8.7 problem 9

Internal problem ID [5710]
Internal file name [OUTPUT/4958_Sunday_June_05_2022_03_15_00_PM_82102922/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 6.4, page 230
Problem number: 9.
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]]

\[ \boxed {y^{\prime \prime }+4 y^{\prime }+5 y=\left (1-\operatorname {Heaviside}\left (-10+t \right )\right ) {\mathrm e}^{t}-{\mathrm e}^{10} \delta \left (-10+t \right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 1] \end {align*}

8.7.1 Existence and uniqueness analysis

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) &=5\\ F &=-{\mathrm e}^{t} \operatorname {Heaviside}\left (-10+t \right )-{\mathrm e}^{10} \delta \left (-10+t \right )+{\mathrm e}^{t} \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+4 y^{\prime }+5 y = -{\mathrm e}^{t} \operatorname {Heaviside}\left (-10+t \right )-{\mathrm e}^{10} \delta \left (-10+t \right )+{\mathrm e}^{t} \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 )+5 Y \left (s \right ) = \frac {-{\mathrm e}^{-10 s +10} s +1}{s -1}\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+4 s Y \left (s \right )+5 Y \left (s \right ) = \frac {-{\mathrm e}^{-10 s +10} s +1}{s -1} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = -\frac {s \left ({\mathrm e}^{-10 s +10}-1\right )}{\left (s -1\right ) \left (s^{2}+4 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 {s \left ({\mathrm e}^{-10 s +10}-1\right )}{\left (s -1\right ) \left (s^{2}+4 s +5\right )}\right )\\ &= \frac {{\mathrm e}^{t} \operatorname {Heaviside}\left (10-t \right )}{10}+\frac {{\mathrm e}^{30-2 t} \operatorname {Heaviside}\left (-10+t \right ) \left (\cos \left (-10+t \right )-7 \sin \left (-10+t \right )\right )}{10}+\frac {\left (-\cos \left (t \right )+7 \sin \left (t \right )\right ) {\mathrm e}^{-2 t}}{10} \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} \frac {\left (-\cos \left (t \right )+7 \sin \left (t \right )\right ) {\mathrm e}^{-2 t}}{10}+\frac {{\mathrm e}^{t}}{10} & t <10 \\ \frac {\left (-\cos \left (10\right )+7 \sin \left (10\right )\right ) {\mathrm e}^{-20}}{10}+\frac {{\mathrm e}^{10}}{5} & t =10 \\ \frac {\left (-\cos \left (t \right )+7 \sin \left (t \right )\right ) {\mathrm e}^{-2 t}}{10}+\frac {{\mathrm e}^{30-2 t} \left (\cos \left (-10+t \right )-7 \sin \left (-10+t \right )\right )}{10} & 10

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\left (\left \{\begin {array}{cc} \left (-{\mathrm e}^{3 t}+\cos \left (t \right )-7 \sin \left (t \right )\right ) {\mathrm e}^{-2 t} & t <10 \\ \left (-2 \,{\mathrm e}^{30}+\cos \left (10\right )-7 \sin \left (10\right )\right ) {\mathrm e}^{-20} & t &=10 \\ \left (-\cos \left (-10+t \right )+7 \sin \left (-10+t \right )\right ) {\mathrm e}^{30-2 t}+\left (\cos \left (t \right )-7 \sin \left (t \right )\right ) {\mathrm e}^{-2 t} & 10

Verification of solutions

\[ y = -\frac {\left (\left \{\begin {array}{cc} \left (-{\mathrm e}^{3 t}+\cos \left (t \right )-7 \sin \left (t \right )\right ) {\mathrm e}^{-2 t} & t <10 \\ \left (-2 \,{\mathrm e}^{30}+\cos \left (10\right )-7 \sin \left (10\right )\right ) {\mathrm e}^{-20} & t =10 \\ \left (-\cos \left (-10+t \right )+7 \sin \left (-10+t \right )\right ) {\mathrm e}^{30-2 t}+\left (\cos \left (t \right )-7 \sin \left (t \right )\right ) {\mathrm e}^{-2 t} & 10

