Internal problem ID [5692]
Internal file name [OUTPUT/4940_Sunday_June_05_2022_03_11_07_PM_29162432/index.tex]
Book: ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section: Chapter 6. Laplace Transforms. Problem set 6.2, page 216
Problem number: 14.
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, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime }+2 y^{\prime }+5 y=50 t -100} \] With initial conditions \begin {align*} [y \left (2\right ) = -4, y^{\prime }\left (2\right ) = 14] \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) &=2\\ q(t) &=5\\ F &=50 t -100 \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+2 y^{\prime }+5 y = 50 t -100 \end {align*}
The domain of \(p(t)=2\) is \[
\{-\infty Since both initial conditions are not at zero, then let \begin {align*} y(0) &= c_{1}\\ y'(0) &= c_{2} \end {align*}
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 )+2 s Y \left (s \right )-2 y \left (0\right )+5 Y \left (s \right ) = \frac {50}{s^{2}}-\frac {100}{s}\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=c_{1}\\ y'(0) &=c_{2} \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-c_{2} -s c_{1} +2 s Y \left (s \right )-2 c_{1} +5 Y \left (s \right ) = \frac {50}{s^{2}}-\frac {100}{s} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {c_{1} s^{3}+2 c_{1} s^{2}+c_{2} s^{2}-100 s +50}{s^{2} \left (s^{2}+2 s +5\right )} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= \frac {\left (-1+2 i\right ) \left (-\frac {c_{1}}{8}-\frac {c_{2}}{8}-\frac {7}{4}\right )+\frac {3 c_{1}}{8}-\frac {c_{2}}{8}+\frac {41}{4}}{s +1-2 i}+\frac {\left (-1-2 i\right ) \left (-\frac {c_{1}}{8}-\frac {c_{2}}{8}-\frac {7}{4}\right )+\frac {3 c_{1}}{8}-\frac {c_{2}}{8}+\frac {41}{4}}{s +1+2 i}-\frac {24}{s}+\frac {10}{s^{2}} \] The inverse Laplace of each term above
is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {\left (-1+2 i\right ) \left (-\frac {c_{1}}{8}-\frac {c_{2}}{8}-\frac {7}{4}\right )+\frac {3 c_{1}}{8}-\frac {c_{2}}{8}+\frac {41}{4}}{s +1-2 i}\right ) &= \frac {{\mathrm e}^{\left (-1+2 i\right ) t} \left (48-14 i-i c_{2} +\left (2-i\right ) c_{1} \right )}{4}\\ \mathcal {L}^{-1}\left (\frac {\left (-1-2 i\right ) \left (-\frac {c_{1}}{8}-\frac {c_{2}}{8}-\frac {7}{4}\right )+\frac {3 c_{1}}{8}-\frac {c_{2}}{8}+\frac {41}{4}}{s +1+2 i}\right ) &= \frac {{\mathrm e}^{\left (-1-2 i\right ) t} \left (48+14 i+i c_{2} +\left (2+i\right ) c_{1} \right )}{4}\\ \mathcal {L}^{-1}\left (-\frac {24}{s}\right ) &= -24\\ \mathcal {L}^{-1}\left (\frac {10}{s^{2}}\right ) &= 10 t \end {align*}
Adding the above results and simplifying gives \[ y=-24+10 t +\frac {\left (2 \cos \left (2 t \right ) \left (c_{1} +24\right )+\sin \left (2 t \right ) \left (c_{1} +c_{2} +14\right )\right ) {\mathrm e}^{-t}}{2} \] Since both initial conditions given are not at
zero, then we need to setup two equations to solve for \(c_{1},c_{1}\). At \(t=2\) the first equation becomes, using
the above solution \begin {align*} -4 &= -4+\frac {\left (2 \cos \left (4\right ) \left (c_{1} +24\right )+\sin \left (4\right ) \left (c_{1} +c_{2} +14\right )\right ) {\mathrm e}^{-2}}{2} \end {align*}
