20.4 problem 5

Internal problem ID [13233]
Internal file name [OUTPUT/11889_Tuesday_December_05_2023_12_12_46_PM_81295945/index.tex]

Book: DIFFERENTIAL EQUATIONS by Paul Blanchard, Robert L. Devaney, Glen R. Hall. 4th edition. Brooks/Cole. Boston, USA. 2012
Section: Chapter 6. Laplace transform. Section 6.4. page 608
Problem number: 5.
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]]

\[ \boxed {y^{\prime \prime }+2 y^{\prime }+3 y=\delta \left (-1+t \right )-3 \delta \left (t -4\right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 0] \end {align*}

20.4.1 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) &=2\\ q(t) &=3\\ F &=\delta \left (-1+t \right )-3 \delta \left (t -4\right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+2 y^{\prime }+3 y = \delta \left (-1+t \right )-3 \delta \left (t -4\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 )+3 Y \left (s \right ) = {\mathrm e}^{-s}-3 \,{\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 )+2 s Y \left (s \right )+3 Y \left (s \right ) = {\mathrm e}^{-s}-3 \,{\mathrm e}^{-4 s} \end {align*}

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

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} 0 & t <1 \\ \frac {\sqrt {2}\, {\mathrm e}^{1-t} \sin \left (\sqrt {2}\, \left (-1+t \right )\right )}{2} & t <4 \\ \frac {\sqrt {2}\, {\mathrm e}^{1-t} \sin \left (\sqrt {2}\, \left (-1+t \right )\right )}{2}-\frac {3 \sqrt {2}\, {\mathrm e}^{4-t} \sin \left (\sqrt {2}\, \left (t -4\right )\right )}{2} & 4\le t \end {array}\right . \] Simplifying the solution gives \[ y = \frac {\sqrt {2}\, \left (\left \{\begin {array}{cc} 0 & t <1 \\ {\mathrm e}^{1-t} \sin \left (\sqrt {2}\, \left (-1+t \right )\right ) & t <4 \\ {\mathrm e}^{1-t} \sin \left (\sqrt {2}\, \left (-1+t \right )\right )-3 \,{\mathrm e}^{4-t} \sin \left (\sqrt {2}\, \left (t -4\right )\right ) & 4\le t \end {array}\right .\right )}{2} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\sqrt {2}\, \left (\left \{\begin {array}{cc} 0 & t <1 \\ {\mathrm e}^{1-t} \sin \left (\sqrt {2}\, \left (-1+t \right )\right ) & t <4 \\ {\mathrm e}^{1-t} \sin \left (\sqrt {2}\, \left (-1+t \right )\right )-3 \,{\mathrm e}^{4-t} \sin \left (\sqrt {2}\, \left (t -4\right )\right ) & 4\le t \end {array}\right .\right )}{2} \\ \end{align*}

Verification of solutions

\[ y = \frac {\sqrt {2}\, \left (\left \{\begin {array}{cc} 0 & t <1 \\ {\mathrm e}^{1-t} \sin \left (\sqrt {2}\, \left (-1+t \right )\right ) & t <4 \\ {\mathrm e}^{1-t} \sin \left (\sqrt {2}\, \left (-1+t \right )\right )-3 \,{\mathrm e}^{4-t} \sin \left (\sqrt {2}\, \left (t -4\right )\right ) & 4\le t \end {array}\right .\right )}{2} \] Verified OK.

