Internal problem ID [2889]
Internal file name [OUTPUT/2381_Sunday_June_05_2022_03_03_19_AM_18570981/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 6.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_laplace", "second_order_linear_constant_coeff", "second_order_ode_can_be_made_integrable"
Maple gives the following as the ode type
[[_2nd_order, _linear, _nonhomogeneous]]
\[ \boxed {y^{\prime \prime }-4 y=\delta \left (t -3\right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 1] \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) &=0\\ q(t) &=-4\\ F &=\delta \left (t -3\right ) \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }-4 y = \delta \left (t -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 )-4 Y \left (s \right ) = {\mathrm e}^{-3 s}\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 )-1-4 Y \left (s \right ) = {\mathrm e}^{-3 s} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {{\mathrm e}^{-3 s}+1}{s^{2}-4} \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}^{-3 s}+1}{s^{2}-4}\right )\\ &= \frac {\operatorname {Heaviside}\left (t -3\right ) \sinh \left (2 t -6\right )}{2}+\frac {\sinh \left (2 t \right )}{2} \end {align*}
Converting the above solution to piecewise it becomes \[
y = \left \{\begin {array}{cc} \frac {\sinh \left (2 t \right )}{2} & t <3 \\ \frac {\sinh \left (2 t \right )}{2}+\frac {\sinh \left (2 t -6\right )}{2} & 3\le t \end {array}\right .
\] Simplifying the solution gives \[
y = \frac {\sinh \left (2 t \right )}{2}+\left (\left \{\begin {array}{cc} 0 & t <3 \\ \frac {\sinh \left (2 t -6\right )}{2} & 3\le t \end {array}\right .\right )
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {\sinh \left (2 t \right )}{2}+\left (\left \{\begin {array}{cc} 0 & t <3 \\ \frac {\sinh \left (2 t -6\right )}{2} & 3\le t \end {array}\right .\right ) \\
\end{align*} Verification of solutions
\[
y = \frac {\sinh \left (2 t \right )}{2}+\left (\left \{\begin {array}{cc} 0 & t <3 \\ \frac {\sinh \left (2 t -6\right )}{2} & 3\le t \end {array}\right .\right )
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-4 y=\mathit {Dirac}\left (t -3\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}-4=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -2\right ) \left (r +2\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-2, 2\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-2 t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{2 t} \\ \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}+c_{2} {\mathrm e}^{2 t}+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 -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} {\mathrm e}^{-2 t} & {\mathrm e}^{2 t} \\ -2 \,{\mathrm e}^{-2 t} & 2 \,{\mathrm e}^{2 t} \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 {\left (\int \mathit {Dirac}\left (t -3\right )d t \right ) \left ({\mathrm e}^{-2 t +6}-{\mathrm e}^{2 t -6}\right )}{4} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\frac {\mathit {Heaviside}\left (t -3\right ) \left ({\mathrm e}^{-2 t +6}-{\mathrm e}^{2 t -6}\right )}{4} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{2 t}-\frac {\mathit {Heaviside}\left (t -3\right ) \left ({\mathrm e}^{-2 t +6}-{\mathrm e}^{2 t -6}\right )}{4} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{2 t}-\frac {\mathit {Heaviside}\left (t -3\right ) \left ({\mathrm e}^{-2 t +6}-{\mathrm e}^{2 t -6}\right )}{4} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} +c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-2 c_{1} {\mathrm e}^{-2 t}+2 c_{2} {\mathrm e}^{2 t}-\frac {\mathit {Dirac}\left (t -3\right ) \left ({\mathrm e}^{-2 t +6}-{\mathrm e}^{2 t -6}\right )}{4}-\frac {\mathit {Heaviside}\left (t -3\right ) \left (-2 \,{\mathrm e}^{-2 t +6}-2 \,{\mathrm e}^{2 t -6}\right )}{4} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=-2 c_{1} +2 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\frac {1}{4}, c_{2} =\frac {1}{4}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {{\mathrm e}^{-2 t}}{4}+\frac {{\mathrm e}^{2 t}}{4}-\frac {\mathit {Heaviside}\left (t -3\right ) {\mathrm e}^{-2 t +6}}{4}+\frac {\mathit {Heaviside}\left (t -3\right ) {\mathrm e}^{2 t -6}}{4} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {{\mathrm e}^{-2 t}}{4}+\frac {{\mathrm e}^{2 t}}{4}-\frac {\mathit {Heaviside}\left (t -3\right ) {\mathrm e}^{-2 t +6}}{4}+\frac {\mathit {Heaviside}\left (t -3\right ) {\mathrm e}^{2 t -6}}{4} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 2.594 (sec). Leaf size: 23
\[
y \left (t \right ) = \frac {\operatorname {Heaviside}\left (-3+t \right ) \sinh \left (2 t -6\right )}{2}+\frac {\sinh \left (2 t \right )}{2}
\]
✓ Solution by Mathematica
Time used: 0.032 (sec). Leaf size: 44
\[
y(t)\to \frac {1}{4} e^{-2 (t+3)} \left (\left (e^{4 t}-e^{12}\right ) \theta (t-3)+e^6 \left (e^{4 t}-1\right )\right )
\]
15.6.2 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*y(t)=Dirac(t-3),y(0) = 0, D(y)(0) = 1],y(t), singsol=all)
DSolve[{y''[t]-4*y[t]==DiracDelta[t-3],{y[0]==0,y'[0]==1}},y[t],t,IncludeSingularSolutions -> True]