19.7 problem 33

Internal problem ID [13228]
Internal file name [OUTPUT/11884_Tuesday_December_05_2023_12_12_43_PM_19494184/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 6.3 page 600
Problem number: 33.
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]]

\[ \boxed {y^{\prime \prime }+4 y^{\prime }+9 y=20 \operatorname {Heaviside}\left (-2+t \right ) \sin \left (-2+t \right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = 2] \end {align*}

19.7.1 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) &=9\\ F &=20 \operatorname {Heaviside}\left (-2+t \right ) \sin \left (-2+t \right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+4 y^{\prime }+9 y = 20 \operatorname {Heaviside}\left (-2+t \right ) \sin \left (-2+t \right ) \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 )+9 Y \left (s \right ) = \frac {20 \,{\mathrm e}^{-2 s}}{s^{2}+1}\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 )-6-s +4 s Y \left (s \right )+9 Y \left (s \right ) = \frac {20 \,{\mathrm e}^{-2 s}}{s^{2}+1} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {s^{3}+6 s^{2}+20 \,{\mathrm e}^{-2 s}+s +6}{\left (s^{2}+1\right ) \left (s^{2}+4 s +9\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^{3}+6 s^{2}+20 \,{\mathrm e}^{-2 s}+s +6}{\left (s^{2}+1\right ) \left (s^{2}+4 s +9\right )}\right )\\ &= \frac {{\mathrm e}^{-2 t} \left (4 \sqrt {5}\, \sin \left (\sqrt {5}\, t \right )+5 \cos \left (\sqrt {5}\, t \right )\right )}{5}+\left ({\mathrm e}^{4-2 t} \cos \left (\sqrt {5}\, \left (-2+t \right )\right )-\cos \left (-2+t \right )+2 \sin \left (-2+t \right )\right ) \operatorname {Heaviside}\left (-2+t \right ) \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} \frac {{\mathrm e}^{-2 t} \left (4 \sqrt {5}\, \sin \left (\sqrt {5}\, t \right )+5 \cos \left (\sqrt {5}\, t \right )\right )}{5} & t <2 \\ \frac {{\mathrm e}^{-2 t} \left (4 \sqrt {5}\, \sin \left (\sqrt {5}\, t \right )+5 \cos \left (\sqrt {5}\, t \right )\right )}{5}+{\mathrm e}^{4-2 t} \cos \left (\sqrt {5}\, \left (-2+t \right )\right )-\cos \left (-2+t \right )+2 \sin \left (-2+t \right ) & 2\le t \end {array}\right . \] Simplifying the solution gives \[ y = \frac {\left (\left \{\begin {array}{cc} {\mathrm e}^{-2 t} \left (4 \sqrt {5}\, \sin \left (\sqrt {5}\, t \right )+5 \cos \left (\sqrt {5}\, t \right )\right ) & t <2 \\ 4 \sin \left (\sqrt {5}\, t \right ) {\mathrm e}^{-2 t} \sqrt {5}+5 \cos \left (\sqrt {5}\, t \right ) {\mathrm e}^{-2 t}+5 \,{\mathrm e}^{4-2 t} \cos \left (\sqrt {5}\, \left (-2+t \right )\right )+10 \sin \left (-2+t \right )-5 \cos \left (-2+t \right ) & 2\le t \end {array}\right .\right )}{5} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (\left \{\begin {array}{cc} {\mathrm e}^{-2 t} \left (4 \sqrt {5}\, \sin \left (\sqrt {5}\, t \right )+5 \cos \left (\sqrt {5}\, t \right )\right ) & t <2 \\ 4 \sin \left (\sqrt {5}\, t \right ) {\mathrm e}^{-2 t} \sqrt {5}+5 \cos \left (\sqrt {5}\, t \right ) {\mathrm e}^{-2 t}+5 \,{\mathrm e}^{4-2 t} \cos \left (\sqrt {5}\, \left (-2+t \right )\right )+10 \sin \left (-2+t \right )-5 \cos \left (-2+t \right ) & 2\le t \end {array}\right .\right )}{5} \\ \end{align*}

Verification of solutions

\[ y = \frac {\left (\left \{\begin {array}{cc} {\mathrm e}^{-2 t} \left (4 \sqrt {5}\, \sin \left (\sqrt {5}\, t \right )+5 \cos \left (\sqrt {5}\, t \right )\right ) & t <2 \\ 4 \sin \left (\sqrt {5}\, t \right ) {\mathrm e}^{-2 t} \sqrt {5}+5 \cos \left (\sqrt {5}\, t \right ) {\mathrm e}^{-2 t}+5 \,{\mathrm e}^{4-2 t} \cos \left (\sqrt {5}\, \left (-2+t \right )\right )+10 \sin \left (-2+t \right )-5 \cos \left (-2+t \right ) & 2\le t \end {array}\right .\right )}{5} \] Verified OK.

