Internal problem ID [13225]
Internal file name [OUTPUT/11881_Tuesday_December_05_2023_12_12_41_PM_10656506/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 6.3 page 600
Problem number: 30.
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]]
\[ \boxed {y^{\prime \prime }+6 y^{\prime }+13 y=13 \operatorname {Heaviside}\left (t -4\right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 3, 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) &=6\\ q(t) &=13\\ F &=13 \operatorname {Heaviside}\left (t -4\right ) \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+6 y^{\prime }+13 y = 13 \operatorname {Heaviside}\left (t -4\right ) \end {align*}
The domain of \(p(t)=6\) 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 )+6 s Y \left (s \right )-6 y \left (0\right )+13 Y \left (s \right ) = \frac {13 \,{\mathrm e}^{-4 s}}{s}\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=3\\ y'(0) &=1 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-19-3 s +6 s Y \left (s \right )+13 Y \left (s \right ) = \frac {13 \,{\mathrm e}^{-4 s}}{s} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {3 s^{2}+13 \,{\mathrm e}^{-4 s}+19 s}{s \left (s^{2}+6 s +13\right )} \end {align*}
Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (\frac {3 s^{2}+13 \,{\mathrm e}^{-4 s}+19 s}{s \left (s^{2}+6 s +13\right )}\right )\\ &= {\mathrm e}^{-3 t} \left (3 \cos \left (2 t \right )+5 \sin \left (2 t \right )\right )+\left (\frac {1}{26}+\frac {3 i}{52}\right ) \left (8-12 i-13 \,{\mathrm e}^{\left (-3-2 i\right ) \left (t -4\right )}+\left (5+12 i\right ) {\mathrm e}^{\left (-3+2 i\right ) \left (t -4\right )}\right ) \operatorname {Heaviside}\left (t -4\right ) \end {align*}
Converting the above solution to piecewise it becomes \[
y = \left \{\begin {array}{cc} {\mathrm e}^{-3 t} \left (3 \cos \left (2 t \right )+5 \sin \left (2 t \right )\right ) & t <4 \\ {\mathrm e}^{-3 t} \left (3 \cos \left (2 t \right )+5 \sin \left (2 t \right )\right )+\left (\frac {1}{26}+\frac {3 i}{52}\right ) \left (8-12 i-13 \,{\mathrm e}^{\left (-3-2 i\right ) \left (t -4\right )}+\left (5+12 i\right ) {\mathrm e}^{\left (-3+2 i\right ) \left (t -4\right )}\right ) & 4\le t \end {array}\right .
\] Simplifying the solution gives \[
y = {\mathrm e}^{-3 t} \left (3 \cos \left (2 t \right )+5 \sin \left (2 t \right )\right )-\left (\left \{\begin {array}{cc} 0 & t <4 \\ -1+\left (\frac {1}{2}+\frac {3 i}{4}\right ) {\mathrm e}^{\left (-3-2 i\right ) \left (t -4\right )}+\left (\frac {1}{2}-\frac {3 i}{4}\right ) {\mathrm e}^{\left (-3+2 i\right ) \left (t -4\right )} & 4\le t \end {array}\right .\right )
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= {\mathrm e}^{-3 t} \left (3 \cos \left (2 t \right )+5 \sin \left (2 t \right )\right )-\left (\left \{\begin {array}{cc} 0 & t <4 \\ -1+\left (\frac {1}{2}+\frac {3 i}{4}\right ) {\mathrm e}^{\left (-3-2 i\right ) \left (t -4\right )}+\left (\frac {1}{2}-\frac {3 i}{4}\right ) {\mathrm e}^{\left (-3+2 i\right ) \left (t -4\right )} & 4\le t \end {array}\right .\right ) \\
\end{align*} Verification of solutions
\[
