Internal problem ID [13230]
Internal file name [OUTPUT/11886_Tuesday_December_05_2023_12_12_45_PM_51119002/index.tex]
Book: DIFFERENTIAL EQUATIONS by Paul Blanchard, Robert L. Devaney, Glen R. Hall. 4th
edition. Brooks/Cole. Boston, USA. 2012
Section: Chapter 6. Laplace transform. Section 6.4. page 608
Problem number: 2.
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]]
\[ \boxed {y^{\prime \prime }+3 y=5 \delta \left (-2+t \right )} \] 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) &=3\\ F &=5 \delta \left (-2+t \right ) \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+3 y = 5 \delta \left (-2+t \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 )+3 Y \left (s \right ) = 5 \,{\mathrm e}^{-2 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 )+3 Y \left (s \right ) = 5 \,{\mathrm e}^{-2 s} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {5 \,{\mathrm e}^{-2 s}}{s^{2}+3} \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}^{-2 s}}{s^{2}+3}\right )\\ &= \frac {5 \operatorname {Heaviside}\left (-2+t \right ) \sin \left (\sqrt {3}\, \left (-2+t \right )\right ) \sqrt {3}}{3} \end {align*}
Converting the above solution to piecewise it becomes \[
y = \left \{\begin {array}{cc} 0 & t <2 \\ \frac {5 \sin \left (\sqrt {3}\, \left (-2+t \right )\right ) \sqrt {3}}{3} & 2\le t \end {array}\right .
\] Simplifying the solution gives \[
y = \left \{\begin {array}{cc} 0 & t <2 \\ \frac {5 \sin \left (\sqrt {3}\, \left (-2+t \right )\right ) \sqrt {3}}{3} & 2\le t \end {array}\right .
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= \left \{\begin {array}{cc} 0 & t <2 \\ \frac {5 \sin \left (\sqrt {3}\, \left (-2+t \right )\right ) \sqrt {3}}{3} & 2\le t \end {array}\right . \\
\end{align*} Verification of solutions
\[
y = \left \{\begin {array}{cc} 0 & t <2 \\ \frac {5 \sin \left (\sqrt {3}\, \left (-2+t \right )\right ) \sqrt {3}}{3} & 2\le t \end {array}\right .
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+3 y=5 \mathit {Dirac}\left (-2+t \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}+3=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {-12}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (\mathrm {-I} \sqrt {3}, \mathrm {I} \sqrt {3}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )=\cos \left (\sqrt {3}\, t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=\sin \left (\sqrt {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} \cos \left (\sqrt {3}\, t \right )+c_{2} \sin \left (\sqrt {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 )=5 \mathit {Dirac}\left (-2+t \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 (\sqrt {3}\, t \right ) & \sin \left (\sqrt {3}\, t \right ) \\ -\sqrt {3}\, \sin \left (\sqrt {3}\, t \right ) & \sqrt {3}\, \cos \left (\sqrt {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 )=\sqrt {3} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=\frac {5 \sqrt {3}\, \left (\int \mathit {Dirac}\left (-2+t \right )d t \right ) \left (\sin \left (\sqrt {3}\, t \right ) \cos \left (2 \sqrt {3}\right )-\cos \left (\sqrt {3}\, t \right ) \sin \left (2 \sqrt {3}\right )\right )}{3} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\frac {5 \sqrt {3}\, \mathit {Heaviside}\left (-2+t \right ) \left (\sin \left (\sqrt {3}\, t \right ) \cos \left (2 \sqrt {3}\right )-\cos \left (\sqrt {3}\, t \right ) \sin \left (2 \sqrt {3}\right )\right )}{3} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} \cos \left (\sqrt {3}\, t \right )+c_{2} \sin \left (\sqrt {3}\, t \right )+\frac {5 \sqrt {3}\, \mathit {Heaviside}\left (-2+t \right ) \left (\sin \left (\sqrt {3}\, t \right ) \cos \left (2 \sqrt {3}\right )-\cos \left (\sqrt {3}\, t \right ) \sin \left (2 \sqrt {3}\right )\right )}{3} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} \cos \left (\sqrt {3}\, t \right )+c_{2} \sin \left (\sqrt {3}\, t \right )+\frac {5 \sqrt {3}\, \mathit {Heaviside}\left (-2+t \right ) \left (\sin \left (\sqrt {3}\, t \right ) \cos \left (2 \sqrt {3}\right )-\cos \left (\sqrt {3}\, t \right ) \sin \left (2 \sqrt {3}\right )\right )}{3} \\ {} & \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 }=-c_{1} \sqrt {3}\, \sin \left (\sqrt {3}\, t \right )+c_{2} \sqrt {3}\, \cos \left (\sqrt {3}\, t \right )+\frac {5 \sqrt {3}\, \mathit {Dirac}\left (-2+t \right ) \left (\sin \left (\sqrt {3}\, t \right ) \cos \left (2 \sqrt {3}\right )-\cos \left (\sqrt {3}\, t \right ) \sin \left (2 \sqrt {3}\right )\right )}{3}+\frac {5 \sqrt {3}\, \mathit {Heaviside}\left (-2+t \right ) \left (\sqrt {3}\, \cos \left (\sqrt {3}\, t \right ) \cos \left (2 \sqrt {3}\right )+\sqrt {3}\, \sin \left (\sqrt {3}\, t \right ) \sin \left (2 \sqrt {3}\right )\right )}{3} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=c_{2} \sqrt {3} \\ {} & \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 {5 \sqrt {3}\, \mathit {Heaviside}\left (-2+t \right ) \left (\sin \left (\sqrt {3}\, t \right ) \cos \left (2 \sqrt {3}\right )-\cos \left (\sqrt {3}\, t \right ) \sin \left (2 \sqrt {3}\right )\right )}{3} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {5 \sqrt {3}\, \mathit {Heaviside}\left (-2+t \right ) \left (\sin \left (\sqrt {3}\, t \right ) \cos \left (2 \sqrt {3}\right )-\cos \left (\sqrt {3}\, t \right ) \sin \left (2 \sqrt {3}\right )\right )}{3} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 15.422 (sec). Leaf size: 21
\[
y \left (t \right ) = \frac {5 \sqrt {3}\, \operatorname {Heaviside}\left (t -2\right ) \sin \left (\sqrt {3}\, \left (t -2\right )\right )}{3}
\]
✓ Solution by Mathematica
Time used: 0.288 (sec). Leaf size: 36
\[
y(t)\to \frac {\theta (t-2) \sin \left (\sqrt {3} (t-2)\right )}{\sqrt {3}}+2 \cos \left (\sqrt {3} t\right )
\]
20.1.2 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)+3*y(t)=5*Dirac(t-2),y(0) = 0, D(y)(0) = 0],y(t), singsol=all)
DSolve[{y''[t]+3*y[t]==DiracDelta[t-2],{y[0]==2,y'[0]==0}},y[t],t,IncludeSingularSolutions -> True]