Internal problem ID [2881]
Internal file name [OUTPUT/2373_Sunday_June_05_2022_03_02_12_AM_772570/index.tex]
Book: Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth
edition, 2015
Section: Chapter 10, The Laplace Transform and Some Elementary Applications. Exercises for
10.7. page 704
Problem number: Problem 41.
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]]
\[ \boxed {y^{\prime \prime }-2 y^{\prime }+5 y=2 \sin \left (t \right )+\operatorname {Heaviside}\left (t -\frac {\pi }{2}\right ) \left (1+\cos \left (t \right )\right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 0] \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 &=\operatorname {Heaviside}\left (t -\frac {\pi }{2}\right ) \cos \left (t \right )+2 \sin \left (t \right )+\operatorname {Heaviside}\left (t -\frac {\pi }{2}\right ) \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }-2 y^{\prime }+5 y = \operatorname {Heaviside}\left (t -\frac {\pi }{2}\right ) \cos \left (t \right )+2 \sin \left (t \right )+\operatorname {Heaviside}\left (t -\frac {\pi }{2}\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 )+5 Y \left (s \right ) = \frac {{\mathrm e}^{-\frac {s \pi }{2}}}{s}+\frac {2-{\mathrm e}^{-\frac {s \pi }{2}}}{s^{2}+1}\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=0\\ y'(0) &=0 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-2 s Y \left (s \right )+5 Y \left (s \right ) = \frac {{\mathrm e}^{-\frac {s \pi }{2}}}{s}+\frac {2-{\mathrm e}^{-\frac {s \pi }{2}}}{s^{2}+1} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {{\mathrm e}^{-\frac {s \pi }{2}} s^{2}-s \,{\mathrm e}^{-\frac {s \pi }{2}}+{\mathrm e}^{-\frac {s \pi }{2}}+2 s}{s \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 {{\mathrm e}^{-\frac {s \pi }{2}} s^{2}-s \,{\mathrm e}^{-\frac {s \pi }{2}}+{\mathrm e}^{-\frac {s \pi }{2}}+2 s}{s \left (s^{2}+1\right ) \left (s^{2}-2 s +5\right )}\right )\\ &= \frac {\cos \left (t \right )}{5}+\frac {2 \sin \left (t \right )}{5}-\frac {{\mathrm e}^{t} \left (2 \cos \left (2 t \right )+\sin \left (2 t \right )\right )}{10}+\frac {\left (-2 \cos \left (2 t \right )+3 \sin \left (2 t \right )\right ) \left (\operatorname {Heaviside}\left (\frac {\pi }{2}-t \right )-1\right ) {\mathrm e}^{t -\frac {\pi }{2}}}{20}+\frac {\operatorname {Heaviside}\left (t -\frac {\pi }{2}\right ) \left (-\sin \left (t \right )+2 \cos \left (t \right )+2\right )}{10} \end {align*}
Converting the above solution to piecewise it becomes \[
y = \left \{\begin {array}{cc} \frac {\cos \left (t \right )}{5}+\frac {2 \sin \left (t \right )}{5}-\frac {{\mathrm e}^{t} \left (2 \cos \left (2 t \right )+\sin \left (2 t \right )\right )}{10} & t <\frac {\pi }{2} \\ \frac {1}{2}+\frac {{\mathrm e}^{\frac {\pi }{2}}}{5} & t =\frac {\pi }{2} \\ \frac {2 \cos \left (t \right )}{5}+\frac {3 \sin \left (t \right )}{10}-\frac {{\mathrm e}^{t} \left (2 \cos \left (2 t \right )+\sin \left (2 t \right )\right )}{10}-\frac {\left (-2 \cos \left (2 t \right )+3 \sin \left (2 t \right )\right ) {\mathrm e}^{t -\frac {\pi }{2}}}{20}+\frac {1}{5} & \frac {\pi }{2} The solution(s) found are the following \begin{align*}
