Internal problem ID [846]
Internal file name [OUTPUT/846_Sunday_June_05_2022_01_51_09_AM_23570421/index.tex]
Book: Elementary differential equations and boundary value problems, 11th ed., Boyce, DiPrima,
Meade
Section: Chapter 6.4, The Laplace Transform. Differential equations with discontinuous forcing
functions. page 268
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"
Maple gives the following as the ode type
[[_2nd_order, _linear, _nonhomogeneous]]
\[ \boxed {y^{\prime \prime }+2 y^{\prime }+2 y=\left \{\begin {array}{cc} 1 & \pi \le t <2 \pi \\ 0 & \operatorname {otherwise} \end {array}\right .} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, 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) &=2\\ q(t) &=2\\ F &=\left \{\begin {array}{cc} 0 & t <\pi \\ 1 & 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 }+2 y = \left \{\begin {array}{cc} 0 & t <\pi \\ 1 & t <2 \pi \\ 0 & 2 \pi \le t \end {array}\right . \end {align*}
The domain of \(p(t)=2\) 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 )+2 s Y \left (s \right )-2 y \left (0\right )+2 Y \left (s \right ) = \frac {{\mathrm e}^{-\pi s}-{\mathrm e}^{-2 \pi s}}{s}\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=0\\ y'(0) &=1 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-1+2 s Y \left (s \right )+2 Y \left (s \right ) = \frac {{\mathrm e}^{-\pi s}-{\mathrm e}^{-2 \pi s}}{s} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {{\mathrm e}^{-\pi s}-{\mathrm e}^{-2 \pi s}+s}{s \left (s^{2}+2 s +2\right )} \end {align*}
Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (\frac {{\mathrm e}^{-\pi s}-{\mathrm e}^{-2 \pi s}+s}{s \left (s^{2}+2 s +2\right )}\right )\\ &= {\mathrm e}^{-t} \sin \left (t \right )+\frac {\left (-1+{\mathrm e}^{2 \pi -t} \left (\cos \left (t \right )+\sin \left (t \right )\right )\right ) \operatorname {Heaviside}\left (t -2 \pi \right )}{2}+\frac {\left (1+{\mathrm e}^{\pi -t} \left (\cos \left (t \right )+\sin \left (t \right )\right )\right ) \operatorname {Heaviside}\left (t -\pi \right )}{2} \end {align*}
Converting the above solution to piecewise it becomes \[
y = \left \{\begin {array}{cc} {\mathrm e}^{-t} \sin \left (t \right ) & t <\pi \\ {\mathrm e}^{-t} \sin \left (t \right )+\frac {1}{2}+\frac {{\mathrm e}^{\pi -t} \left (\cos \left (t \right )+\sin \left (t \right )\right )}{2} & t <2 \pi \\ {\mathrm e}^{-t} \sin \left (t \right )+\frac {{\mathrm e}^{2 \pi -t} \left (\cos \left (t \right )+\sin \left (t \right )\right )}{2}+\frac {{\mathrm e}^{\pi -t} \left (\cos \left (t \right )+\sin \left (t \right )\right )}{2} & 2 \pi \le t \end {array}\right .
