Internal problem ID [13858]
Internal file name [OUTPUT/13030_Friday_February_23_2024_06_46_46_AM_12123920/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 27. Differentiation and the Laplace transform. Additional Exercises. page
496
Problem number: 27.1 (k).
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]]
\[ \boxed {y^{\prime \prime }-4 y^{\prime }+13 y={\mathrm e}^{2 t} \sin \left (3 t \right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 4, y^{\prime }\left (0\right ) = 3] \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) &=13\\ F &={\mathrm e}^{2 t} \sin \left (3 t \right ) \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }-4 y^{\prime }+13 y = {\mathrm e}^{2 t} \sin \left (3 t \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 )+13 Y \left (s \right ) = \frac {3}{\left (s -2\right )^{2}+9}\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=4\\ y'(0) &=3 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )+13-4 s -4 s Y \left (s \right )+13 Y \left (s \right ) = \frac {3}{\left (s -2\right )^{2}+9} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {4 s^{3}-29 s^{2}+104 s -166}{\left (s^{2}-4 s +13\right )^{2}} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= -\frac {1}{12 \left (s -2-3 i\right )^{2}}-\frac {1}{12 \left (s -2+3 i\right )^{2}}+\frac {2+\frac {29 i}{36}}{s -2-3 i}+\frac {2-\frac {29 i}{36}}{s -2+3 i} \] The inverse Laplace of each term above
is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {1}{12 \left (s -2-3 i\right )^{2}}\right ) &= -\frac {t \,{\mathrm e}^{\left (2+3 i\right ) t}}{12}\\ \mathcal {L}^{-1}\left (-\frac {1}{12 \left (s -2+3 i\right )^{2}}\right ) &= -\frac {t \,{\mathrm e}^{\left (2-3 i\right ) t}}{12}\\ \mathcal {L}^{-1}\left (\frac {2+\frac {29 i}{36}}{s -2-3 i}\right ) &= \left (2+\frac {29 i}{36}\right ) {\mathrm e}^{\left (2+3 i\right ) t}\\ \mathcal {L}^{-1}\left (\frac {2-\frac {29 i}{36}}{s -2+3 i}\right ) &= \left (2-\frac {29 i}{36}\right ) {\mathrm e}^{\left (2-3 i\right ) t} \end {align*}
Adding the above results and simplifying gives \[ y=-\frac {{\mathrm e}^{2 t} \left (29 \sin \left (3 t \right )+3 \cos \left (3 t \right ) \left (-24+t \right )\right )}{18} \] Simplifying the solution gives \[
y = -\frac {{\mathrm e}^{2 t} \left (-24+t \right ) \cos \left (3 t \right )}{6}-\frac {29 \,{\mathrm e}^{2 t} \sin \left (3 t \right )}{18}
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= -\frac {{\mathrm e}^{2 t} \left (-24+t \right ) \cos \left (3 t \right )}{6}-\frac {29 \,{\mathrm e}^{2 t} \sin \left (3 t \right )}{18} \\
\end{align*} Verification of solutions
\[
y = -\frac {{\mathrm e}^{2 t} \left (-24+t \right ) \cos \left (3 t \right )}{6}-\frac {29 \,{\mathrm e}^{2 t} \sin \left (3 t \right )}{18}
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }-4 y^{\prime }+13 y={\mathrm e}^{2 t} \sin \left (3 t \right ), y \left (0\right )=4, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=3\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}-4 r +13=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {4\pm \left (\sqrt {-36}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (2-3 \,\mathrm {I}, 2+3 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{2 t} \cos \left (3 t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{2 t} \sin \left (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} {\mathrm e}^{2 t} \cos \left (3 t \right )+c_{2} {\mathrm e}^{2 t} \sin \left (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 )={\mathrm e}^{2 t} \sin \left (3 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} {\mathrm e}^{2 t} \cos \left (3 t \right ) & {\mathrm e}^{2 t} \sin \left (3 t \right ) \\ 2 \,{\mathrm e}^{2 t} \cos \left (3 t \right )-3 \,{\mathrm e}^{2 t} \sin \left (3 t \right ) & 2 \,{\mathrm e}^{2 t} \sin \left (3 t \right )+3 \,{\mathrm e}^{2 t} \cos \left (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 )=3 \,{\mathrm e}^{4 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=\frac {{\mathrm e}^{2 t} \left (\sin \left (3 t \right ) \left (\int \sin \left (6 t \right )d t \right )-2 \cos \left (3 t \right ) \left (\int \sin \left (3 t \right )^{2}d t \right )\right )}{6} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\frac {\left (\sin \left (3 t \right )-6 \cos \left (3 t \right ) t \right ) {\mathrm e}^{2 t}}{36} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{2 t} \cos \left (3 t \right )+c_{2} {\mathrm e}^{2 t} \sin \left (3 t \right )+\frac {\left (\sin \left (3 t \right )-6 \cos \left (3 t \right ) t \right ) {\mathrm e}^{2 t}}{36} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{2 t} \cos \left (3 t \right )+c_{2} {\mathrm e}^{2 t} \sin \left (3 t \right )+\frac {\left (\sin \left (3 t \right )-6 \cos \left (3 t \right ) t \right ) {\mathrm e}^{2 t}}{36} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=4 \\ {} & {} & 4=c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=2 c_{1} {\mathrm e}^{2 t} \cos \left (3 t \right )-3 c_{1} {\mathrm e}^{2 t} \sin \left (3 t \right )+2 c_{2} {\mathrm e}^{2 t} \sin \left (3 t \right )+3 c_{2} {\mathrm e}^{2 t} \cos \left (3 t \right )+\frac {\left (-3 \cos \left (3 t \right )+18 \sin \left (3 t \right ) t \right ) {\mathrm e}^{2 t}}{36}+\frac {\left (\sin \left (3 t \right )-6 \cos \left (3 t \right ) t \right ) {\mathrm e}^{2 t}}{18} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=3 \\ {} & {} & 3=2 c_{1} -\frac {1}{12}+3 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =4, c_{2} =-\frac {59}{36}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {{\mathrm e}^{2 t} \left (-24+t \right ) \cos \left (3 t \right )}{6}-\frac {29 \,{\mathrm e}^{2 t} \sin \left (3 t \right )}{18} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {{\mathrm e}^{2 t} \left (-24+t \right ) \cos \left (3 t \right )}{6}-\frac {29 \,{\mathrm e}^{2 t} \sin \left (3 t \right )}{18} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 5.0 (sec). Leaf size: 26
\[
y \left (t \right ) = -\frac {\left (-24+t \right ) {\mathrm e}^{2 t} \cos \left (3 t \right )}{6}-\frac {29 \,{\mathrm e}^{2 t} \sin \left (3 t \right )}{18}
\]
✓ Solution by Mathematica
Time used: 0.139 (sec). Leaf size: 30
\[
y(t)\to -\frac {1}{18} e^{2 t} (29 \sin (3 t)+3 (t-24) \cos (3 t))
\]
18.11.2 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)+13*y(t)=exp(2*t)*sin(3*t),y(0) = 4, D(y)(0) = 3],y(t), singsol=all)
DSolve[{y''[t]-4*y'[t]+13*y[t]==Exp[2*t]*Sin[3*t],{y[0]==4,y'[0]==3}},y[t],t,IncludeSingularSolutions -> True]