And taking derivative of the solution and evaluating at \(t=2\) gives the second equation as
\begin {align*} 14 &= 10+\frac {\left (-4 \sin \left (4\right ) \left (c_{1} +24\right )+2 \cos \left (4\right ) \left (c_{1} +c_{2} +14\right )\right ) {\mathrm e}^{-2}}{2}-\frac {\left (2 \cos \left (4\right ) \left (c_{1} +24\right )+\sin \left (4\right ) \left (c_{1} +c_{2} +14\right )\right ) {\mathrm e}^{-2}}{2} \end {align*}
Solving gives \begin {align*} c_{1} &= -\frac {2 \,{\mathrm e}^{2} \left (12 \cos \left (4\right )^{2} {\mathrm e}^{-2}+12 \sin \left (4\right )^{2} {\mathrm e}^{-2}+\sin \left (4\right )\right )}{\cos \left (4\right )^{2}+\sin \left (4\right )^{2}}\\ c_{2} &= \frac {2 \left (5 \cos \left (4\right )^{2} {\mathrm e}^{-2}+5 \sin \left (4\right )^{2} {\mathrm e}^{-2}+2 \cos \left (4\right )+\sin \left (4\right )\right ) {\mathrm e}^{2}}{\cos \left (4\right )^{2}+\sin \left (4\right )^{2}} \end {align*}
Subtituting these in the solution obtained above gives \begin {align*} y &= -24+10 t +\frac {\left (2 \cos \left (2 t \right ) \left (-\frac {2 \,{\mathrm e}^{2} \left (12 \cos \left (4\right )^{2} {\mathrm e}^{-2}+12 \sin \left (4\right )^{2} {\mathrm e}^{-2}+\sin \left (4\right )\right )}{\cos \left (4\right )^{2}+\sin \left (4\right )^{2}}+24\right )+\sin \left (2 t \right ) \left (-\frac {2 \,{\mathrm e}^{2} \left (12 \cos \left (4\right )^{2} {\mathrm e}^{-2}+12 \sin \left (4\right )^{2} {\mathrm e}^{-2}+\sin \left (4\right )\right )}{\cos \left (4\right )^{2}+\sin \left (4\right )^{2}}+\frac {2 \left (5 \cos \left (4\right )^{2} {\mathrm e}^{-2}+5 \sin \left (4\right )^{2} {\mathrm e}^{-2}+2 \cos \left (4\right )+\sin \left (4\right )\right ) {\mathrm e}^{2}}{\cos \left (4\right )^{2}+\sin \left (4\right )^{2}}+14\right )\right ) {\mathrm e}^{-t}}{2}\\ &= 2 \cos \left (4\right ) {\mathrm e}^{2-t} \sin \left (2 t \right )-2 \,{\mathrm e}^{2-t} \sin \left (4\right ) \cos \left (2 t \right )+10 t -24 \end {align*}
Simplifying the solution gives \[
y = 2 \cos \left (4\right ) {\mathrm e}^{2-t} \sin \left (2 t \right )-2 \,{\mathrm e}^{2-t} \sin \left (4\right ) \cos \left (2 t \right )+10 t -24
\] The solution(s) found are the following \begin{align*}
\tag{1} y &= 2 \cos \left (4\right ) {\mathrm e}^{2-t} \sin \left (2 t \right )-2 \,{\mathrm e}^{2-t} \sin \left (4\right ) \cos \left (2 t \right )+10 t -24 \\
\end{align*} Verification of solutions
\[
y = 2 \cos \left (4\right ) {\mathrm e}^{2-t} \sin \left (2 t \right )-2 \,{\mathrm e}^{2-t} \sin \left (4\right ) \cos \left (2 t \right )+10 t -24
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+2 y^{\prime }+5 y=50 t -100, y \left (2\right )=-4, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}2\right \}}}}=14\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}+2 r +5=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-2\right )\pm \left (\sqrt {-16}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-1-2 \,\mathrm {I}, -1+2 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )=\cos \left (2 t \right ) {\mathrm e}^{-t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=\sin \left (2 t \right ) {\mathrm