4.5 problem 5

Internal problem ID [849]
Internal file name [OUTPUT/849_Sunday_June_05_2022_01_51_22_AM_5247366/index.tex]

Book: Elementary differential equations and boundary value problems, 11th ed., Boyce, DiPrima, Meade
Section: Chapter 6.4, The Laplace Transform. Differential equations with discontinuous forcing functions. page 268
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]]

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

4.5.1 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) &=1\\ q(t) &={\frac {5}{4}}\\ F &=\frac {\left (-2 t +\pi \right ) \operatorname {Heaviside}\left (t -\frac {\pi }{2}\right )}{2}+t \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+y^{\prime }+\frac {5 y}{4} = \frac {\left (-2 t +\pi \right ) \operatorname {Heaviside}\left (t -\frac {\pi }{2}\right )}{2}+t \end {align*}

The domain of \(p(t)=1\) 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 )+s Y \left (s \right )-y \left (0\right )+\frac {5 Y \left (s \right )}{4} = \frac {-{\mathrm e}^{-\frac {s \pi }{2}}+1}{s^{2}}\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 )+s Y \left (s \right )+\frac {5 Y \left (s \right )}{4} = \frac {-{\mathrm e}^{-\frac {s \pi }{2}}+1}{s^{2}} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = -\frac {4 \left ({\mathrm e}^{-\frac {s \pi }{2}}-1\right )}{s^{2} \left (4 s^{2}+4 s +5\right )} \end {align*}

Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (-\frac {4 \left ({\mathrm e}^{-\frac {s \pi }{2}}-1\right )}{s^{2} \left (4 s^{2}+4 s +5\right )}\right )\\ &= \frac {4 \,{\mathrm e}^{-\frac {t}{2}} \left (4 \cos \left (t \right )-3 \sin \left (t \right )\right )}{25}+\frac {4 \operatorname {Heaviside}\left (-t +\frac {\pi }{2}\right ) \left (-4+5 t \right )}{25}+\frac {2 \left (5 \pi -2 \,{\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}} \left (3 \cos \left (t \right )+4 \sin \left (t \right )\right )\right ) \operatorname {Heaviside}\left (t -\frac {\pi }{2}\right )}{25} \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} \frac {4 \,{\mathrm e}^{-\frac {t}{2}} \left (4 \cos \left (t \right )-3 \sin \left (t \right )\right )}{25}-\frac {16}{25}+\frac {4 t}{5} & t <\frac {\pi }{2} \\ -\frac {12 \,{\mathrm e}^{-\frac {\pi }{4}}}{25}-\frac {32}{25}+\frac {4 \pi }{5} & t =\frac {\pi }{2} \\ \frac {4 \,{\mathrm e}^{-\frac {t}{2}} \left (4 \cos \left (t \right )-3 \sin \left (t \right )\right )}{25}+\frac {2 \pi }{5}-\frac {4 \,{\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}} \left (3 \cos \left (t \right )+4 \sin \left (t \right )\right )}{25} & \frac {\pi }{2}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {2 \left (\left \{\begin {array}{cc} 8-10 t +\left (-8 \cos \left (t \right )+6 \sin \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}} & t <\frac {\pi }{2} \\ 6 \,{\mathrm e}^{-\frac {\pi }{4}}+16-10 \pi & t &=\frac {\pi }{2} \\ -5 \pi +\left (-8 \cos \left (t \right )+6 \sin \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}}+\left (6 \cos \left (t \right )+8 \sin \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}} & \frac {\pi }{2}

Verification of solutions

\[ y = -\frac {2 \left (\left \{\begin {array}{cc} 8-10 t +\left (-8 \cos \left (t \right )+6 \sin \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}} & t <\frac {\pi }{2} \\ 6 \,{\mathrm e}^{-\frac {\pi }{4}}+16-10 \pi & t =\frac {\pi }{2} \\ -5 \pi +\left (-8 \cos \left (t \right )+6 \sin \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}}+\left (6 \cos \left (t \right )+8 \sin \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}} & \frac {\pi }{2}

