(c).
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]]
\[ \boxed {y^{\prime \prime }+4 y^{\prime }+29 y=5 \delta \left (t -\pi \right )-5 \delta \left (t -2 \pi \right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, 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) &=29\\ F &=5 \delta \left (t -\pi \right )-5 \delta \left (t -2 \pi \right ) \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+4 y^{\prime }+29 y = 5 \delta \left (t -\pi \right )-5 \delta \left (t -2 \pi \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 )+29 Y \left (s \right ) = 5 \,{\mathrm e}^{-s \pi }-5 \,{\mathrm e}^{-2 s \pi }\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 )+4 s Y \left (s \right )+29 Y \left (s \right ) = 5 \,{\mathrm e}^{-s \pi }-5 \,{\mathrm e}^{-2 s \pi } \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {5 \,{\mathrm e}^{-s \pi }-5 \,{\mathrm e}^{-2 s \pi }}{s^{2}+4 s +29} \end {align*}
Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (\frac {5 \,{\mathrm e}^{-s \pi }-5 \,{\mathrm e}^{-2 s \pi }}{s^{2}+4 s +29}\right )\\ &= -\sin \left (5 t \right ) \left (\operatorname {Heaviside}\left (t -\pi \right ) {\mathrm e}^{-2 t +2 \pi }+\operatorname {Heaviside}\left (t -2 \pi \right ) {\mathrm e}^{-2 t +4 \pi }\right ) \end {align*}
Converting the above solution to piecewise it becomes \[
y = \left \{\begin {array}{cc} 0 & t <\pi \\ -\sin \left (5 t \right ) {\mathrm e}^{-2 t +2 \pi } & t <2 \pi \\ -\sin \left (5 t \right ) \left ({\mathrm e}^{-2 t +2 \pi }+{\mathrm e}^{-2 t +4 \pi }\right ) & 2 \pi \le t \end {array}\right .
\] Simplifying the solution gives \[
y = -\sin \left (5 t \right ) \left (\left \{\begin {array}{cc} 0 & t <\pi \\ {\mathrm e}^{-2 t +2 \pi } & t <2 \pi \\ {\mathrm e}^{-2 t +2 \pi }+{\mathrm e}^{-2 t +4 \pi } & 2 \pi \le t \end {array}\right .\right )
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= -\sin \left (5 t \right ) \left (\left \{\begin {array}{cc} 0 & t <\pi \\ {\mathrm e}^{-2 t +2 \pi } & t <2 \pi \\ {\mathrm e}^{-2 t +2 \pi }+{\mathrm e}^{-2 t +4 \pi } & 2 \pi \le t \end {array}\right .\right ) \\
\end{align*} Verification of solutions
\[
y = -\sin \left (5 t \right ) \left (\left \{\begin {array}{cc} 0 & t <\pi \\ {\mathrm e}^{-2 t +2 \pi } & t <2 \pi \\ {\mathrm e}^{-2 t +2 \pi }+{\mathrm e}^{-2 t +4 \pi } & 2 \pi \le t \end {array}\right .\right )
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+4 y^{\prime }+29 y=5 \mathit {Dirac}\left (t -\pi \right )-5 \mathit {Dirac}\left (t -2 \pi \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 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+4 r +29=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-4\right )\pm \left (\sqrt {-100}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-2-5 \,\mathrm {I}, -2+5 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-2 t} \cos \left (5 t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-2 t} \sin \left (5 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 (5 t \right )+c_{2} {\mathrm e}^{-2 t} \sin \left (5 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 )=5 \mathit {Dirac}\left (t -\pi \right )-5 \mathit {Dirac}\left (t -2 \pi \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 (5 t \right ) & {\mathrm e}^{-2 t} \sin \left (5 t \right ) \\ -2 \,{\mathrm e}^{-2 t} \cos \left (5 t \right )-5 \,{\mathrm e}^{-2 t} \sin \left (5 t \right ) & -2 \,{\mathrm e}^{-2 t} \sin \left (5 