Internal problem ID [13902]
Internal file name [OUTPUT/13074_Friday_February_23_2024_06_54_32_AM_85574514/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 31. Delta Functions. Additional Exercises. page 572
Problem number: 31.7 (e).
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]]
\[ \boxed {y^{\prime \prime }-16 y=\delta \left (t -10\right )} \] 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) &=0\\ q(t) &=-16\\ F &=\delta \left (t -10\right ) \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }-16 y = \delta \left (t -10\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 ) = {\mathrm e}^{-10 s}\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 ) = {\mathrm e}^{-10 s} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {{\mathrm e}^{-10 s}}{s^{2}-16} \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}^{-10 s}}{s^{2}-16}\right )\\ &= \frac {\operatorname {Heaviside}\left (t -10\right ) \sinh \left (4 t -40\right )}{4} \end {align*}
Converting the above solution to piecewise it becomes \[
y = \left \{\begin {array}{cc} 0 & t <10 \\ \frac {\sinh \left (4 t -40\right )}{4} & 10\le t \end {array}\right .
\] Simplifying the solution gives \[
y = \left \{\begin {array}{cc} 0 & t <10 \\ \frac {\sinh \left (4 t -40\right )}{4} & 10\le t \end {array}\right .
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= \left \{\begin {array}{cc} 0 & t <10 \\ \frac {\sinh \left (4 t -40\right )}{4} & 10\le t \end {array}\right . \\
\end{align*} Verification of solutions
\[
y = \left \{\begin {array}{cc} 0 & t <10 \\ \frac {\sinh \left (4 t -40\right )}{4} & 10\le t \end {array}\right .
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-16 y=\mathit {Dirac}\left (t -10\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}-16=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -4\right ) \left (r +4\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-4, 4\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-4 t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{4 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}^{-4 t}+c_{2} {\mathrm e}^{4 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 -10\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}^{-4 t} & {\mathrm e}^{4 t} \\ -4 \,{\mathrm e}^{-4 t} & 4 \,{\mathrm e}^{4 t} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=8 \\ {} & \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 -10\right )d t \right ) \left ({\mathrm e}^{-4 t +40}-{\mathrm e}^{4 t -40}\right )}{8} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\frac {\mathit {Heaviside}\left (t -10\right ) \left ({\mathrm e}^{-4 t +40}-{\mathrm e}^{4 t -40}\right )}{8} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-4 t}+c_{2} {\mathrm e}^{4 t}-\frac {\mathit {Heaviside}\left (t -10\right ) \left ({\mathrm e}^{-4 t +40}-{\mathrm e}^{4 t -40}\right )}{8} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-4 t}+c_{2} {\mathrm e}^{4 t}-\frac {\mathit {Heaviside}\left (t -10\right ) \left ({\mathrm e}^{-4 t +40}-{\mathrm e}^{4 t -40}\right )}{8} \\ {} & \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 }=-4 c_{1} {\mathrm e}^{-4 t}+4 c_{2} {\mathrm e}^{4 t}-\frac {\mathit {Dirac}\left (t -10\right ) \left ({\mathrm e}^{-4 t +40}-{\mathrm e}^{4 t -40}\right )}{8}-\frac {\mathit {Heaviside}\left (t -10\right ) \left (-4 \,{\mathrm e}^{-4 t +40}-4 \,{\mathrm e}^{4 t -40}\right )}{8} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-4 c_{1} +4 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=-\frac {\mathit {Heaviside}\left (t -10\right ) \left ({\mathrm e}^{-4 t +40}-{\mathrm e}^{4 t -40}\right )}{8} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {\mathit {Heaviside}\left (t -10\right ) \left ({\mathrm e}^{-4 t +40}-{\mathrm e}^{4 t -40}\right )}{8} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 5.656 (sec). Leaf size: 16
\[
y \left (t \right ) = \frac {\operatorname {Heaviside}\left (t -10\right ) \sinh \left (-40+4 t \right )}{4}
\]
✓ Solution by Mathematica
Time used: 0.028 (sec). Leaf size: 31
\[
y(t)\to \frac {1}{8} e^{-4 (t+10)} \left (e^{8 t}-e^{80}\right ) \theta (t-10)
\]
22.12.2 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)-16*y(t)=Dirac(t-10),y(0) = 0, D(y)(0) = 0],y(t), singsol=all)
DSolve[{y''[t]-16*y[t]==DiracDelta[t-10],{y[0]==0,y'[0]==0}},y[t],t,IncludeSingularSolutions -> True]