y = {\mathrm e}^{-3 t} \left (3 \cos \left (2 t \right )+5 \sin \left (2 t \right )\right )-\left (\left \{\begin {array}{cc} 0 & t <4 \\ -1+\left (\frac {1}{2}+\frac {3 i}{4}\right ) {\mathrm e}^{\left (-3-2 i\right ) \left (t -4\right )}+\left (\frac {1}{2}-\frac {3 i}{4}\right ) {\mathrm e}^{\left (-3+2 i\right ) \left (t -4\right )} & 4\le t \end {array}\right .\right )
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+6 y^{\prime }+13 y=13 \mathit {Heaviside}\left (t -4\right ), y \left (0\right )=3, 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}+6 r +13=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-6\right )\pm \left (\sqrt {-16}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-3-2 \,\mathrm {I}, -3+2 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-3 t} \cos \left (2 t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-3 t} \sin \left (2 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}^{-3 t} \cos \left (2 t \right )+c_{2} {\mathrm e}^{-3 t} \sin \left (2 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 )=13 \mathit {Heaviside}\left (t -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}^{-3 t} \cos \left (2 t \right ) & {\mathrm e}^{-3 t} \sin \left (2 t \right ) \\ -3 \,{\mathrm e}^{-3 t} \cos \left (2 t \right )-2 \,{\mathrm e}^{-3 t} \sin \left (2 t \right ) & -3 \,{\mathrm e}^{-3 t} \sin \left (2 t \right )+2 \,{\mathrm e}^{-3 t} \cos \left (2 t \right ) \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=2 \,{\mathrm e}^{-6 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\frac {13 \,{\mathrm e}^{-3 t} \left (\cos \left (2 t \right ) \left (\int \sin \left (2 t \right ) \mathit {Heaviside}\left (t -4\right ) {\mathrm e}^{3 t}d t \right )-\sin \left (2 t \right ) \left (\int \cos \left (2 t \right ) \mathit {Heaviside}\left (t -4\right ) {\mathrm e}^{3 t}d t \right )\right )}{2} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\mathit {Heaviside}\left (t -4\right ) \left (-1+\left (\left (\cos \left (8\right )-\frac {3 \sin \left (8\right )}{2}\right ) \cos \left (2 t \right )+\frac {3 \sin \left (2 t \right ) \left (\cos \left (8\right )+\frac {2 \sin \left (8\right )}{3}\right )}{2}\right ) {\mathrm e}^{-3 t +12}\right ) \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-3 t} \cos \left (2 t \right )+c_{2} {\mathrm e}^{-3 t} \sin \left (2 t \right )-\mathit {Heaviside}\left (t -4\right ) \left (-1+\left (\left (\cos \left (8\right )-\frac {3 \sin \left (8\right )}{2}\right ) \cos \left (2 t \right )+\frac {3 \sin \left (2 t \right ) \left (\cos \left (8\right )+\frac {2 \sin \left (8\right )}{3}\right )}{2}\right ) {\mathrm e}^{-3 t +12}\right ) \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-3 t} \cos \left (2 t \right )+c_{2} {\mathrm e}^{-3 t} \sin \left (2 t \right )-\mathit {Heaviside}\left (t -4\right ) \left (-1+\left (\left (\cos \left (8\right )-\frac {3 \sin \left (8\right )}{2}\right ) \cos \left (2 t \right )+\frac {3 \sin \left (2 t \right ) \left (\cos \left (8\right )+\frac {2 \sin \left (8\right )}{3}\right )}{2}\right ) {\mathrm e}^{-3 t +12}\right ) \\ {} & \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 }=-3 c_{1} {\mathrm e}^{-3 t} \cos \left (2 t \right )-2 c_{1} {\mathrm