t \right )+5 \,{\mathrm e}^{-2 t} \cos \left (5 t \right ) \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=5 \,{\mathrm e}^{-4 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )={\mathrm e}^{-2 t} \sin \left (5 t \right ) \left (\int \left (-{\mathrm e}^{4 \pi } \mathit {Dirac}\left (t -2 \pi \right )-{\mathrm e}^{2 \pi } \mathit {Dirac}\left (t -\pi \right )\right )d t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )={\mathrm e}^{-2 t} \sin \left (5 t \right ) \left (-{\mathrm e}^{2 \pi } \mathit {Heaviside}\left (t -\pi \right )-{\mathrm e}^{4 \pi } \mathit {Heaviside}\left (t -2 \pi \right )\right ) \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-2 t} \cos \left (5 t \right )+c_{2} {\mathrm e}^{-2 t} \sin \left (5 t \right )+{\mathrm e}^{-2 t} \sin \left (5 t \right ) \left (-{\mathrm e}^{2 \pi } \mathit {Heaviside}\left (t -\pi \right )-{\mathrm e}^{4 \pi } \mathit {Heaviside}\left (t -2 \pi \right )\right ) \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-2 t} \cos \left (5 t \right )+c_{2} {\mathrm e}^{-2 t} \sin \left (5 t \right )+{\mathrm e}^{-2 t} \sin \left (5 t \right ) \left (-{\mathrm e}^{2 \pi } \mathit {Heaviside}\left (t -\pi \right )-{\mathrm e}^{4 \pi } \mathit {Heaviside}\left (t -2 \pi \right )\right ) \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-2 c_{1} {\mathrm e}^{-2 t} \cos \left (5 t \right )-5 c_{1} {\mathrm e}^{-2 t} \sin \left (5 t \right )-2 c_{2} {\mathrm e}^{-2 t} \sin \left (5 t \right )+5 c_{2} {\mathrm e}^{-2 t} \cos \left (5 t \right )-2 \,{\mathrm e}^{-2 t} \sin \left (5 t \right ) \left (-{\mathrm e}^{2 \pi } \mathit {Heaviside}\left (t -\pi \right )-{\mathrm e}^{4 \pi } \mathit {Heaviside}\left (t -2 \pi \right )\right )+5 \,{\mathrm e}^{-2 t} \cos \left (5 t \right ) \left (-{\mathrm e}^{2 \pi } \mathit {Heaviside}\left (t -\pi \right )-{\mathrm e}^{4 \pi } \mathit {Heaviside}\left (t -2 \pi \right )\right )+{\mathrm e}^{-2 t} \sin \left (5 t \right ) \left (-{\mathrm e}^{4 \pi } \mathit {Dirac}\left (t -2 \pi \right )-{\mathrm e}^{2 \pi } \mathit {Dirac}\left (t -\pi \right )\right ) \\ {} & \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} +5 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =0, c_{2} =0\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-2 t} \sin \left (5 t \right ) \left (-{\mathrm e}^{2 \pi } \mathit {Heaviside}\left (t -\pi \right )-{\mathrm e}^{4 \pi } \mathit {Heaviside}\left (t -2 \pi \right )\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-2 t} \sin \left (5 t \right ) \left (-{\mathrm e}^{2 \pi } \mathit {Heaviside}\left (t -\pi \right )-{\mathrm e}^{4 \pi } \mathit {Heaviside}\left (t -2 \pi \right )\right ) \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 5.234 (sec). Leaf size: 41
\[
y \left (t \right ) = -\sin \left (5 t \right ) \left ({\mathrm e}^{-2 t +2 \pi } \operatorname {Heaviside}\left (t -\pi \right )+\operatorname {Heaviside}\left (-2 \pi +t \right ) {\mathrm e}^{4 \pi -2 t}\right )
\]
✓ Solution by Mathematica
Time used: 0.143 (sec). Leaf size: 39
\[
y(t)\to -e^{2 \pi -2 t} \left (e^{2 \pi } \theta (t-2 \pi )+\theta (t-\pi )\right ) \sin (5 t)
\]
4.32.2 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)+29*y(t)=5*Dirac(t-Pi)-5*Dirac(t-2*Pi),y(0) = 0, D(y)(0) = 0],y(t), singsol=all)
DSolve[{y''[t]+4*y'[t]+29*y[t]==5*DiracDelta[t-Pi]-5*DiracDelta[t-2*Pi],{y[0]==0,y'[0]==0}},y[t],t,IncludeSingularSolutions -> True]