8.12 problem 12

Internal problem ID [6708]
Internal file name [OUTPUT/5956_Sunday_June_05_2022_04_04_11_PM_56284687/index.tex]

Book: DIFFERENTIAL EQUATIONS with Boundary Value Problems. DENNIS G. ZILL, WARREN S. WRIGHT, MICHAEL R. CULLEN. Brooks/Cole. Boston, MA. 2013. 8th edition.
Section: CHAPTER 7 THE LAPLACE TRANSFORM. EXERCISES 7.5. Page 315
Problem number: 12.
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]]

\[ \boxed {y^{\prime \prime }-7 y^{\prime }+6 y={\mathrm e}^{t}+\delta \left (-2+t \right )+\delta \left (t -4\right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 0] \end {align*}

8.12.1 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) &=-7\\ q(t) &=6\\ F &={\mathrm e}^{t}+\delta \left (-2+t \right )+\delta \left (t -4\right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }-7 y^{\prime }+6 y = {\mathrm e}^{t}+\delta \left (-2+t \right )+\delta \left (t -4\right ) \end {align*}

The domain of \(p(t)=-7\) 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 )-7 s Y \left (s \right )+7 y \left (0\right )+6 Y \left (s \right ) = \frac {1}{s -1}+{\mathrm e}^{-2 s}+{\mathrm e}^{-4 s}\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 )-7 s Y \left (s \right )+6 Y \left (s \right ) = \frac {1}{s -1}+{\mathrm e}^{-2 s}+{\mathrm e}^{-4 s} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {{\mathrm e}^{-2 s} s +{\mathrm e}^{-4 s} s -{\mathrm e}^{-2 s}-{\mathrm e}^{-4 s}+1}{\left (s -1\right ) \left (s^{2}-7 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 {{\mathrm e}^{-2 s} s +{\mathrm e}^{-4 s} s -{\mathrm e}^{-2 s}-{\mathrm e}^{-4 s}+1}{\left (s -1\right ) \left (s^{2}-7 s +6\right )}\right )\\ &= \frac {{\mathrm e}^{6 t}}{25}+\frac {{\mathrm e}^{-12+6 t}}{5}-\frac {{\mathrm e}^{-2+t}}{5}-\frac {{\mathrm e}^{t -4}}{5}+\frac {{\mathrm e}^{6 t -24}}{5}-\frac {{\mathrm e}^{t} \left (5 t +1\right )}{25}+\frac {\left (-{\mathrm e}^{-12+6 t}+{\mathrm e}^{-2+t}\right ) \operatorname {Heaviside}\left (2-t \right )}{5}+\frac {\left (-{\mathrm e}^{6 t -24}+{\mathrm e}^{t -4}\right ) \operatorname {Heaviside}\left (-t +4\right )}{5} \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} \frac {{\mathrm e}^{6 t}}{25}-\frac {{\mathrm e}^{t} \left (5 t +1\right )}{25} & t \le 2 \\ \frac {{\mathrm e}^{6 t}}{25}+\frac {{\mathrm e}^{-12+6 t}}{5}-\frac {{\mathrm e}^{-2+t}}{5}-\frac {{\mathrm e}^{t} \left (5 t +1\right )}{25} & t \le 4 \\ \frac {{\mathrm e}^{6 t}}{25}+\frac {{\mathrm e}^{-12+6 t}}{5}-\frac {{\mathrm e}^{-2+t}}{5}-\frac {{\mathrm e}^{t -4}}{5}+\frac {{\mathrm e}^{6 t -24}}{5}-\frac {{\mathrm e}^{t} \left (5 t +1\right )}{25} & 4

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (\left \{\begin {array}{cc} 0 & t \le 2 \\ {\mathrm e}^{-12+6 t}-{\mathrm e}^{-2+t} & t \le 4 \\ {\mathrm e}^{-12+6 t}-{\mathrm e}^{-2+t}-{\mathrm e}^{t -4}+{\mathrm e}^{6 t -24} & 4

Verification of solutions

\[ y = \frac {\left (\left \{\begin {array}{cc} 0 & t \le 2 \\ {\mathrm e}^{-12+6 t}-{\mathrm e}^{-2+t} & t \le 4 \\ {\mathrm e}^{-12+6 t}-{\mathrm e}^{-2+t}-{\mathrm e}^{t -4}+{\mathrm e}^{6 t -24} & 4