8.7.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+4 y^{\prime }+5 y=-{\mathrm e}^{t} \mathit {Heaviside}\left (-10+t \right )-{\mathrm e}^{10} \mathit {Dirac}\left (-10+t \right )+{\mathrm e}^{t}, 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 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+4 r +5=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-4\right )\pm \left (\sqrt {-4}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-2-\mathrm {I}, -2+\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-2 t} \cos \left (t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-2 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}^{-2 t} \cos \left (t \right )+c_{2} {\mathrm e}^{-2 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 )=-{\mathrm e}^{t} \mathit {Heaviside}\left (-10+t \right )-{\mathrm e}^{10} \mathit {Dirac}\left (-10+t \right )+{\mathrm e}^{t}\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 (t \right ) & {\mathrm e}^{-2 t} \sin \left (t \right ) \\ -2 \,{\mathrm e}^{-2 t} \cos \left (t \right )-{\mathrm e}^{-2 t} \sin \left (t \right ) & -2 \,{\mathrm e}^{-2 t} \sin \left (t \right )+{\mathrm e}^{-2 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}^{-4 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )={\mathrm e}^{-2 t} \left (\cos \left (t \right ) \left (\int \left ({\mathrm e}^{2 t +10} \mathit {Dirac}\left (-10+t \right )+{\mathrm e}^{3 t} \left (-1+\mathit {Heaviside}\left (-10+t \right )\right )\right ) \sin \left (t \right )d t \right )-\sin \left (t \right ) \left (\int \left ({\mathrm e}^{2 t +10} \mathit {Dirac}\left (-10+t \right )+{\mathrm e}^{3 t} \left (-1+\mathit {Heaviside}\left (-10+t \right )\right )\right ) \cos \left (t \right )d t \right )\right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\frac {{\mathrm e}^{-2 t} \left (\left (-{\mathrm e}^{3 t}+\left (\cos \left (10\right ) \left (\cos \left (t \right )-7 \sin \left (t \right )\right )+\left (\sin \left (t \right )+7 \cos \left (t \right )\right ) \sin \left (10\right )\right ) {\mathrm e}^{30}\right ) \mathit {Heaviside}\left (-10+t \right )+{\mathrm e}^{3 t}\right )}{10} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-2 t} \cos \left (t \right )+c_{2} {\mathrm e}^{-2 t} \sin \left (t \right )+\frac {{\mathrm e}^{-2 t} \left (\left (-{\mathrm e}^{3 t}+\left (\cos \left (10\right ) \left (\cos \left (t \right )-7 \sin \left (t \right )\right )+\left (\sin \left (t \right )+7 \cos \left (t \right )\right ) \sin \left (10\right )\right ) {\mathrm e}^{30}\right ) \mathit {Heaviside}\left (-10+t \right )+{\mathrm e}^{3 t}\right )}{10} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-2 t} \cos \left (t \right )+c_{2} {\mathrm e}^{-2 t} \sin \left (t \right )+\frac {{\mathrm e}^{-2 t} \left (\left (-{\mathrm e}^{3 t}+\left (\cos \left (10\right ) \left (\cos \left (t \right )-7 \sin \left (t \right )\right )+\left (\sin \left (t \right )+7 \cos \left (t \right )\right ) \sin \left (10\right )\right ) {\mathrm e}^{30}\right ) \mathit {Heaviside}\left (-10+t \right )+{\mathrm e}^{3 t}\right )}{10} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} +\frac {1}{10} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-2 c_{1} {\mathrm e}^{-2 t} \cos \left (t \right )-c_{1} {\mathrm e}^{-2 t} \sin \left (t \right )-2 c_{2} {\mathrm e}^{-2 t} \sin \left (t \right )+c_{2} {\mathrm e}^{-2 t} \cos \left (t \right )-\frac {{\mathrm e}^{-2 t} \left (\left (-{\mathrm e}^{3 t}+\left (\cos \left (10\right ) \left (\cos \left (t \right )-7 \sin \left (t \right )\right )+\left (\sin \left (t \right )+7 \cos \left (t \right )\right ) \sin \left (10\right )\right ) {\mathrm e}^{30}\right ) \mathit {Heaviside}\left (-10+t \right )+{\mathrm e}^{3 t}\right )}{5}+\frac {{\mathrm e}^{-2 t} \left (\left (-3 \,{\mathrm e}^{3 t}+\left (\cos \left (10\right ) \left (-\sin \left (t \right )-7 \cos \left (t \right )\right )+\left (\cos \left (t \right )-7 \sin \left (t \right )\right ) \sin \left (10\right )\right ) {\mathrm e}^{30}\right ) \mathit {Heaviside}\left (-10+t \right )+\left (-{\mathrm e}^{3 t}+\left (\cos \left (10\right ) \left (\cos \left (t \right )-7 \sin \left (t \right )\right )+\left (\sin \left (t \right )+7 \cos \left (t \right )\right ) \sin \left (10\right )\right ) {\mathrm e}^{30}\right ) \mathit {Dirac}\left (-10+t \right )+3 \,{\mathrm e}^{3 t}\right )}{10} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=-2 c_{1} +\frac {1}{10}+c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\frac {1}{10}, c_{2} =\frac {7}{10}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {{\mathrm e}^{-2 t} \left (\left (-{\mathrm e}^{3 t}+\left (\left (\cos \left (10\right )+7 \sin \left (10\right )\right ) \cos \left (t \right )+\left (-7 \cos \left (10\right )+\sin \left (10\right )\right ) \sin \left (t \right )\right ) {\mathrm e}^{30}\right ) \mathit {Heaviside}\left (-10+t \right )-\cos \left (t \right )+7 \sin \left (t \right )+{\mathrm e}^{3 t}\right )}{10} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {{\mathrm e}^{-2 t} \left (\left (-{\mathrm e}^{3 t}+\left (\left (\cos \left (10\right )+7 \sin \left (10\right )\right ) \cos \left (t \right )+\left (-7 \cos \left (10\right )+\sin \left (10\right )\right ) \sin \left (t \right )\right ) {\mathrm e}^{30}\right ) \mathit {Heaviside}\left (-10+t \right )-\cos \left (t \right )+7 \sin \left (t \right )+{\mathrm e}^{3 t}\right )}{10} \end {array} \]