e}^{-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} \cos \left (2 t \right ) {\mathrm e}^{-t}+c_{2} \sin \left (2 t \right ) {\mathrm e}^{-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 )=50 t -100\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 (2 t \right ) {\mathrm e}^{-t} & \sin \left (2 t \right ) {\mathrm e}^{-t} \\ -2 \sin \left (2 t \right ) {\mathrm e}^{-t}-\cos \left (2 t \right ) {\mathrm e}^{-t} & 2 \cos \left (2 t \right ) {\mathrm e}^{-t}-\sin \left (2 t \right ) {\mathrm e}^{-t} \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}^{-2 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=25 \,{\mathrm e}^{-t} \left (-\cos \left (2 t \right ) \left (\int \left (-2+t \right ) \sin \left (2 t \right ) {\mathrm e}^{t}d t \right )+\sin \left (2 t \right ) \left (\int \left (-2+t \right ) \cos \left (2 t \right ) {\mathrm e}^{t}d t \right )\right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-24+10 t \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} \cos \left (2 t \right ) {\mathrm e}^{-t}+c_{2} \sin \left (2 t \right ) {\mathrm e}^{-t}-24+10 t \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} \cos \left (2 t \right ) {\mathrm e}^{-t}+c_{2} \sin \left (2 t \right ) {\mathrm e}^{-t}-24+10 t \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (2\right )=-4 \\ {} & {} & -4=c_{1} \cos \left (4\right ) {\mathrm e}^{-2}+c_{2} \sin \left (4\right ) {\mathrm e}^{-2}-4 \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-2 c_{1} \sin \left (2 t \right ) {\mathrm e}^{-t}-c_{1} \cos \left (2 t \right ) {\mathrm e}^{-t}+2 c_{2} \cos \left (2 t \right ) {\mathrm e}^{-t}-c_{2} \sin \left (2 t \right ) {\mathrm e}^{-t}+10 \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}2\right \}}}}=14 \\ {} & {} & 14=-2 c_{1} \sin \left (4\right ) {\mathrm e}^{-2}-c_{1} \cos \left (4\right ) {\mathrm e}^{-2}+2 c_{2} \cos \left (4\right ) {\mathrm e}^{-2}-c_{2} \sin \left (4\right ) {\mathrm e}^{-2}+10 \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\frac {2 \sin \left (4\right )}{{\mathrm e}^{-2} \left (\cos \left (4\right )^{2}+\sin \left (4\right )^{2}\right )}, c_{2} =\frac {2 \cos \left (4\right )}{{\mathrm e}^{-2} \left (\cos \left (4\right )^{2}+\sin \left (4\right )^{2}\right )}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=2 \cos \left (4\right ) {\mathrm e}^{2-t} \sin \left (2 t \right )-2 \,{\mathrm e}^{2-t} \sin \left (4\right ) \cos \left (2 t \right )+10 t -24 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=2 \cos \left (4\right ) {\mathrm e}^{2-t} \sin \left (2 t \right )-2 \,{\mathrm e}^{2-t} \sin \left (4\right ) \cos \left (2 t \right )+10 t -24 \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.672 (sec). Leaf size: 23
\[
y \left (t \right ) = 2 \sin \left (2 t -4\right ) {\mathrm e}^{-t +2}-24+10 t
\]
✓ Solution by Mathematica
Time used: 0.02 (sec). Leaf size: 25
\[
y(t)\to 10 t-2 e^{2-t} \sin (4-2 t)-24
\]
6.14.2 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)+2*diff(y(t),t)+5*y(t)=50*t-100,y(2) = -4, D(y)(2) = 14],y(t), singsol=all)
DSolve[{y''[t]+2*y'[t]+5*y[t]==50*t-100,{y[2]==-4,y'[2]==14}},y[t],t,IncludeSingularSolutions -> True]