4.22 problem Problem 3(h)

Internal problem ID [12329]
Internal file name [OUTPUT/10982_Monday_October_02_2023_02_47_42_AM_54582995/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(h).
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]]

\[ \boxed {y^{\prime \prime }+4 y=3 \operatorname {Heaviside}\left (t \right )-3 \operatorname {Heaviside}\left (t -4\right )+\left (2 t -5\right ) \operatorname {Heaviside}\left (t -4\right )} \] With initial conditions \begin {align*} \left [y \left (0\right ) = {\frac {3}{4}}, y^{\prime }\left (0\right ) = 2\right ] \end {align*}

4.22.1 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) &=0\\ q(t) &=4\\ F &=\left (2 t -8\right ) \operatorname {Heaviside}\left (t -4\right )+3 \operatorname {Heaviside}\left (t \right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+4 y = \left (2 t -8\right ) \operatorname {Heaviside}\left (t -4\right )+3 \operatorname {Heaviside}\left (t \right ) \end {align*}

The domain of \(p(t)=0\) 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 Y \left (s \right ) = \frac {2 \,{\mathrm e}^{-4 s}}{s^{2}}+\frac {3}{s}\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&={\frac {3}{4}}\\ y'(0) &=2 \end {align*}

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

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {3 s^{3}+8 s^{2}+8 \,{\mathrm e}^{-4 s}+12 s}{4 s^{2} \left (s^{2}+4\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}+8 s^{2}+8 \,{\mathrm e}^{-4 s}+12 s}{4 s^{2} \left (s^{2}+4\right )}\right )\\ &= \sin \left (2 t \right )+\frac {\operatorname {Heaviside}\left (t -4\right ) \left (2 t -8-\sin \left (2 t -8\right )\right )}{4}+\frac {3}{4} \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} \sin \left (2 t \right )+\frac {3}{4} & t <4 \\ \sin \left (2 t \right )-\frac {5}{4}+\frac {t}{2}-\frac {\sin \left (2 t -8\right )}{4} & 4\le t \end {array}\right . \] Simplifying the solution gives \[ y = \sin \left (2 t \right )-\frac {\left (\left \{\begin {array}{cc} -3 & t <4 \\ 5-2 t +\sin \left (2 t -8\right ) & 4\le t \end {array}\right .\right )}{4} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \sin \left (2 t \right )-\frac {\left (\left \{\begin {array}{cc} -3 & t <4 \\ 5-2 t +\sin \left (2 t -8\right ) & 4\le t \end {array}\right .\right )}{4} \\ \end{align*}

Verification of solutions

\[ y = \sin \left (2 t \right )-\frac {\left (\left \{\begin {array}{cc} -3 & t <4 \\ 5-2 t +\sin \left (2 t -8\right ) & 4\le t \end {array}\right .\right )}{4} \] Verified OK.