Maple trace

`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`
 

✓ Solution by Maple

Time used: 1.328 (sec). Leaf size: 53

dsolve([diff(y(t),t$2)+4*diff(y(t),t)+5*y(t)=(1-Heaviside(t-10))*exp(t)-exp(10)*Dirac(t-10),y(0) = 0, D(y)(0) = 1],y(t), singsol=all)
 

\[ y \left (t \right ) = \frac {{\mathrm e}^{-2 t} \left (\left (-{\mathrm e}^{3 t}+\left (\left (-7 \cos \left (10\right )+\sin \left (10\right )\right ) \sin \left (t \right )+\left (\cos \left (10\right )+7 \sin \left (10\right )\right ) \cos \left (t \right )\right ) {\mathrm e}^{30}\right ) \operatorname {Heaviside}\left (t -10\right )-\cos \left (t \right )+7 \sin \left (t \right )+{\mathrm e}^{3 t}\right )}{10} \]

✓ Solution by Mathematica

Time used: 0.571 (sec). Leaf size: 94

DSolve[{y''[t]+4*y'[t]+5*y[t]==(1-UnitStep[t-10])*Exp[t]-Exp[10]*DiracDelta[t-10],{y[0]==0,y'[0]==1}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {1}{10} e^{-2 t} \left (10 e^{30} \theta (t-10) \sin (10-t)+\theta (10-t) \left (e^{3 t}+3 e^{30} \sin (10-t)-e^{30} \cos (10-t)\right )-3 e^{30} \sin (10-t)+7 \sin (t)+e^{30} \cos (10-t)-\cos (t)\right ) \]