14.9 problem Problem 35

Internal problem ID [2875]
Internal file name [OUTPUT/2367_Sunday_June_05_2022_03_01_27_AM_4958839/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 35.
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]]

\[ \boxed {y^{\prime \prime }-y^{\prime }-2 y=1-3 \operatorname {Heaviside}\left (t -2\right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = -2] \end {align*}

14.9.1 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) &=-1\\ q(t) &=-2\\ F &=1-3 \operatorname {Heaviside}\left (t -2\right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }-y^{\prime }-2 y = 1-3 \operatorname {Heaviside}\left (t -2\right ) \end {align*}

The domain of \(p(t)=-1\) 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 )-s Y \left (s \right )+y \left (0\right )-2 Y \left (s \right ) = \frac {1-3 \,{\mathrm e}^{-2 s}}{s}\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&=1\\ y'(0) &=-2 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )+3-s -s Y \left (s \right )-2 Y \left (s \right ) = \frac {1-3 \,{\mathrm e}^{-2 s}}{s} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = -\frac {-s^{2}+3 \,{\mathrm e}^{-2 s}+3 s -1}{s \left (s^{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 {-s^{2}+3 \,{\mathrm e}^{-2 s}+3 s -1}{s \left (s^{2}-s -2\right )}\right )\\ &= -\frac {1}{2}+\frac {5 \,{\mathrm e}^{-t}}{3}-\frac {{\mathrm e}^{2 t}}{6}+\frac {{\mathrm e}^{2 t -4} \left (-1+\operatorname {Heaviside}\left (-t +2\right )\right )}{2}+\frac {\operatorname {Heaviside}\left (t -2\right ) \left (3-2 \,{\mathrm e}^{-t +2}\right )}{2} \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} -\frac {1}{2}+\frac {5 \,{\mathrm e}^{-t}}{3}-\frac {{\mathrm e}^{2 t}}{6} & t <2 \\ \frac {5 \,{\mathrm e}^{-2}}{3}-\frac {{\mathrm e}^{4}}{6} & t =2 \\ 1+\frac {5 \,{\mathrm e}^{-t}}{3}-\frac {{\mathrm e}^{2 t}}{6}-\frac {{\mathrm e}^{2 t -4}}{2}-{\mathrm e}^{-t +2} & 2

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (\left \{\begin {array}{cc} -3+10 \,{\mathrm e}^{-t}-{\mathrm e}^{2 t} & t <2 \\ 10 \,{\mathrm e}^{-2}-{\mathrm e}^{4} & t &=2 \\ 6+10 \,{\mathrm e}^{-t}-{\mathrm e}^{2 t}-3 \,{\mathrm e}^{2 t -4}-6 \,{\mathrm e}^{-t +2} & 2

Verification of solutions

\[ y = \frac {\left (\left \{\begin {array}{cc} -3+10 \,{\mathrm e}^{-t}-{\mathrm e}^{2 t} & t <2 \\ 10 \,{\mathrm e}^{-2}-{\mathrm e}^{4} & t =2 \\ 6+10 \,{\mathrm e}^{-t}-{\mathrm e}^{2 t}-3 \,{\mathrm e}^{2 t -4}-6 \,{\mathrm e}^{-t +2} & 2

14.9.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-y^{\prime }-2 y=1-3 \mathit {Heaviside}\left (t -2\right ), y \left (0\right )=1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-2\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}-r -2=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +1\right ) \left (r -2\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-1, 2\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{2 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} {\mathrm e}^{-t}+c_{2} {\mathrm e}^{2 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 )=1-3 \mathit {Heaviside}\left (t -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} {\mathrm e}^{-t} & {\mathrm e}^{2 t} \\ -{\mathrm e}^{-t} & 2 \,{\mathrm e}^{2 t} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=3 \,{\mathrm e}^{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 (\int {\mathrm e}^{t} \left (-1+3 \mathit {Heaviside}\left (t -2\right )\right )d t \right )}{3}-\frac {{\mathrm e}^{2 t} \left (\int {\mathrm e}^{-2 t} \left (-1+3 \mathit {Heaviside}\left (t -2\right )\right )d t \right )}{3} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\frac {3 \mathit {Heaviside}\left (t -2\right )}{2}-\mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{-t +2}-\frac {1}{2}-\frac {\mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{2 t -4}}{2} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-t}+c_{2} {\mathrm e}^{2 t}+\frac {3 \mathit {Heaviside}\left (t -2\right )}{2}-\mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{-t +2}-\frac {1}{2}-\frac {\mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{2 t -4}}{2} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-t}+c_{2} {\mathrm e}^{2 t}+\frac {3 \mathit {Heaviside}\left (t -2\right )}{2}-\mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{-t +2}-\frac {1}{2}-\frac {\mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{2 t -4}}{2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=c_{1} +c_{2} -\frac {1}{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-c_{1} {\mathrm e}^{-t}+2 c_{2} {\mathrm e}^{2 t}+\frac {3 \mathit {Dirac}\left (t -2\right )}{2}-\mathit {Dirac}\left (t -2\right ) {\mathrm e}^{-t +2}+\mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{-t +2}-\frac {\mathit {Dirac}\left (t -2\right ) {\mathrm e}^{2 t -4}}{2}-\mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{2 t -4} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-2 \\ {} & {} & -2=-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 {5}{3}, c_{2} =-\frac {1}{6}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {5 \,{\mathrm e}^{-t}}{3}-\frac {{\mathrm e}^{2 t}}{6}+\frac {3 \mathit {Heaviside}\left (t -2\right )}{2}-\mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{-t +2}-\frac {1}{2}-\frac {\mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{2 t -4}}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {5 \,{\mathrm e}^{-t}}{3}-\frac {{\mathrm e}^{2 t}}{6}+\frac {3 \mathit {Heaviside}\left (t -2\right )}{2}-\mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{-t +2}-\frac {1}{2}-\frac {\mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{2 t -4}}{2} \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: 2.218 (sec). Leaf size: 50

dsolve([diff(y(t),t$2)-diff(y(t),t)-2*y(t)=1-3*Heaviside(t-2),y(0) = 1, D(y)(0) = -2],y(t), singsol=all)
 

\[ y \left (t \right ) = -\frac {1}{2}+\frac {5 \,{\mathrm e}^{-t}}{3}-\frac {{\mathrm e}^{2 t}}{6}-\frac {\operatorname {Heaviside}\left (t -2\right ) {\mathrm e}^{-4+2 t}}{2}-\operatorname {Heaviside}\left (t -2\right ) {\mathrm e}^{2-t}+\frac {3 \operatorname {Heaviside}\left (t -2\right )}{2} \]

✓ Solution by Mathematica

Time used: 0.029 (sec). Leaf size: 70

DSolve[{y''[t]-y'[t]-2*y[t]==1-3*UnitStep[t-2],{y[0]==1,y'[0]==-2}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} -\frac {1}{6} e^{-t} \left (-10+3 e^t+e^{3 t}\right ) & t\leq 2 \\ \frac {1}{6} \left (6-6 e^{2-t}+10 e^{-t}-e^{2 t}-3 e^{2 t-4}\right ) & \text {True} \\ \end {array} \\ \end {array} \]