4.21 problem Problem 3(g)

Internal problem ID [12328]
Internal file name [OUTPUT/10981_Monday_October_02_2023_02_47_41_AM_74918722/index.tex]

Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A. Dobrushkin. CRC Press 2015
Section: Chapter 5.6 Laplace transform. Nonhomogeneous equations. Problems page 368
Problem number: Problem 3(g).
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]]

\[ \boxed {y^{\prime \prime }+4 y^{\prime }+13 y=39 \operatorname {Heaviside}\left (t \right )-507 \left (t -2\right ) \operatorname {Heaviside}\left (t -2\right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 3, y^{\prime }\left (0\right ) = 1] \end {align*}

4.21.1 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) &=13\\ F &=\left (-507 t +1014\right ) \operatorname {Heaviside}\left (t -2\right )+39 \operatorname {Heaviside}\left (t \right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+4 y^{\prime }+13 y = \left (-507 t +1014\right ) \operatorname {Heaviside}\left (t -2\right )+39 \operatorname {Heaviside}\left (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 )+13 Y \left (s \right ) = -\frac {507 \,{\mathrm e}^{-2 s}}{s^{2}}+\frac {39}{s}\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&=3\\ y'(0) &=1 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-13-3 s +4 s Y \left (s \right )+13 Y \left (s \right ) = -\frac {507 \,{\mathrm e}^{-2 s}}{s^{2}}+\frac {39}{s} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = -\frac {-3 s^{3}-13 s^{2}+507 \,{\mathrm e}^{-2 s}-39 s}{s^{2} \left (s^{2}+4 s +13\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^{3}-13 s^{2}+507 \,{\mathrm e}^{-2 s}-39 s}{s^{2} \left (s^{2}+4 s +13\right )}\right )\\ &= 3+\frac {{\mathrm e}^{-2 t} \sin \left (3 t \right )}{3}+\left (90-39 t +\left (-12 \cos \left (3 t -6\right )+5 \sin \left (3 t -6\right )\right ) {\mathrm e}^{-2 t +4}\right ) \operatorname {Heaviside}\left (t -2\right ) \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} 3+\frac {{\mathrm e}^{-2 t} \sin \left (3 t \right )}{3} & t <2 \\ 93+\frac {{\mathrm e}^{-2 t} \sin \left (3 t \right )}{3}-39 t +\left (-12 \cos \left (3 t -6\right )+5 \sin \left (3 t -6\right )\right ) {\mathrm e}^{-2 t +4} & 2\le t \end {array}\right . \] Simplifying the solution gives \[ y = \frac {{\mathrm e}^{-2 t} \sin \left (3 t \right )}{3}+\left (\left \{\begin {array}{cc} 3 & t <2 \\ 93-39 t +\left (-12 \cos \left (3 t -6\right )+5 \sin \left (3 t -6\right )\right ) {\mathrm e}^{-2 t +4} & 2\le t \end {array}\right .\right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {{\mathrm e}^{-2 t} \sin \left (3 t \right )}{3}+\left (\left \{\begin {array}{cc} 3 & t <2 \\ 93-39 t +\left (-12 \cos \left (3 t -6\right )+5 \sin \left (3 t -6\right )\right ) {\mathrm e}^{-2 t +4} & 2\le t \end {array}\right .\right ) \\ \end{align*}

Verification of solutions

\[ y = \frac {{\mathrm e}^{-2 t} \sin \left (3 t \right )}{3}+\left (\left \{\begin {array}{cc} 3 & t <2 \\ 93-39 t +\left (-12 \cos \left (3 t -6\right )+5 \sin \left (3 t -6\right )\right ) {\mathrm e}^{-2 t +4} & 2\le t \end {array}\right .\right ) \] Verified OK.