19.7.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+4 y^{\prime }+9 y=20 \mathit {Heaviside}\left (-2+t \right ) \sin \left (-2+t \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 \\ {} & {} & \frac {d}{d t}y^{\prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+4 r +9=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-4\right )\pm \left (\sqrt {-20}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-2-\mathrm {I} \sqrt {5}, -2+\mathrm {I} \sqrt {5}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )=\cos \left (\sqrt {5}\, t \right ) {\mathrm e}^{-2 t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-2 t} \sin \left (\sqrt {5}\, 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} \cos \left (\sqrt {5}\, t \right ) {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{-2 t} \sin \left (\sqrt {5}\, 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 )=20 \mathit {Heaviside}\left (-2+t \right ) \sin \left (-2+t \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} \cos \left (\sqrt {5}\, t \right ) {\mathrm e}^{-2 t} & {\mathrm e}^{-2 t} \sin \left (\sqrt {5}\, t \right ) \\ -\sin \left (\sqrt {5}\, t \right ) {\mathrm e}^{-2 t} \sqrt {5}-2 \cos \left (\sqrt {5}\, t \right ) {\mathrm e}^{-2 t} & -2 \,{\mathrm e}^{-2 t} \sin \left (\sqrt {5}\, t \right )+{\mathrm e}^{-2 t} \cos \left (\sqrt {5}\, t \right ) \sqrt {5} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=\sqrt {5}\, {\mathrm e}^{-4 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=4 \sqrt {5}\, {\mathrm e}^{-2 t} \left (-\cos \left (\sqrt {5}\, t \right ) \left (\int {\mathrm e}^{2 t} \sin \left (-2+t \right ) \sin \left (\sqrt {5}\, t \right ) \mathit {Heaviside}\left (-2+t \right )d t \right )+\sin \left (\sqrt {5}\, t \right ) \left (\int {\mathrm e}^{2 t} \sin \left (-2+t \right ) \cos \left (\sqrt {5}\, t \right ) \mathit {Heaviside}\left (-2+t \right )d t \right )\right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\left (-{\mathrm e}^{4-2 t} \cos \left (\sqrt {5}\, \left (-2+t \right )\right )+\cos \left (-2+t \right )-2 \sin \left (-2+t \right )\right ) \mathit {Heaviside}\left (-2+t \right ) \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} \cos \left (\sqrt {5}\, t \right ) {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{-2 t} \sin \left (\sqrt {5}\, t \right )-\left (-{\mathrm e}^{4-2 t} \cos \left (\sqrt {5}\, \left (-2+t \right )\right )+\cos \left (-2+t \right )-2 \sin \left (-2+t \right )\right ) \mathit {Heaviside}\left (-2+t \right ) \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} \cos \left (\sqrt {5}\, t \right ) {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{-2 t} \sin \left (\sqrt {5}\, t \right )-\left (-{\mathrm e}^{4-2 t} \cos \left (\sqrt {5}\, \left (-2+t \right )\right )+\cos \left (-2+t \right )-2 \sin \left (-2+t \right )\right ) \mathit {Heaviside}\left (-2+t \right ) \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-c_{1} \sqrt {5}\, \sin \left (\sqrt {5}\, t \right ) {\mathrm e}^{-2 t}-2 c_{1} \cos \left (\sqrt {5}\, t \right ) {\mathrm e}^{-2 t}-2 c_{2} {\mathrm e}^{-2 t} \sin \left (\sqrt {5}\, t \right )+c_{2} {\mathrm e}^{-2 t} \sqrt {5}\, \cos \left (\sqrt {5}\, t \right )-\left (2 \,{\mathrm e}^{4-2 t} \cos \left (\sqrt {5}\, \left (-2+t \right )\right )+{\mathrm e}^{4-2 t} \sqrt {5}\, \sin \left (\sqrt {5}\, \left (-2+t \right )\right )-\sin \left (-2+t \right )-2 \cos \left (-2+t \right )\right ) \mathit {Heaviside}\left (-2+t \right )-\left (-{\mathrm e}^{4-2 t} \cos \left (\sqrt {5}\, \left (-2+t \right )\right )+\cos \left (-2+t \right )-2 \sin \left (-2+t \right )\right ) \mathit {Dirac}\left (-2+t \right ) \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=2 \\ {} & {} & 2=-2 c_{1} +c_{2} \sqrt {5} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =1, c_{2} =\frac {4 \sqrt {5}}{5}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\mathit {Heaviside}\left (-2+t \right ) \cos \left (\sqrt {5}\, \left (-2+t \right )\right ) {\mathrm e}^{4-2 t}+\cos \left (\sqrt {5}\, t \right ) {\mathrm e}^{-2 t}+\frac {4 \sin \left (\sqrt {5}\, t \right ) {\mathrm e}^{-2 t} \sqrt {5}}{5}-\mathit {Heaviside}\left (-2+t \right ) \left (-2 \sin \left (-2+t \right )+\cos \left (-2+t \right )\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\mathit {Heaviside}\left (-2+t \right ) \cos \left (\sqrt {5}\, \left (-2+t \right )\right ) {\mathrm e}^{4-2 t}+\cos \left (\sqrt {5}\, t \right ) {\mathrm e}^{-2 t}+\frac {4 \sin \left (\sqrt {5}\, t \right ) {\mathrm e}^{-2 t} \sqrt {5}}{5}-\mathit {Heaviside}\left (-2+t \right ) \left (-2 \sin \left (-2+t \right )+\cos \left (-2+t \right )\right ) \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: 6.953 (sec). Leaf size: 64

dsolve([diff(y(t),t$2)+4*diff(y(t),t)+9*y(t)=20*Heaviside(t-2)*sin(t-2),y(0) = 1, D(y)(0) = 2],y(t), singsol=all)
 

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

✓ Solution by Mathematica

Time used: 2.391 (sec). Leaf size: 115

DSolve[{y''[t]+4*y'[t]+9*y[t]==20*UnitStep[t-2]*Sin[t-2],{y[0]==1,y'[0]==2}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} -\cos (2-t)+e^{4-2 t} \cos \left (\sqrt {5} (t-2)\right )+e^{-2 t} \cos \left (\sqrt {5} t\right )-2 \sin (2-t)+\frac {4 e^{-2 t} \sin \left (\sqrt {5} t\right )}{\sqrt {5}} & t>2 \\ \frac {1}{5} e^{-2 t} \left (5 \cos \left (\sqrt {5} t\right )+4 \sqrt {5} \sin \left (\sqrt {5} t\right )\right ) & \text {True} \\ \end {array} \\ \end {array} \]