8.12.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-7 y^{\prime }+6 y={\mathrm e}^{t}+\mathit {Dirac}\left (-2+t \right )+\mathit {Dirac}\left (t -4\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}-7 r +6=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -1\right ) \left (r -6\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (1, 6\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}^{6 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}^{6 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 )={\mathrm e}^{t}+\mathit {Dirac}\left (-2+t \right )+\mathit {Dirac}\left (t -4\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}^{6 t} \\ {\mathrm e}^{t} & 6 \,{\mathrm e}^{6 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}^{7 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 \left (1+{\mathrm e}^{-2} \mathit {Dirac}\left (-2+t \right )+{\mathrm e}^{-4} \mathit {Dirac}\left (t -4\right )\right )d t \right )}{5}+\frac {{\mathrm e}^{6 t} \left (\int \left ({\mathrm e}^{-12} \mathit {Dirac}\left (-2+t \right )+{\mathrm e}^{-24} \mathit {Dirac}\left (t -4\right )+{\mathrm e}^{-5 t}\right )d t \right )}{5} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\frac {\mathit {Heaviside}\left (-2+t \right ) {\mathrm e}^{-12+6 t}}{5}+\frac {\mathit {Heaviside}\left (t -4\right ) {\mathrm e}^{6 t -24}}{5}-\frac {\mathit {Heaviside}\left (-2+t \right ) {\mathrm e}^{-2+t}}{5}-\frac {\mathit {Heaviside}\left (t -4\right ) {\mathrm e}^{t -4}}{5}+\frac {\left (-5 t -1\right ) {\mathrm e}^{t}}{25} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{t}+c_{2} {\mathrm e}^{6 t}+\frac {\mathit {Heaviside}\left (-2+t \right ) {\mathrm e}^{-12+6 t}}{5}+\frac {\mathit {Heaviside}\left (t -4\right ) {\mathrm e}^{6 t -24}}{5}-\frac {\mathit {Heaviside}\left (-2+t \right ) {\mathrm e}^{-2+t}}{5}-\frac {\mathit {Heaviside}\left (t -4\right ) {\mathrm e}^{t -4}}{5}+\frac {\left (-5 t -1\right ) {\mathrm e}^{t}}{25} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{t}+c_{2} {\mathrm e}^{6 t}+\frac {\mathit {Heaviside}\left (-2+t \right ) {\mathrm e}^{-12+6 t}}{5}+\frac {\mathit {Heaviside}\left (t -4\right ) {\mathrm e}^{6 t -24}}{5}-\frac {\mathit {Heaviside}\left (-2+t \right ) {\mathrm e}^{-2+t}}{5}-\frac {\mathit {Heaviside}\left (t -4\right ) {\mathrm e}^{t -4}}{5}+\frac {\left (-5 t -1\right ) {\mathrm e}^{t}}{25} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} +c_{2} -\frac {1}{25} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=c_{1} {\mathrm e}^{t}+6 c_{2} {\mathrm e}^{6 t}+\frac {\mathit {Dirac}\left (-2+t \right ) {\mathrm e}^{-12+6 t}}{5}+\frac {6 \mathit {Heaviside}\left (-2+t \right ) {\mathrm e}^{-12+6 t}}{5}+\frac {\mathit {Dirac}\left (t -4\right ) {\mathrm e}^{6 t -24}}{5}+\frac {6 \mathit {Heaviside}\left (t -4\right ) {\mathrm e}^{6 t -24}}{5}-\frac {\mathit {Dirac}\left (-2+t \right ) {\mathrm e}^{-2+t}}{5}-\frac {\mathit {Heaviside}\left (-2+t \right ) {\mathrm e}^{-2+t}}{5}-\frac {\mathit {Dirac}\left (t -4\right ) {\mathrm e}^{t -4}}{5}-\frac {\mathit {Heaviside}\left (t -4\right ) {\mathrm e}^{t -4}}{5}-\frac {{\mathrm e}^{t}}{5}+\frac {\left (-5 t -1\right ) {\mathrm e}^{t}}{25} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=c_{1} +6 c_{2} -\frac {6}{25} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =0, c_{2} =\frac {1}{25}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {{\mathrm e}^{6 t}}{25}+\frac {\mathit {Heaviside}\left (-2+t \right ) {\mathrm e}^{-12+6 t}}{5}+\frac {\mathit {Heaviside}\left (t -4\right ) {\mathrm e}^{6 t -24}}{5}-\frac {\mathit {Heaviside}\left (-2+t \right ) {\mathrm e}^{-2+t}}{5}-\frac {\mathit {Heaviside}\left (t -4\right ) {\mathrm e}^{t -4}}{5}+\frac {\left (-5 t -1\right ) {\mathrm e}^{t}}{25} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {{\mathrm e}^{6 t}}{25}+\frac {\mathit {Heaviside}\left (-2+t \right ) {\mathrm e}^{-12+6 t}}{5}+\frac {\mathit {Heaviside}\left (t -4\right ) {\mathrm e}^{6 t -24}}{5}-\frac {\mathit {Heaviside}\left (-2+t \right ) {\mathrm e}^{-2+t}}{5}-\frac {\mathit {Heaviside}\left (t -4\right ) {\mathrm e}^{t -4}}{5}+\frac {\left (-5 t -1\right ) {\mathrm e}^{t}}{25} \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: 2.266 (sec). Leaf size: 90

dsolve([diff(y(t),t$2)-7*diff(y(t),t)+6*y(t)=exp(t)+Dirac(t-2)+Dirac(t-4),y(0) = 0, D(y)(0) = 0],y(t), singsol=all)
 

\[ y \left (t \right ) = \frac {{\mathrm e}^{-24+6 t} \operatorname {Heaviside}\left (t -4\right )}{5}+\frac {{\mathrm e}^{-12+6 t} \operatorname {Heaviside}\left (t -2\right )}{5}-\frac {{\mathrm e}^{t -4} \operatorname {Heaviside}\left (t -4\right )}{5}-\frac {{\mathrm e}^{t -2} \operatorname {Heaviside}\left (t -2\right )}{5}+\frac {{\mathrm e}^{6 t}}{25}+\frac {\left (-5 t -1\right ) {\mathrm e}^{t}}{25} \]

✓ Solution by Mathematica

Time used: 0.115 (sec). Leaf size: 67

DSolve[{y''[t]-7*y'[t]+6*y[t]==Exp[t]+DiracDelta[t-2]+DiracDelta[t-4],{y[0]==9,y'[0]==0}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {1}{25} e^{t-24} \left (5 \left (e^{5 t}-e^{20}\right ) \theta (t-4)+5 \left (e^{5 t+12}-e^{22}\right ) \theta (t-2)+e^{24} \left (-5 t-44 e^{5 t}+269\right )\right ) \]