4.21.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+4 y^{\prime }+13 y=\left (-507 t +1014\right ) \mathit {Heaviside}\left (t -2\right )+39 \mathit {Heaviside}\left (t \right ), y \left (0\right )=3, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d t}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y^{\prime }=-13 y-507 t \mathit {Heaviside}\left (t -2\right )-4 y^{\prime }+1014 \mathit {Heaviside}\left (t -2\right )+39 \mathit {Heaviside}\left (t \right ) \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y^{\prime }+4 y^{\prime }+13 y=-507 t \mathit {Heaviside}\left (t -2\right )+39 \mathit {Heaviside}\left (t \right )+1014 \mathit {Heaviside}\left (t -2\right ) \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+4 r +13=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-4\right )\pm \left (\sqrt {-36}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-2-3 \,\mathrm {I}, -2+3 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-2 t} \cos \left (3 t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-2 t} \sin \left (3 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}^{-2 t} \cos \left (3 t \right )+c_{2} {\mathrm e}^{-2 t} \sin \left (3 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 )=-507 t \mathit {Heaviside}\left (t -2\right )+39 \mathit {Heaviside}\left (t \right )+1014 \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}^{-2 t} \cos \left (3 t \right ) & {\mathrm e}^{-2 t} \sin \left (3 t \right ) \\ -2 \,{\mathrm e}^{-2 t} \cos \left (3 t \right )-3 \,{\mathrm e}^{-2 t} \sin \left (3 t \right ) & -2 \,{\mathrm e}^{-2 t} \sin \left (3 t \right )+3 \,{\mathrm e}^{-2 t} \cos \left (3 t \right ) \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}^{-4 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=13 \,{\mathrm e}^{-2 t} \left (\cos \left (3 t \right ) \left (\int {\mathrm e}^{2 t} \sin \left (3 t \right ) \left (13 t \mathit {Heaviside}\left (t -2\right )-26 \mathit {Heaviside}\left (t -2\right )-\mathit {Heaviside}\left (t \right )\right )d t \right )-\sin \left (3 t \right ) \left (\int {\mathrm e}^{2 t} \cos \left (3 t \right ) \left (13 t \mathit {Heaviside}\left (t -2\right )-26 \mathit {Heaviside}\left (t -2\right )-\mathit {Heaviside}\left (t \right )\right )d t \right )\right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-12 \left (\left (\cos \left (6\right )+\frac {5 \sin \left (6\right )}{12}\right ) \cos \left (3 t \right )-\frac {5 \left (\cos \left (6\right )-\frac {12 \sin \left (6\right )}{5}\right ) \sin \left (3 t \right )}{12}\right ) \mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{-2 t +4}+\left (-39 t +90\right ) \mathit {Heaviside}\left (t -2\right )-3 \left ({\mathrm e}^{-2 t} \cos \left (3 t \right )+\frac {2 \,{\mathrm e}^{-2 t} \sin \left (3 t \right )}{3}-1\right ) \mathit {Heaviside}\left (t \right ) \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-2 t} \cos \left (3 t \right )+c_{2} {\mathrm e}^{-2 t} \sin \left (3 t \right )-12 \left (\left (\cos \left (6\right )+\frac {5 \sin \left (6\right )}{12}\right ) \cos \left (3 t \right )-\frac {5 \left (\cos \left (6\right )-\frac {12 \sin \left (6\right )}{5}\right ) \sin \left (3 t \right )}{12}\right ) \mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{-2 t +4}+\left (-39 t +90\right ) \mathit {Heaviside}\left (t -2\right )-3 \left ({\mathrm e}^{-2 t} \cos \left (3 t \right )+\frac {2 \,{\mathrm e}^{-2 t} \sin \left (3 t \right )}{3}-1\right ) \mathit {Heaviside}\left (t \right ) \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-2 t} \cos \left (3 t \right )+c_{2} {\mathrm e}^{-2 t} \sin \left (3 t \right )-12 \left (\left (\cos \left (6\right )+\frac {5 \sin \left (6\right )}{12}\right ) \cos \left (3 t \right )-\frac {5 \left (\cos \left (6\right )-\frac {12 \sin \left (6\right )}{5}\right ) \sin \left (3 t \right )}{12}\right ) \mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{-2 t +4}+\left (-39 t +90\right ) \mathit {Heaviside}\left (t -2\right )-3 \left ({\mathrm e}^{-2 t} \cos \left (3 t \right )+\frac {2 {\mathrm e}^{-2 t} \sin \left (3 t \right )}{3}-1\right ) \mathit {Heaviside}\left (t \right ) \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=3 \\ {} & {} & 3=c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-2 c_{1} {\mathrm e}^{-2 t} \cos \left (3 t \right )-3 c_{1} {\mathrm e}^{-2 t} \sin \left (3 t \right )-2 c_{2} {\mathrm e}^{-2 t} \sin \left (3 t \right )+3 c_{2} {\mathrm e}^{-2 t} \cos \left (3 t \right )-12 \left (-3 \left (\cos \left (6\right )+\frac {5 \sin \left (6\right )}{12}\right ) \sin \left (3 t \right )-\frac {5 \left (\cos \left (6\right )-\frac {12 \sin \left (6\right )}{5}\right ) \cos \left (3 t \right )}{4}\right ) \mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{-2 t +4}-12 \left (\left (\cos \left (6\right )+\frac {5 \sin \left (6\right )}{12}\right ) \cos \left (3 t \right )-\frac {5 \left (\cos \left (6\right )-\frac {12 \sin \left (6\right )}{5}\right ) \sin \left (3 t \right )}{12}\right ) \mathit {Dirac}\left (t -2\right ) {\mathrm e}^{-2 t +4}+24 \left (\left (\cos \left (6\right )+\frac {5 \sin \left (6\right )}{12}\right ) \cos \left (3 t \right )-\frac {5 \left (\cos \left (6\right )-\frac {12 \sin \left (6\right )}{5}\right ) \sin \left (3 t \right )}{12}\right ) \mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{-2 t +4}-39 \mathit {Heaviside}\left (t -2\right )+\left (-39 t +90\right ) \mathit {Dirac}\left (t -2\right )+13 \sin \left (3 t \right ) \mathit {Heaviside}\left (t \right ) {\mathrm e}^{-2 t}-3 \left ({\mathrm e}^{-2 t} \cos \left (3 t \right )+\frac {2 \,{\mathrm e}^{-2 t} \sin \left (3 t \right )}{3}-1\right ) \mathit {Dirac}\left (t \right ) \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=-2 c_{1} +3 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =3, c_{2} =\frac {7}{3}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-12 \left (\left (\cos \left (6\right )+\frac {5 \sin \left (6\right )}{12}\right ) \cos \left (3 t \right )-\frac {5 \left (\cos \left (6\right )-\frac {12 \sin \left (6\right )}{5}\right ) \sin \left (3 t \right )}{12}\right ) \mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{-2 t +4}+3 \left (30-13 t \right ) \mathit {Heaviside}\left (t -2\right )-3 \,{\mathrm e}^{-2 t} \left (\mathit {Heaviside}\left (t \right )-1\right ) \cos \left (3 t \right )+\frac {\left (-6 \mathit {Heaviside}\left (t \right )+7\right ) \sin \left (3 t \right ) {\mathrm e}^{-2 t}}{3}+3 \mathit {Heaviside}\left (t \right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-12 \left (\left (\cos \left (6\right )+\frac {5 \sin \left (6\right )}{12}\right ) \cos \left (3 t \right )-\frac {5 \left (\cos \left (6\right )-\frac {12 \sin \left (6\right )}{5}\right ) \sin \left (3 t \right )}{12}\right ) \mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{-2 t +4}+3 \left (30-13 t \right ) \mathit {Heaviside}\left (t -2\right )-3 \,{\mathrm e}^{-2 t} \left (\mathit {Heaviside}\left (t \right )-1\right ) \cos \left (3 t \right )+\frac {\left (-6 \mathit {Heaviside}\left (t \right )+7\right ) \sin \left (3 t \right ) {\mathrm e}^{-2 t}}{3}+3 \mathit {Heaviside}\left (t \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: 5.844 (sec). Leaf size: 50