\] Simplifying the solution gives \[
y = {\mathrm e}^{-t} \sin \left (t \right )+\frac {\left (\left \{\begin {array}{cc} 0 & t <\pi \\ 1+{\mathrm e}^{\pi -t} \left (\cos \left (t \right )+\sin \left (t \right )\right ) & t <2 \pi \\ \left ({\mathrm e}^{2 \pi -t}+{\mathrm e}^{\pi -t}\right ) \left (\cos \left (t \right )+\sin \left (t \right )\right ) & 2 \pi \le t \end {array}\right .\right )}{2}
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= {\mathrm e}^{-t} \sin \left (t \right )+\frac {\left (\left \{\begin {array}{cc} 0 & t <\pi \\ 1+{\mathrm e}^{\pi -t} \left (\cos \left (t \right )+\sin \left (t \right )\right ) & t <2 \pi \\ \left ({\mathrm e}^{2 \pi -t}+{\mathrm e}^{\pi -t}\right ) \left (\cos \left (t \right )+\sin \left (t \right )\right ) & 2 \pi \le t \end {array}\right .\right )}{2} \\
\end{align*} Verification of solutions
\[
y = {\mathrm e}^{-t} \sin \left (t \right )+\frac {\left (\left \{\begin {array}{cc} 0 & t <\pi \\ 1+{\mathrm e}^{\pi -t} \left (\cos \left (t \right )+\sin \left (t \right )\right ) & t <2 \pi \\ \left ({\mathrm e}^{2 \pi -t}+{\mathrm e}^{\pi -t}\right ) \left (\cos \left (t \right )+\sin \left (t \right )\right ) & 2 \pi \le t \end {array}\right .\right )}{2}
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+2 y^{\prime }+2 y=\left \{\begin {array}{cc} 0 & t <\pi \\ 1 & t <2 \pi \\ 0 & 2 \pi \le t \end {array}\right ., y \left (0\right )=0, 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 \\ {} & {} & \frac {d}{d t}y^{\prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+2 r +2=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-2\right )\pm \left (\sqrt {-4}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-1-\mathrm {I}, -1+\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-t} \cos \left (t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-t} \sin \left (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}^{-t} \cos \left (t \right )+c_{2} {\mathrm e}^{-t} \sin \left (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 )=\left \{\begin {array}{cc} 0 & t <\pi \\ 1 & 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} {\mathrm e}^{-t} \cos \left (t \right ) & {\mathrm e}^{-t} \sin \left (t \right ) \\ -{\mathrm e}^{-t} \cos \left (t \right )-{\mathrm e}^{-t} \sin \left (t \right ) & -{\mathrm e}^{-t} \sin \left (t \right )+{\mathrm e}^{-t} \cos \left (t \right ) \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )={\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 (t \right ) \left (\int \left (\left \{\begin {array}{cc} 0 & t <\pi \\ {\mathrm e}^{t} \sin \left (t \right ) & t <2 \pi \\ 0 & 2 \pi \le t \end {array}\right .\right )d t \right )-\sin \left (t \right ) \left (\int \left (\left \{\begin {array}{cc} 0 & t <\pi \\ {\mathrm e}^{t} \cos \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 )=\frac {\left (\left \{\begin {array}{cc} 0 & t \le \pi \\ 1+{\mathrm e}^{\pi -t} \left (\cos \left (t \right )+\sin \left (t \right )\right ) & t \le 2 \pi \\ {\mathrm e}^{\pi -t} \left ({\mathrm e}^{\pi }+1\right ) \left (\cos \left (t \right )+\sin \left (t \right )\right ) & 2 \pi Maple trace
✓ Solution by Maple
Time used: 0.797 (sec). Leaf size: 83
\[
y \left (t \right ) = \sin \left (t \right ) {\mathrm e}^{-t}+\frac {\left (\left \{\begin {array}{cc} 0 & t <\pi \\ 1+{\mathrm e}^{\pi -t} \left (\cos \left (t \right )+\sin \left (t \right )\right ) & t <2 \pi \\ \left (\cos \left (t \right )+\sin \left (t \right )\right ) \left ({\mathrm e}^{\pi -t}+{\mathrm e}^{2 \pi -t}\right ) & 2 \pi \le t \end {array}\right .\right )}{2}
\]
✓ Solution by Mathematica
Time used: 0.047 (sec). Leaf size: 89
\[
y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} e^{-t} \sin (t) & t\leq \pi \\ \frac {1}{2} e^{-t} \left (e^{\pi } \cos (t)+e^t+\left (2+e^{\pi }\right ) \sin (t)\right ) & \pi 4.2.2 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)+2*y(t)=piecewise(Pi<=t and t<2*Pi,1,true,0),y(0) = 0, D(y)(0) = 1],y(t), singsol=all)
DSolve[{y''[t]+2*y'[t]+2*y[t]==Piecewise[{{1,Pi<=t<2*Pi},{0,True}}],{y[0]==0,y'[0]==1}},y[t],t,IncludeSingularSolutions -> True]