20.4.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+2 y^{\prime }+3 y=\mathit {Dirac}\left (-1+t \right )-3 \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 \\ {} & {} & \frac {d}{d t}y^{\prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+2 r +3=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-2\right )\pm \left (\sqrt {-8}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-1-\mathrm {I} \sqrt {2}, -1+\mathrm {I} \sqrt {2}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-t} \cos \left (\sqrt {2}\, t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-t} \sin \left (\sqrt {2}\, 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}^{-t} \cos \left (\sqrt {2}\, t \right )+c_{2} {\mathrm e}^{-t} \sin \left (\sqrt {2}\, 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 )=\mathit {Dirac}\left (-1+t \right )-3 \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} \cos \left (\sqrt {2}\, t \right ) & {\mathrm e}^{-t} \sin \left (\sqrt {2}\, t \right ) \\ -{\mathrm e}^{-t} \cos \left (\sqrt {2}\, t \right )-{\mathrm e}^{-t} \sin \left (\sqrt {2}\, t \right ) \sqrt {2} & -{\mathrm e}^{-t} \sin \left (\sqrt {2}\, t \right )+{\mathrm e}^{-t} \cos \left (\sqrt {2}\, t \right ) \sqrt {2} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=\sqrt {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 {\sqrt {2}\, {\mathrm e}^{1-t} \left (\cos \left (\sqrt {2}\, t \right ) \left (\int \left (3 \sin \left (4 \sqrt {2}\right ) \mathit {Dirac}\left (t -4\right ) {\mathrm e}^{3}-\mathit {Dirac}\left (-1+t \right ) \sin \left (\sqrt {2}\right )\right )d t \right )-\sin \left (\sqrt {2}\, t \right ) \left (\int \left (3 \cos \left (4 \sqrt {2}\right ) \mathit {Dirac}\left (t -4\right ) {\mathrm e}^{3}-\mathit {Dirac}\left (-1+t \right ) \cos \left (\sqrt {2}\right )\right )d t \right )\right )}{2} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\frac {\sqrt {2}\, {\mathrm e}^{1-t} \left (3 \cos \left (4 \sqrt {2}\right ) {\mathrm e}^{3} \sin \left (\sqrt {2}\, t \right ) \mathit {Heaviside}\left (t -4\right )-3 \sin \left (4 \sqrt {2}\right ) {\mathrm e}^{3} \cos \left (\sqrt {2}\, t \right ) \mathit {Heaviside}\left (t -4\right )-\cos \left (\sqrt {2}\right ) \sin \left (\sqrt {2}\, t \right ) \mathit {Heaviside}\left (-1+t \right )+\sin \left (\sqrt {2}\right ) \cos \left (\sqrt {2}\, t \right ) \mathit {Heaviside}\left (-1+t \right )\right )}{2} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-t} \cos \left (\sqrt {2}\, t \right )+c_{2} {\mathrm e}^{-t} \sin \left (\sqrt {2}\, t \right )-\frac {\sqrt {2}\, {\mathrm e}^{1-t} \left (3 \cos \left (4 \sqrt {2}\right ) {\mathrm e}^{3} \sin \left (\sqrt {2}\, t \right ) \mathit {Heaviside}\left (t -4\right )-3 \sin \left (4 \sqrt {2}\right ) {\mathrm e}^{3} \cos \left (\sqrt {2}\, t \right ) \mathit {Heaviside}\left (t -4\right )-\cos \left (\sqrt {2}\right ) \sin \left (\sqrt {2}\, t \right ) \mathit {Heaviside}\left (-1+t \right )+\sin \left (\sqrt {2}\right ) \cos \left (\sqrt {2}\, t \right ) \mathit {Heaviside}\left (-1+t \right )\right )}{2} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-t} \cos \left (\sqrt {2}\, t \right )+c_{2} {\mathrm e}^{-t} \sin \left (\sqrt {2}\, t \right )-\frac {\sqrt {2}\, {\mathrm e}^{1-t} \left (3 \cos \left (4 \sqrt {2}\right ) {\mathrm e}^{3} \sin \left (\sqrt {2}\, t \right ) \mathit {Heaviside}\left (t -4\right )-3 \sin \left (4 \sqrt {2}\right ) {\mathrm e}^{3} \cos \left (\sqrt {2}\, t \right ) \mathit {Heaviside}\left (t -4\right )-\cos \left (\sqrt {2}\right ) \sin \left (\sqrt {2}\, t \right ) \mathit {Heaviside}\left (-1+t \right )+\sin \left (\sqrt {2}\right ) \cos \left (\sqrt {2}\, t \right ) \mathit {Heaviside}\left (-1+t \right )\right )}{2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-c_{1} {\mathrm e}^{-t} \cos \left (\sqrt {2}\, t \right )-c_{1} {\mathrm e}^{-t} \sin \left (\sqrt {2}\, t \right ) \sqrt {2}-c_{2} {\mathrm e}^{-t} \sin \left (\sqrt {2}\, t \right )+c_{2} {\mathrm e}^{-t} \cos \left (\sqrt {2}\, t \right ) \sqrt {2}+\frac {\sqrt {2}\, {\mathrm e}^{1-t} \left (3 \cos \left (4 \sqrt {2}\right ) {\mathrm e}^{3} \sin \left (\sqrt {2}\, t \right ) \mathit {Heaviside}\left (t -4\right )-3 \sin \left (4 \sqrt {2}\right ) {\mathrm e}^{3} \cos \left (\sqrt {2}\, t \right ) \mathit {Heaviside}\left (t -4\right )-\cos \left (\sqrt {2}\right ) \sin \left (\sqrt {2}\, t \right ) \mathit {Heaviside}\left (-1+t \right )+\sin \left (\sqrt {2}\right ) \cos \left (\sqrt {2}\, t \right ) \mathit {Heaviside}\left (-1+t \right )\right )}{2}-\frac {\sqrt {2}\, {\mathrm e}^{1-t} \left (3 \cos \left (4 \sqrt {2}\right ) {\mathrm e}^{3} \cos \left (\sqrt {2}\, t \right ) \sqrt {2}\, \mathit {Heaviside}\left (t -4\right )+3 \cos \left (4 \sqrt {2}\right ) {\mathrm e}^{3} \sin \left (\sqrt {2}\, t \right ) \mathit {Dirac}\left (t -4\right )+3 \sin \left (4 \sqrt {2}\right ) {\mathrm e}^{3} \sin \left (\sqrt {2}\, t \right ) \sqrt {2}\, \mathit {Heaviside}\left (t -4\right )-3 \sin \left (4 \sqrt {2}\right ) {\mathrm e}^{3} \cos \left (\sqrt {2}\, t \right ) \mathit {Dirac}\left (t -4\right )-\cos \left (\sqrt {2}\right ) \cos \left (\sqrt {2}\, t \right ) \sqrt {2}\, \mathit {Heaviside}\left (-1+t \right )-\cos \left (\sqrt {2}\right ) \sin \left (\sqrt {2}\, t \right ) \mathit {Dirac}\left (-1+t \right )-\sin \left (\sqrt {2}\right ) \sin \left (\sqrt {2}\, t \right ) \sqrt {2}\, \mathit {Heaviside}\left (-1+t \right )+\sin \left (\sqrt {2}\right ) \cos \left (\sqrt {2}\, t \right ) \mathit {Dirac}\left (-1+t \right )\right )}{2} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-c_{1} +c_{2} \sqrt {2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =0, c_{2} =0\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {\sqrt {2}\, {\mathrm e}^{1-t} \left (3 \cos \left (4 \sqrt {2}\right ) {\mathrm e}^{3} \sin \left (\sqrt {2}\, t \right ) \mathit {Heaviside}\left (t -4\right )-3 \sin \left (4 \sqrt {2}\right ) {\mathrm e}^{3} \cos \left (\sqrt {2}\, t \right ) \mathit {Heaviside}\left (t -4\right )-\cos \left (\sqrt {2}\right ) \sin \left (\sqrt {2}\, t \right ) \mathit {Heaviside}\left (-1+t \right )+\sin \left (\sqrt {2}\right ) \cos \left (\sqrt {2}\, t \right ) \mathit {Heaviside}\left (-1+t \right )\right )}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {\sqrt {2}\, {\mathrm e}^{1-t} \left (3 \cos \left (4 \sqrt {2}\right ) {\mathrm e}^{3} \sin \left (\sqrt {2}\, t \right ) \mathit {Heaviside}\left (t -4\right )-3 \sin \left (4 \sqrt {2}\right ) {\mathrm e}^{3} \cos \left (\sqrt {2}\, t \right ) \mathit {Heaviside}\left (t -4\right )-\cos \left (\sqrt {2}\right ) \sin \left (\sqrt {2}\, t \right ) \mathit {Heaviside}\left (-1+t \right )+\sin \left (\sqrt {2}\right ) \cos \left (\sqrt {2}\, t \right ) \mathit {Heaviside}\left (-1+t \right )\right )}{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: 6.562 (sec). Leaf size: 51

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

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

✓ Solution by Mathematica

Time used: 0.371 (sec). Leaf size: 53

DSolve[{y''[t]+2*y'[t]+3*y[t]==DiracDelta[t-1]-3*DiracDelta[t-4],{y[0]==0,y'[0]==0}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {e^{1-t} \left (\theta (t-1) \sin \left (\sqrt {2} (t-1)\right )-3 e^3 \theta (t-4) \sin \left (\sqrt {2} (t-4)\right )\right )}{\sqrt {2}} \]