dsolve([diff(y(t),t$2)+4*diff(y(t),t)+13*y(t)=39*Heaviside(t)-507*(t-2)*Heaviside(t-2),y(0) = 3, D(y)(0) = 1],y(t), singsol=all)
 

\[ y \left (t \right ) = 3-12 \left (\left (-\frac {5 \cos \left (6\right )}{12}+\sin \left (6\right )\right ) \sin \left (3 t \right )+\cos \left (3 t \right ) \left (\cos \left (6\right )+\frac {5 \sin \left (6\right )}{12}\right )\right ) \operatorname {Heaviside}\left (t -2\right ) {\mathrm e}^{-2 t +4}+3 \left (30-13 t \right ) \operatorname {Heaviside}\left (t -2\right )+\frac {{\mathrm e}^{-2 t} \sin \left (3 t \right )}{3} \]

✓ Solution by Mathematica

Time used: 0.097 (sec). Leaf size: 103

DSolve[{y''[t]+4*y'[t]+13*y[t]==39*UnitStep[t]-507*(t-2)*UnitStep[t-2],{y[0]==3,y'[0]==1}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} -39 t-12 e^{4-2 t} \cos (6-3 t)-5 e^{4-2 t} \sin (6-3 t)+\frac {1}{3} e^{-2 t} \sin (3 t)+93 & t>2 \\ \frac {1}{3} e^{-2 t} \sin (3 t)+3 & 0\leq t\leq 2 \\ \frac {1}{3} e^{-2 t} (9 \cos (3 t)+7 \sin (3 t)) & \text {True} \\ \end {array} \\ \end {array} \]