4.5.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+y^{\prime }+\frac {5 y}{4}=\frac {\left (-2 t +\pi \right ) \mathit {Heaviside}\left (t -\frac {\pi }{2}\right )}{2}+t , 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 {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y^{\prime }=-\frac {5 y}{4}-y^{\prime }+\frac {\mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) \pi }{2}-\mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) t +t \\ \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 }+y^{\prime }+\frac {5 y}{4}=t -\mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) t +\frac {\mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) \pi }{2} \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+r +\frac {5}{4}=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-1\right )\pm \left (\sqrt {-4}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-\frac {1}{2}-\mathrm {I}, -\frac {1}{2}+\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-\frac {t}{2}} \cos \left (t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-\frac {t}{2}} \sin \left (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}^{-\frac {t}{2}} \cos \left (t \right )+c_{2} {\mathrm e}^{-\frac {t}{2}} \sin \left (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 )=t -\mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) t +\frac {\mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) \pi }{2}\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}^{-\frac {t}{2}} \cos \left (t \right ) & {\mathrm e}^{-\frac {t}{2}} \sin \left (t \right ) \\ -\frac {{\mathrm e}^{-\frac {t}{2}} \cos \left (t \right )}{2}-{\mathrm e}^{-\frac {t}{2}} \sin \left (t \right ) & -\frac {{\mathrm e}^{-\frac {t}{2}} \sin \left (t \right )}{2}+{\mathrm e}^{-\frac {t}{2}} \cos \left (t \right ) \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )={\mathrm e}^{-t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\frac {{\mathrm e}^{-\frac {t}{2}} \left (\cos \left (t \right ) \left (\int -2 \,{\mathrm e}^{\frac {t}{2}} \sin \left (t \right ) \left (\mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) \left (t -\frac {\pi }{2}\right )-t \right )d t \right )-\sin \left (t \right ) \left (\int -2 \,{\mathrm e}^{\frac {t}{2}} \cos \left (t \right ) \left (\mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) \left (t -\frac {\pi }{2}\right )-t \right )d t \right )\right )}{2} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\frac {16}{25}-\frac {12 \left (\cos \left (t \right )+\frac {4 \sin \left (t \right )}{3}\right ) \mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) {\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}}}{25}+\frac {2 \left (8-10 t +5 \pi \right ) \mathit {Heaviside}\left (t -\frac {\pi }{2}\right )}{25}+\frac {4 t}{5} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-\frac {t}{2}} \cos \left (t \right )+c_{2} {\mathrm e}^{-\frac {t}{2}} \sin \left (t \right )-\frac {16}{25}-\frac {12 \left (\cos \left (t \right )+\frac {4 \sin \left (t \right )}{3}\right ) \mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) {\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}}}{25}+\frac {2 \left (8-10 t +5 \pi \right ) \mathit {Heaviside}\left (t -\frac {\pi }{2}\right )}{25}+\frac {4 t}{5} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-\frac {t}{2}} \cos \left (t \right )+c_{2} {\mathrm e}^{-\frac {t}{2}} \sin \left (t \right )-\frac {16}{25}-\frac {12 \left (\cos \left (t \right )+\frac {4 \sin \left (t \right )}{3}\right ) \mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) {\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}}}{25}+\frac {2 \left (8-10 t +5 \pi \right ) \mathit {Heaviside}\left (t -\frac {\pi }{2}\right )}{25}+\frac {4 t}{5} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} -\frac {16}{25} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {c_{1} {\mathrm e}^{-\frac {t}{2}} \cos \left (t \right )}{2}-c_{1} {\mathrm e}^{-\frac {t}{2}} \sin \left (t \right )-\frac {c_{2} {\mathrm e}^{-\frac {t}{2}} \sin \left (t \right )}{2}+c_{2} {\mathrm e}^{-\frac {t}{2}} \cos \left (t \right )-\frac {12 \left (-\sin \left (t \right )+\frac {4 \cos \left (t \right )}{3}\right ) \mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) {\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}}}{25}-\frac {12 \left (\cos \left (t \right )+\frac {4 \sin \left (t \right )}{3}\right ) \mathit {Dirac}\left (t -\frac {\pi }{2}\right ) {\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}}}{25}+\frac {6 \left (\cos \left (t \right )+\frac {4 \sin \left (t \right )}{3}\right ) \mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) {\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}}}{25}-\frac {4 \mathit {Heaviside}\left (t -\frac {\pi }{2}\right )}{5}+\frac {2 \left (8-10 t +5 \pi \right ) \mathit {Dirac}\left (t -\frac {\pi }{2}\right )}{25}+\frac {4}{5} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-\frac {c_{1}}{2}+\frac {4}{5}+c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =\frac {16}{25}, c_{2} =-\frac {12}{25}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {16}{25}-\frac {12 \left (\cos \left (t \right )+\frac {4 \sin \left (t \right )}{3}\right ) \mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) {\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}}}{25}+\frac {2 \left (8-10 t +5 \pi \right ) \mathit {Heaviside}\left (t -\frac {\pi }{2}\right )}{25}+\frac {4 \,{\mathrm e}^{-\frac {t}{2}} \left (4 \cos \left (t \right )-3 \sin \left (t \right )\right )}{25}+\frac {4 t}{5} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {16}{25}-\frac {12 \left (\cos \left (t \right )+\frac {4 \sin \left (t \right )}{3}\right ) \mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) {\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}}}{25}+\frac {2 \left (8-10 t +5 \pi \right ) \mathit {Heaviside}\left (t -\frac {\pi }{2}\right )}{25}+\frac {4 \,{\mathrm e}^{-\frac {t}{2}} \left (4 \cos \left (t \right )-3 \sin \left (t \right )\right )}{25}+\frac {4 t}{5} \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: 0.422 (sec). Leaf size: 66

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

\[ y \left (t \right ) = -\frac {16}{25}-\frac {12 \operatorname {Heaviside}\left (t -\frac {\pi }{2}\right ) \left (\cos \left (t \right )+\frac {4 \sin \left (t \right )}{3}\right ) {\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}}}{25}+\frac {2 \left (8-10 t +5 \pi \right ) \operatorname {Heaviside}\left (t -\frac {\pi }{2}\right )}{25}+\frac {4 \left (4 \cos \left (t \right )-3 \sin \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}}}{25}+\frac {4 t}{5} \]

✓ Solution by Mathematica

Time used: 0.036 (sec). Leaf size: 96

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

\[ y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {4}{25} e^{-t/2} \left (e^{t/2} (5 t-4)+4 \cos (t)-3 \sin (t)\right ) & 2 t\leq \pi \\ -\frac {2}{25} e^{-t/2} \left (\left (-8+6 e^{\pi /4}\right ) \cos (t)+\left (6+8 e^{\pi /4}\right ) \sin (t)-5 e^{t/2} \pi \right ) & \text {True} \\ \end {array} \\ \end {array} \]