e}^{-3 t} \sin \left (2 t \right )-3 c_{2} {\mathrm e}^{-3 t} \sin \left (2 t \right )+2 c_{2} {\mathrm e}^{-3 t} \cos \left (2 t \right )-\mathit {Dirac}\left (t -4\right ) \left (-1+\left (\left (\cos \left (8\right )-\frac {3 \sin \left (8\right )}{2}\right ) \cos \left (2 t \right )+\frac {3 \sin \left (2 t \right ) \left (\cos \left (8\right )+\frac {2 \sin \left (8\right )}{3}\right )}{2}\right ) {\mathrm e}^{-3 t +12}\right )-\mathit {Heaviside}\left (t -4\right ) \left (\left (-2 \left (\cos \left (8\right )-\frac {3 \sin \left (8\right )}{2}\right ) \sin \left (2 t \right )+3 \cos \left (2 t \right ) \left (\cos \left (8\right )+\frac {2 \sin \left (8\right )}{3}\right )\right ) {\mathrm e}^{-3 t +12}-3 \left (\left (\cos \left (8\right )-\frac {3 \sin \left (8\right )}{2}\right ) \cos \left (2 t \right )+\frac {3 \sin \left (2 t \right ) \left (\cos \left (8\right )+\frac {2 \sin \left (8\right )}{3}\right )}{2}\right ) {\mathrm e}^{-3 t +12}\right ) \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=-3 c_{1} +2 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =3, c_{2} =5\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\mathit {Heaviside}\left (t -4\right ) \left (\left (\cos \left (8\right )-\frac {3 \sin \left (8\right )}{2}\right ) \cos \left (2 t \right )+\frac {3 \sin \left (2 t \right ) \left (\cos \left (8\right )+\frac {2 \sin \left (8\right )}{3}\right )}{2}\right ) {\mathrm e}^{-3 t +12}+3 \,{\mathrm e}^{-3 t} \cos \left (2 t \right )+5 \,{\mathrm e}^{-3 t} \sin \left (2 t \right )+\mathit {Heaviside}\left (t -4\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\mathit {Heaviside}\left (t -4\right ) \left (\left (\cos \left (8\right )-\frac {3 \sin \left (8\right )}{2}\right ) \cos \left (2 t \right )+\frac {3 \sin \left (2 t \right ) \left (\cos \left (8\right )+\frac {2 \sin \left (8\right )}{3}\right )}{2}\right ) {\mathrm e}^{-3 t +12}+3 \,{\mathrm e}^{-3 t} \cos \left (2 t \right )+5 \,{\mathrm e}^{-3 t} \sin \left (2 t \right )+\mathit {Heaviside}\left (t -4\right ) \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 7.156 (sec). Leaf size: 57
\[
y \left (t \right ) = \left (-\frac {1}{2}-\frac {3 i}{4}\right ) \operatorname {Heaviside}\left (t -4\right ) {\mathrm e}^{\left (-3-2 i\right ) \left (t -4\right )}+\left (-\frac {1}{2}+\frac {3 i}{4}\right ) \operatorname {Heaviside}\left (t -4\right ) {\mathrm e}^{\left (-3+2 i\right ) \left (t -4\right )}+\operatorname {Heaviside}\left (t -4\right )+{\mathrm e}^{-3 t} \left (3 \cos \left (2 t \right )+5 \sin \left (2 t \right )\right )
\]
✓ Solution by Mathematica
Time used: 0.057 (sec). Leaf size: 82
\[
y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} e^{2 t} (3 \cos (t)-5 \sin (t)) & t\leq 4 \\ -\frac {1}{5} e^{2 t-8} \cos (4-t)+3 e^{2 t} \cos (t)-\frac {2}{5} e^{2 t-8} \sin (4-t)-5 e^{2 t} \sin (t)+\frac {1}{5} & \text {True} \\ \end {array} \\ \end {array}
\]
19.4.2 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)+6*diff(y(t),t)+13*y(t)=13*Heaviside(t-4),y(0) = 3, D(y)(0) = 1],y(t), singsol=all)
DSolve[{y''[t]-4*y'[t]+5*y[t]==UnitStep[t-4],{y[0]==3,y'[0]==1}},y[t],t,IncludeSingularSolutions -> True]