14.13 problem Problem 39

Internal problem ID [2879]
Internal file name [OUTPUT/2371_Sunday_June_05_2022_03_01_56_AM_85053664/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 39.
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]]

\[ \boxed {y^{\prime \prime }+y^{\prime }-6 y=30 \operatorname {Heaviside}\left (t -1\right ) {\mathrm e}^{1-t}} \] With initial conditions \begin {align*} [y \left (0\right ) = 3, y^{\prime }\left (0\right ) = -4] \end {align*}

14.13.1 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) &=-6\\ F &=30 \operatorname {Heaviside}\left (t -1\right ) {\mathrm e}^{1-t} \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+y^{\prime }-6 y = 30 \operatorname {Heaviside}\left (t -1\right ) {\mathrm e}^{1-t} \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 )-6 Y \left (s \right ) = \frac {30 \,{\mathrm e}^{-s}}{1+s}\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&=3\\ y'(0) &=-4 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )+1-3 s +s Y \left (s \right )-6 Y \left (s \right ) = \frac {30 \,{\mathrm e}^{-s}}{1+s} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {3 s^{2}+30 \,{\mathrm e}^{-s}+2 s -1}{\left (1+s \right ) \left (s^{2}+s -6\right )} \end {align*}

Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (\frac {3 s^{2}+30 \,{\mathrm e}^{-s}+2 s -1}{\left (1+s \right ) \left (s^{2}+s -6\right )}\right )\\ &= 2 \,{\mathrm e}^{-3 t}+{\mathrm e}^{2 t}+2 \left (1-\operatorname {Heaviside}\left (1-t \right )\right ) {\mathrm e}^{2 t -2}+\left (3 \,{\mathrm e}^{-3 t +3}-5 \,{\mathrm e}^{1-t}\right ) \operatorname {Heaviside}\left (t -1\right ) \end {align*}

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

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left \{\begin {array}{cc} \left ({\mathrm e}^{5 t}+2\right ) {\mathrm e}^{-3 t} & t <1 \\ \left ({\mathrm e}^{5}-2 \,{\mathrm e}^{3}+2\right ) {\mathrm e}^{-3} & t &=1 \\ \left ({\mathrm e}^{5 t}-5 \,{\mathrm e}^{2 t +1}+2 \,{\mathrm e}^{5 t -2}+3 \,{\mathrm e}^{3}+2\right ) {\mathrm e}^{-3 t} & 1

Verification of solutions

\[ y = \left \{\begin {array}{cc} \left ({\mathrm e}^{5 t}+2\right ) {\mathrm e}^{-3 t} & t <1 \\ \left ({\mathrm e}^{5}-2 \,{\mathrm e}^{3}+2\right ) {\mathrm e}^{-3} & t =1 \\ \left ({\mathrm e}^{5 t}-5 \,{\mathrm e}^{2 t +1}+2 \,{\mathrm e}^{5 t -2}+3 \,{\mathrm e}^{3}+2\right ) {\mathrm e}^{-3 t} & 1

14.13.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+y^{\prime }-6 y=30 \mathit {Heaviside}\left (t -1\right ) {\mathrm e}^{1-t}, y \left (0\right )=3, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-4\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 -6=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +3\right ) \left (r -2\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-3, 2\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-3 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}^{-3 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 )=30 \mathit {Heaviside}\left (t -1\right ) {\mathrm e}^{1-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}^{-3 t} & {\mathrm e}^{2 t} \\ -3 \,{\mathrm e}^{-3 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 )=5 \,{\mathrm e}^{-t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=6 \left ({\mathrm e}^{5 t} \left (\int \mathit {Heaviside}\left (t -1\right ) {\mathrm e}^{-3 t +1}d t \right )-\left (\int \mathit {Heaviside}\left (t -1\right ) {\mathrm e}^{2 t +1}d t \right )\right ) {\mathrm e}^{-3 t} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )={\mathrm e}^{-3 t} \mathit {Heaviside}\left (t -1\right ) \left (3 \,{\mathrm e}^{3}-5 \,{\mathrm e}^{2 t +1}+2 \,{\mathrm e}^{5 t -2}\right ) \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-3 t}+c_{2} {\mathrm e}^{2 t}+{\mathrm e}^{-3 t} \mathit {Heaviside}\left (t -1\right ) \left (3 \,{\mathrm e}^{3}-5 \,{\mathrm e}^{2 t +1}+2 \,{\mathrm e}^{5 t -2}\right ) \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-3 t}+c_{2} {\mathrm e}^{2 t}+{\mathrm e}^{-3 t} \mathit {Heaviside}\left (t -1\right ) \left (3 {\mathrm e}^{3}-5 {\mathrm e}^{2 t +1}+2 {\mathrm e}^{5 t -2}\right ) \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=3 \\ {} & {} & 3=c_{1} +c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-3 c_{1} {\mathrm e}^{-3 t}+2 c_{2} {\mathrm e}^{2 t}-3 \,{\mathrm e}^{-3 t} \mathit {Heaviside}\left (t -1\right ) \left (3 \,{\mathrm e}^{3}-5 \,{\mathrm e}^{2 t +1}+2 \,{\mathrm e}^{5 t -2}\right )+{\mathrm e}^{-3 t} \mathit {Dirac}\left (t -1\right ) \left (3 \,{\mathrm e}^{3}-5 \,{\mathrm e}^{2 t +1}+2 \,{\mathrm e}^{5 t -2}\right )+{\mathrm e}^{-3 t} \mathit {Heaviside}\left (t -1\right ) \left (-10 \,{\mathrm e}^{2 t +1}+10 \,{\mathrm e}^{5 t -2}\right ) \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-4 \\ {} & {} & -4=-3 c_{1} +2 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =2, c_{2} =1\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\left ({\mathrm e}^{5 t}+3 \mathit {Heaviside}\left (t -1\right ) {\mathrm e}^{3}-5 \mathit {Heaviside}\left (t -1\right ) {\mathrm e}^{2 t +1}+2 \mathit {Heaviside}\left (t -1\right ) {\mathrm e}^{5 t -2}+2\right ) {\mathrm e}^{-3 t} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\left ({\mathrm e}^{5 t}+3 \mathit {Heaviside}\left (t -1\right ) {\mathrm e}^{3}-5 \mathit {Heaviside}\left (t -1\right ) {\mathrm e}^{2 t +1}+2 \mathit {Heaviside}\left (t -1\right ) {\mathrm e}^{5 t -2}+2\right ) {\mathrm e}^{-3 t} \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: 2.265 (sec). Leaf size: 55

dsolve([diff(y(t),t$2)+diff(y(t),t)-6*y(t)=30*Heaviside(t-1)*exp(-(t-1)),y(0) = 3, D(y)(0) = -4],y(t), singsol=all)
 

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

✓ Solution by Mathematica

Time used: 0.087 (sec). Leaf size: 66

DSolve[{y''[t]+y'[t]-6*y[t]==30*UnitStep[t-1]*Exp[-(t-1)],{y[0]==3,y'[0]==-4}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} e^{-3 t} \left (2+e^{5 t}\right ) & t\leq 1 \\ e^{-3 t-2} \left (2 e^2+3 e^5+2 e^{5 t}-5 e^{2 t+3}+e^{5 t+2}\right ) & \text {True} \\ \end {array} \\ \end {array} \]