problem Problem 3(i)

Internal problem ID [12330]
Internal ﬁle name [OUTPUT/10983_Monday_October_02_2023_02_47_42_AM_32227872/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(i).
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_laplace", "second_order_linear_constant_coeﬀ"

\[ \boxed {4 y^{\prime \prime }+4 y^{\prime }+5 y=25 t \left (\operatorname {Heaviside}\left (t \right )-\operatorname {Heaviside}\left (t -\frac {\pi }{2}\right )\right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 2, y^{\prime }\left (0\right ) = 2] \end {align*}

4.23.1 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 {25 t \left (\operatorname {Heaviside}\left (t \right )-\operatorname {Heaviside}\left (t -\frac {\pi }{2}\right )\right )}{4} \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+y^{\prime }+\frac {5 y}{4} = \frac {25 t \left (\operatorname {Heaviside}\left (t \right )-\operatorname {Heaviside}\left (t -\frac {\pi }{2}\right )\right )}{4} \end {align*}

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*} 4 s^{2} Y \left (s \right )-4 y^{\prime }\left (0\right )-4 s y \left (0\right )+4 s Y \left (s \right )-4 y \left (0\right )+5 Y \left (s \right ) = \frac {25-\frac {25 \,{\mathrm e}^{-\frac {s \pi }{2}} \left (s \pi +2\right )}{2}}{s^{2}}\tag {1} \end {align*}

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

Substituting these initial conditions in above in Eq (1) gives \begin {align*} 4 s^{2} Y \left (s \right )-16-8 s +4 s Y \left (s \right )+5 Y \left (s \right ) = \frac {25-\frac {25 \,{\mathrm e}^{-\frac {s \pi }{2}} \left (s \pi +2\right )}{2}}{s^{2}} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = -\frac {25 \pi \,{\mathrm e}^{-\frac {s \pi }{2}} s -16 s^{3}-32 s^{2}+50 \,{\mathrm e}^{-\frac {s \pi }{2}}-50}{2 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 {25 \pi \,{\mathrm e}^{-\frac {s \pi }{2}} s -16 s^{3}-32 s^{2}+50 \,{\mathrm e}^{-\frac {s \pi }{2}}-50}{2 s^{2} \left (4 s^{2}+4 s +5\right )}\right )\\ &= 6 \,{\mathrm e}^{-\frac {t}{2}} \cos \left (t \right )+\left (5 t -4\right ) \operatorname {Heaviside}\left (-t +\frac {\pi }{2}\right )+\left (\frac {1}{20}-\frac {i}{40}\right ) \left (25 \pi \,{\mathrm e}^{\left (-\frac {1}{4}+\frac {i}{2}\right ) \left (2 t -\pi \right )}+\left (15+20 i\right ) {\mathrm e}^{\left (-\frac {1}{4}-\frac {i}{2}\right ) \left (2 t -\pi \right )} \pi +\left (-16-8 i\right ) {\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 ) \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} 6 \,{\mathrm e}^{-\frac {t}{2}} \cos \left (t \right )+5 t -4 & t <\frac {\pi }{2} \\ -4+\frac {5 \pi }{2}+\left (\frac {1}{20}-\frac {i}{40}\right ) \left (25 \pi +\left (15+20 i\right ) \pi -64-32 i\right ) & t =\frac {\pi }{2} \\ 6 \,{\mathrm e}^{-\frac {t}{2}} \cos \left (t \right )+\left (\frac {1}{20}-\frac {i}{40}\right ) \left (25 \pi \,{\mathrm e}^{\left (-\frac {1}{4}+\frac {i}{2}\right ) \left (2 t -\pi \right )}+\left (15+20 i\right ) {\mathrm e}^{\left (-\frac {1}{4}-\frac {i}{2}\right ) \left (2 t -\pi \right )} \pi +\left (-16-8 i\right ) {\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}} \left (3 \cos \left (t \right )+4 \sin \left (t \right )\right )\right ) & \frac {\pi }{2}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left \{\begin {array}{cc} 6 \,{\mathrm e}^{-\frac {t}{2}} \cos \left (t \right )+5 t -4 & t <\frac {\pi }{2} \\ -8+5 \pi & t &=\frac {\pi }{2} \\ \left (\frac {5}{4}-\frac {5 i}{8}\right ) \pi \,{\mathrm e}^{\left (\frac {1}{4}-\frac {i}{2}\right ) \left (-2 t +\pi \right )}+\left (\frac {5}{4}+\frac {5 i}{8}\right ) \pi \,{\mathrm e}^{\left (\frac {1}{4}+\frac {i}{2}\right ) \left (-2 t +\pi \right )}+\left (-3 \cos \left (t \right )-4 \sin \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}}+6 \,{\mathrm e}^{-\frac {t}{2}} \cos \left (t \right ) & \frac {\pi }{2}