4.22.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+4 y=\left (2 t -8\right ) \mathit {Heaviside}\left (t -4\right )+3 \mathit {Heaviside}\left (t \right ), y \left (0\right )=\frac {3}{4}, 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 {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y^{\prime }=-4 y+2 \mathit {Heaviside}\left (t -4\right ) t +3 \mathit {Heaviside}\left (t \right )-8 \mathit {Heaviside}\left (t -4\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=2 \mathit {Heaviside}\left (t -4\right ) t +3 \mathit {Heaviside}\left (t \right )-8 \mathit {Heaviside}\left (t -4\right ) \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+4=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {-16}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-2 \,\mathrm {I}, 2 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )=\cos \left (2 t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=\sin \left (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} \cos \left (2 t \right )+c_{2} \sin \left (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 )=2 \mathit {Heaviside}\left (t -4\right ) t +3 \mathit {Heaviside}\left (t \right )-8 \mathit {Heaviside}\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} \cos \left (2 t \right ) & \sin \left (2 t \right ) \\ -2 \sin \left (2 t \right ) & 2 \cos \left (2 t \right ) \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=2 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\frac {\cos \left (2 t \right ) \left (\int \sin \left (2 t \right ) \left (2 \mathit {Heaviside}\left (t -4\right ) t +3 \mathit {Heaviside}\left (t \right )-8 \mathit {Heaviside}\left (t -4\right )\right )d t \right )}{2}+\frac {\sin \left (2 t \right ) \left (\int \cos \left (2 t \right ) \left (2 \mathit {Heaviside}\left (t -4\right ) t +3 \mathit {Heaviside}\left (t \right )-8 \mathit {Heaviside}\left (t -4\right )\right )d t \right )}{2} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\frac {\left (-\sin \left (2 t \right ) \cos \left (8\right )+\cos \left (2 t \right ) \sin \left (8\right )+2 t -8\right ) \mathit {Heaviside}\left (t -4\right )}{4}-\frac {3 \mathit {Heaviside}\left (t \right ) \left (\cos \left (2 t \right )-1\right )}{4} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} \cos \left (2 t \right )+c_{2} \sin \left (2 t \right )+\frac {\left (-\sin \left (2 t \right ) \cos \left (8\right )+\cos \left (2 t \right ) \sin \left (8\right )+2 t -8\right ) \mathit {Heaviside}\left (t -4\right )}{4}-\frac {3 \mathit {Heaviside}\left (t \right ) \left (\cos \left (2 t \right )-1\right )}{4} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} \cos \left (2 t \right )+c_{2} \sin \left (2 t \right )+\frac {\left (-\sin \left (2 t \right ) \cos \left (8\right )+\cos \left (2 t \right ) \sin \left (8\right )+2 t -8\right ) \mathit {Heaviside}\left (t -4\right )}{4}-\frac {3 \mathit {Heaviside}\left (t \right ) \left (\cos \left (2 t \right )-1\right )}{4} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=\frac {3}{4} \\ {} & {} & \frac {3}{4}=c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-2 c_{1} \sin \left (2 t \right )+2 c_{2} \cos \left (2 t \right )+\frac {\left (-2 \cos \left (2 t \right ) \cos \left (8\right )-2 \sin \left (2 t \right ) \sin \left (8\right )+2\right ) \mathit {Heaviside}\left (t -4\right )}{4}+\frac {\left (-\sin \left (2 t \right ) \cos \left (8\right )+\cos \left (2 t \right ) \sin \left (8\right )+2 t -8\right ) \mathit {Dirac}\left (t -4\right )}{4}-\frac {3 \mathit {Dirac}\left (t \right ) \left (\cos \left (2 t \right )-1\right )}{4}+\frac {3 \sin \left (2 t \right ) \mathit {Heaviside}\left (t \right )}{2} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=2 \\ {} & {} & 2=2 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =\frac {3}{4}, c_{2} =1\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {\left (-\sin \left (2 t \right ) \cos \left (8\right )+\cos \left (2 t \right ) \sin \left (8\right )+2 t -8\right ) \mathit {Heaviside}\left (t -4\right )}{4}+\frac {\left (-3 \mathit {Heaviside}\left (t \right )+3\right ) \cos \left (2 t \right )}{4}+\sin \left (2 t \right )+\frac {3 \mathit {Heaviside}\left (t \right )}{4} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\left (-\sin \left (2 t \right ) \cos \left (8\right )+\cos \left (2 t \right ) \sin \left (8\right )+2 t -8\right ) \mathit {Heaviside}\left (t -4\right )}{4}+\frac {\left (-3 \mathit {Heaviside}\left (t \right )+3\right ) \cos \left (2 t \right )}{4}+\sin \left (2 t \right )+\frac {3 \mathit {Heaviside}\left (t \right )}{4} \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: 5.5 (sec). Leaf size: 29

dsolve([diff(y(t),t$2)+4*y(t)=3*(Heaviside(t)-Heaviside(t-4))+(2*t-5)*Heaviside(t-4),y(0) = 3/4, D(y)(0) = 2],y(t), singsol=all)
 

\[ y \left (t \right ) = -\frac {\operatorname {Heaviside}\left (t -4\right ) \sin \left (2 t -8\right )}{4}+\frac {\operatorname {Heaviside}\left (t -4\right ) t}{2}+\sin \left (2 t \right )-2 \operatorname {Heaviside}\left (t -4\right )+\frac {3}{4} \]

✓ Solution by Mathematica

Time used: 0.069 (sec). Leaf size: 60

DSolve[{y''[t]+4*y[t]==3*(UnitStep[t]-UnitStep[t-4])+(2*t-5)*UnitStep[t-4],{y[0]==3/4,y'[0]==2}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} \sin (2 t)+\frac {3}{4} & 0\leq t\leq 4 \\ \frac {3}{4} \cos (2 t)+\sin (2 t) & t<0 \\ \frac {1}{4} (2 t+\sin (8-2 t)+4 \sin (2 t)-5) & \text {True} \\ \end {array} \\ \end {array} \]