Internal problem ID [5702]
Internal file name [OUTPUT/4950_Sunday_June_05_2022_03_14_28_PM_35172071/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 6.3, page 224
Problem number: 26.
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 }+2 y^{\prime }+5 y=\left \{\begin {array}{cc} 10 \sin \left (t \right ) & 0
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 &=\left \{\begin {array}{cc} 0 & t \le 0 \\ 10 \sin \left (t \right ) & t <2 \pi \\ 0 & 2 \pi \le t \end {array}\right . \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+2 y^{\prime }+5 y = \left \{\begin {array}{cc} 0 & t \le 0 \\ 10 \sin \left (t \right ) & t <2 \pi \\ 0 & 2 \pi \le t \end {array}\right . \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 {10-10 \,{\mathrm e}^{-2 \pi s}}{s^{2}+1}\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 {10-10 \,{\mathrm e}^{-2 \pi s}}{s^{2}+1} \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}-s c_{1} +10 \,{\mathrm e}^{-2 \pi s}-2 c_{1} -c_{2} -10}{\left (s^{2}+1\right ) \left (s^{2}+2 s +5\right )} \end {align*}
Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (-\frac {-c_{1} s^{3}-2 c_{1} s^{2}-c_{2} s^{2}-s c_{1} +10 \,{\mathrm e}^{-2 \pi s}-2 c_{1} -c_{2} -10}{\left (s^{2}+1\right ) \left (s^{2}+2 s +5\right )}\right )\\ &= \frac {\left (-2 \cos \left (2 t \right )+\sin \left (2 t \right )\right ) \operatorname {Heaviside}\left (t -2 \pi \right ) {\mathrm e}^{2 \pi -t}}{2}+\frac {\left (2 \cos \left (2 t \right ) \left (1+c_{1} \right )+\sin \left (2 t \right ) \left (-1+c_{1} +c_{2} \right )\right ) {\mathrm e}^{-t}}{2}+\left (-\cos \left (t \right )+2 \sin \left (t \right )\right ) \operatorname {Heaviside}\left (2 \pi -t \right ) \end {align*}
Since both initial conditions given are not at zero, then we need to setup two equations to
solve for \(c_{1},c_{1}\). At \(t=\pi \) the first equation becomes, using the above solution \begin {align*} 1 &= 1+\frac {\left (2 c_{1} +2\right ) {\mathrm e}^{-\pi }}{2} \end {align*}
And taking derivative of the solution and evaluating at \(t=\pi \) gives the second equation as
\begin {align*} 2 \,{\mathrm e}^{-\pi }-2 &= -2+\frac {\left (-2+2 c_{1} +2 c_{2} \right ) {\mathrm e}^{-\pi }}{2}-\frac {\left (2 c_{1} +2\right ) {\mathrm e}^{-\pi }}{2} \end {align*}
Solving gives \begin {align*} c_{1} &= -1\\ c_{2} &= 4 \end {align*}
Subtituting these in the solution obtained above gives \begin {align*} y &= \frac {\left (-2 \cos \left (2 t \right )+\sin \left (2 t \right )\right ) \operatorname {Heaviside}\left (t -2 \pi \right ) {\mathrm e}^{2 \pi -t}}{2}+\sin \left (2 t \right ) {\mathrm e}^{-t}+\left (-\cos \left (t \right )+2 \sin \left (t \right )\right ) \operatorname {Heaviside}\left (2 \pi -t \right )\\ &= \left (-2 \cos \left (t \right )^{2}+\sin \left (t \right ) \cos \left (t \right )+1\right ) \operatorname {Heaviside}\left (t -2 \pi \right ) {\mathrm e}^{2 \pi -t}+\left (\cos \left (t \right )-2 \sin \left (t \right )\right ) \operatorname {Heaviside}\left (t -2 \pi \right )+2 \sin \left (t \right ) \cos \left (t \right ) {\mathrm e}^{-t}-\cos \left (t \right )+2 \sin \left (t \right ) \end {align*}
Converting the above solution to piecewise it becomes \[
y = \left \{\begin {array}{cc} 2 \sin \left (t \right ) \cos \left (t \right ) {\mathrm e}^{-t}-\cos \left (t \right )+2 \sin \left (t \right ) & t <2 \pi \\ 2 \sin \left (t \right ) \cos \left (t \right ) {\mathrm e}^{-t}+\left (-2 \cos \left (t \right )^{2}+\sin \left (t \right ) \cos \left (t \right )+1\right ) {\mathrm e}^{2 \pi -t} & 2 \pi \le t \end {array}\right .