\tag{1} y &= \left \{\begin {array}{cc} \frac {\left (-2 \cos \left (t \right )^{2}-\sin \left (t \right ) \cos \left (t \right )+1\right ) {\mathrm e}^{t}}{5}+\frac {\cos \left (t \right )}{5}+\frac {2 \sin \left (t \right )}{5} & t <\frac {\pi }{2} \\ \frac {1}{2}+\frac {{\mathrm e}^{\frac {\pi }{2}}}{5} & t &=\frac {\pi }{2} \\ \frac {1}{5}+\frac {\left (2 \cos \left (t \right )^{2}-3 \sin \left (t \right ) \cos \left (t \right )-1\right ) {\mathrm e}^{t -\frac {\pi }{2}}}{10}-\frac {2 \,{\mathrm e}^{t} \cos \left (t \right )^{2}}{5}+\frac {\left (-{\mathrm e}^{t} \sin \left (t \right )+2\right ) \cos \left (t \right )}{5}+\frac {{\mathrm e}^{t}}{5}+\frac {3 \sin \left (t \right )}{10} & \frac {\pi }{2} Verification of solutions
\[
y = \left \{\begin {array}{cc} \frac {\left (-2 \cos \left (t \right )^{2}-\sin \left (t \right ) \cos \left (t \right )+1\right ) {\mathrm e}^{t}}{5}+\frac {\cos \left (t \right )}{5}+\frac {2 \sin \left (t \right )}{5} & t <\frac {\pi }{2} \\ \frac {1}{2}+\frac {{\mathrm e}^{\frac {\pi }{2}}}{5} & t =\frac {\pi }{2} \\ \frac {1}{5}+\frac {\left (2 \cos \left (t \right )^{2}-3 \sin \left (t \right ) \cos \left (t \right )-1\right ) {\mathrm e}^{t -\frac {\pi }{2}}}{10}-\frac {2 \,{\mathrm e}^{t} \cos \left (t \right )^{2}}{5}+\frac {\left (-{\mathrm e}^{t} \sin \left (t \right )+2\right ) \cos \left (t \right )}{5}+\frac {{\mathrm e}^{t}}{5}+\frac {3 \sin \left (t \right )}{10} & \frac {\pi }{2}
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-2 y^{\prime }+5 y=\mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) \cos \left (t \right )+2 \sin \left (t \right )+\mathit {Heaviside}\left (t -\frac {\pi }{2}\right ), y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0\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 {2\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 )=\mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) \cos \left (t \right )+2 \sin \left (t \right )+\mathit {Heaviside}\left (t -\frac {\pi }{2}\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 )=\frac {{\mathrm e}^{t} \left (\sin \left (2 t \right ) \left (\int \cos \left (2 t \right ) \left (\mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) \cos \left (t \right )+2 \sin \left (t \right )+\mathit {Heaviside}\left (t -\frac {\pi }{2}\right )\right ) {\mathrm e}^{-t}d t \right )-2 \cos \left (2 t \right ) \left (\int {\mathrm e}^{-t} \sin \left (t \right ) \cos \left (t \right ) \left (\mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) \cos \left (t \right )+2 \sin \left (t \right )+\mathit {Heaviside}\left (t -\frac {\pi }{2}\right )\right )d t \right )\right )}{2} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\frac {\left (\left (2 \cos \left (t \right )^{2}-3 \sin \left (t \right ) \cos \left (t \right )-1\right ) {\mathrm e}^{t -\frac {\pi }{2}}+2 \cos \left (t \right )-\sin \left (t \right )+2\right ) \mathit {Heaviside}\left (t -\frac {\pi }{2}\right )}{10}+\frac {\cos \left (t \right )}{5}+\frac {2 \sin \left (t \right )}{5} \\ \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}+\frac {\left (\left (2 \cos \left (t \right )^{2}-3 \sin \left (t \right ) \cos \left (t \right )-1\right ) {\mathrm e}^{t -\frac {\pi }{2}}+2 \cos \left (t \right )-\sin \left (t \right )+2\right ) \mathit {Heaviside}\left (t -\frac {\pi }{2}\right )}{10}+\frac {\cos \left (t \right )}{5}+\frac {2 \sin \left (t \right )}{5} \\ \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}+\frac {\left (\left (2 \cos \left (t \right )^{2}-3 \sin \left (t \right ) \cos \left (t \right )-1\right ) {\mathrm e}^{t -\frac {\pi }{2}}+2 \cos \left (t \right )-\sin \left (t \right )+2\right ) \mathit {Heaviside}\left (t -\frac {\pi }{2}\right )}{10}+\frac {\cos \left (t \right )}{5}+\frac {2 \sin \left (t \right )}{5} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} +\frac {1}{5} \\ {} & \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}+\frac {\left (\left (-4 \sin \left (t \right ) \cos \left (t \right )-3 \cos \left (t \right )^{2}+3 \sin \left (t \right )^{2}\right ) {\mathrm e}^{t -\frac {\pi }{2}}+\left (2 \cos \left (t \right )^{2}-3 \sin \left (t \right ) \cos \left (t \right )-1\right ) {\mathrm e}^{t -\frac {\pi }{2}}-2 \sin \left (t \right )-\cos \left (t \right )\right ) \mathit {Heaviside}\left (t -\frac {\pi }{2}\right )}{10}+\frac {\left (\left (2 \cos \left (t \right )^{2}-3 \sin \left (t \right ) \cos \left (t \right )-1\right ) {\mathrm e}^{t -\frac {\pi }{2}}+2 \cos \left (t \right )-\sin \left (t \right )+2\right ) \mathit {Dirac}\left (t -\frac {\pi }{2}\right )}{10}-\frac {\sin \left (t \right )}{5}+\frac {2 \cos \left (t \right )}{5} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=\frac {2}{5}+c_{1} +2 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\frac {1}{5}, c_{2} =-\frac {1}{10}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {\left (\left (2 \cos \left (t \right )^{2}-3 \sin \left (t \right ) \cos \left (t \right )-1\right ) {\mathrm e}^{t -\frac {\pi }{2}}+2 \cos \left (t \right )-\sin \left (t \right )+2\right ) \mathit {Heaviside}\left (t -\frac {\pi }{2}\right )}{10}-\frac {2 \,{\mathrm e}^{t} \cos \left (t \right )^{2}}{5}-\frac {\sin \left (t \right ) \cos \left (t \right ) {\mathrm e}^{t}}{5}+\frac {\cos \left (t \right )}{5}+\frac {{\mathrm e}^{t}}{5}+\frac {2 \sin \left (t \right )}{5} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\left (\left (2 \cos \left (t \right )^{2}-3 \sin \left (t \right ) \cos \left (t \right )-1\right ) {\mathrm e}^{t -\frac {\pi }{2}}+2 \cos \left (t \right )-\sin \left (t \right )+2\right ) \mathit {Heaviside}\left (t -\frac {\pi }{2}\right )}{10}-\frac {2 \,{\mathrm e}^{t} \cos \left (t \right )^{2}}{5}-\frac {\sin \left (t \right ) \cos \left (t \right ) {\mathrm e}^{t}}{5}+\frac {\cos \left (t \right )}{5}+\frac {{\mathrm e}^{t}}{5}+\frac {2 \sin \left (t \right )}{5} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 2.515 (sec). Leaf size: 77
\[
y \left (t \right ) = \frac {\left (\left (2 \cos \left (t \right )^{2}-3 \cos \left (t \right ) \sin \left (t \right )-1\right ) {\mathrm e}^{t -\frac {\pi }{2}}+2 \cos \left (t \right )-\sin \left (t \right )+2\right ) \operatorname {Heaviside}\left (t -\frac {\pi }{2}\right )}{10}-\frac {2 \,{\mathrm e}^{t} \cos \left (t \right )^{2}}{5}-\frac {\sin \left (t \right ) \cos \left (t \right ) {\mathrm e}^{t}}{5}+\frac {\cos \left (t \right )}{5}+\frac {{\mathrm e}^{t}}{5}+\frac {2 \sin \left (t \right )}{5}
\]
✓ Solution by Mathematica
Time used: 0.502 (sec). Leaf size: 98
\[
y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {1}{5} \left (-e^t \sin (t) \cos (t)+\cos (t)-e^t \cos (2 t)+2 \sin (t)\right ) & 2 t\leq \pi \\ \frac {1}{20} \left (8 \cos (t)+2 e^t \left (-2+e^{-\pi /2}\right ) \cos (2 t)+6 \sin (t)-2 e^t \sin (2 t)-3 e^{t-\frac {\pi }{2}} \sin (2 t)+4\right ) & \text {True} \\ \end {array} \\ \end {array}
\]
14.15.2 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)=2*sin(t)+Heaviside(t-Pi/2)*(1-sin(t-Pi/2)),y(0) = 0, D(y)(0) = 0],y(t), singsol=all)
DSolve[{y''[t]-2*y'[t]+5*y[t]==2*Sin[t]+UnitStep[t-Pi/2]*(1-Sin[t-Pi/2]),{y[0]==0,y'[0]==0}},y[t],t,IncludeSingularSolutions -> True]