\] Simplifying the solution gives \[
y = \left \{\begin {array}{cc} 2 \sin \left (t \right ) \cos \left (t \right ) {\mathrm e}^{-t}-\cos \left (t \right )+2 \sin \left (t \right ) & t <2 \pi \\ \frac {\left (-2 \cos \left (2 t \right )+\sin \left (2 t \right )\right ) {\mathrm e}^{2 \pi -t}}{2}+\sin \left (2 t \right ) {\mathrm e}^{-t} & 2 \pi \le t \end {array}\right .
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= \left \{\begin {array}{cc} 2 \sin \left (t \right ) \cos \left (t \right ) {\mathrm e}^{-t}-\cos \left (t \right )+2 \sin \left (t \right ) & t <2 \pi \\ \frac {\left (-2 \cos \left (2 t \right )+\sin \left (2 t \right )\right ) {\mathrm e}^{2 \pi -t}}{2}+\sin \left (2 t \right ) {\mathrm e}^{-t} & 2 \pi \le t \end {array}\right . \\
\end{align*} Verification of solutions
\[
y = \left \{\begin {array}{cc} 2 \sin \left (t \right ) \cos \left (t \right ) {\mathrm e}^{-t}-\cos \left (t \right )+2 \sin \left (t \right ) & t <2 \pi \\ \frac {\left (-2 \cos \left (2 t \right )+\sin \left (2 t \right )\right ) {\mathrm e}^{2 \pi -t}}{2}+\sin \left (2 t \right ) {\mathrm e}^{-t} & 2 \pi \le t \end {array}\right .
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+2 y^{\prime }+5 y=\left \{\begin {array}{cc} 0 & t \le 0 \\ 10 \sin \left (t \right ) & t <2 \pi \\ 0 & 2 \pi \le t \end {array}\right ., y \left (\pi \right )=1, y^{\prime }\left (\pi \right )=2 \,{\mathrm e}^{-\pi }-2\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}+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 )=\left \{\begin {array}{cc} 0 & t \le 0 \\ 10 \sin \left (t \right ) & t <2 \pi \\ 0 & 2 \pi \le t \end {array}\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 (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 )={\mathrm e}^{-t} \left (-\cos \left (2 t \right ) \left (\int \left (\left \{\begin {array}{cc} 0 & t \le 0 \\ 5 \sin \left (2 t \right ) {\mathrm e}^{t} \sin \left (t \right ) & t <2 \pi \\ 0 & 2 \pi \le t \end {array}\right .\right )d t \right )+\sin \left (2 t \right ) \left (\int \left (\left \{\begin {array}{cc} 0 & t \le 0 \\ 5 \cos \left (2 t \right ) {\mathrm e}^{t} \sin \left (t \right ) & t <2 \pi \\ 0 & 2 \pi \le t \end {array}\right .\right )d t \right )\right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\left \{\begin {array}{cc} 0 & t \le 0 \\ \left (2 \cos \left (t \right )^{2}-\sin \left (t \right ) \cos \left (t \right )-1\right ) {\mathrm e}^{-t}+2 \sin \left (t \right )-\cos \left (t \right ) & t \le 2 \pi \\ -\left (-1+{\mathrm e}^{2 \pi }\right ) \left (\cos \left (2 t \right )-\frac {\sin \left (2 t \right )}{2}\right ) {\mathrm e}^{-t} & 2 \pi Maple trace
✓ Solution by Maple
Time used: 1.343 (sec). Leaf size: 70
\[
y \left (t \right ) = \left \{\begin {array}{cc} \sin \left (2 t \right ) {\mathrm e}^{-t}-\cos \left (t \right )+2 \sin \left (t \right ) & t <2 \pi \\ -2 & t =2 \pi \\ \sin \left (2 t \right ) {\mathrm e}^{-t}+\frac {\left (-2 \cos \left (2 t \right )+\sin \left (2 t \right )\right ) {\mathrm e}^{2 \pi -t}}{2} & 2 \pi ✓ Solution by Mathematica
Time used: 0.06 (sec). Leaf size: 94
\[
y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {1}{2} e^{-t} (3 \sin (2 t)-2 \cos (2 t)) & t\leq 0 \\ -\cos (t)+2 \sin (t)+e^{-t} \sin (2 t) & 07.9.1 Existence and uniqueness analysis
7.9.2 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)=piecewise(0<t and t<2*Pi,10*sin(t),t>2*Pi,0),y(Pi) = 1, D(y)(Pi) = 2*exp(-Pi)-2],y(t), singsol=all)
DSolve[{y''[t]+2*y'[t]+5*y[t]==Piecewise[{{10*Sin[t],0<t<2*Pi},{0,t>2*Pi}}],{y[Pi]==1,y'[Pi]==2*Exp[-Pi]-2}},y[t],t,IncludeSingularSolutions -> True]