Veriﬁcation of solutions

\[ y = \left \{\begin {array}{cc} 6 \,{\mathrm e}^{-\frac {t}{2}} \cos \left (t \right )+5 t -4 & t <\frac {\pi }{2} \\ -8+5 \pi & t =\frac {\pi }{2} \\ \left (\frac {5}{4}-\frac {5 i}{8}\right ) \pi \,{\mathrm e}^{\left (\frac {1}{4}-\frac {i}{2}\right ) \left (-2 t +\pi \right )}+\left (\frac {5}{4}+\frac {5 i}{8}\right ) \pi \,{\mathrm e}^{\left (\frac {1}{4}+\frac {i}{2}\right ) \left (-2 t +\pi \right )}+\left (-3 \cos \left (t \right )-4 \sin \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}}+6 \,{\mathrm e}^{-\frac {t}{2}} \cos \left (t \right ) & \frac {\pi }{2}

4.23.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [4 \frac {d}{d t}y^{\prime }+4 y^{\prime }+5 y=25 t \left (\mathit {Heaviside}\left (t \right )-\mathit {Heaviside}\left (t -\frac {\pi }{2}\right )\right ), y \left (0\right )=2, 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 }=-\frac {5 y}{4}-y^{\prime }+\frac {25 t \mathit {Heaviside}\left (t \right )}{4}-\frac {25 t \mathit {Heaviside}\left (t -\frac {\pi }{2}\right )}{4} \\ \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}=\frac {25 t \left (\mathit {Heaviside}\left (t \right )-\mathit {Heaviside}\left (t -\frac {\pi }{2}\right )\right )}{4} \\ \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 )=\frac {25 t \left (\mathit {Heaviside}\left (t \right )-\mathit {Heaviside}\left (t -\frac {\pi }{2}\right )\right )}{4}\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 {25 \,{\mathrm e}^{-\frac {t}{2}} \left (\cos \left (t \right ) \left (\int \left (\mathit {Heaviside}\left (t \right )-\mathit {Heaviside}\left (t -\frac {\pi }{2}\right )\right ) t \sin \left (t \right ) {\mathrm e}^{\frac {t}{2}}d t \right )-\sin \left (t \right ) \left (\int \left (\mathit {Heaviside}\left (t \right )-\mathit {Heaviside}\left (t -\frac {\pi }{2}\right )\right ) t \cos \left (t \right ) {\mathrm e}^{\frac {t}{2}}d t \right )\right )}{4} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\frac {5 \mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) \left (\left (\frac {12}{5}+\pi \right ) \cos \left (t \right )-2 \left (-\frac {8}{5}+\pi \right ) \sin \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}}}{4}+\left (4-5 t \right ) \mathit {Heaviside}\left (t -\frac {\pi }{2}\right )+5 \left (-\frac {4}{5}+\frac {\left (-3 \sin \left (t \right )+4 \cos \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}}}{5}+t \right ) \mathit {Heaviside}\left (t \right ) \\ \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 {5 \mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) \left (\left (\frac {12}{5}+\pi \right ) \cos \left (t \right )-2 \left (-\frac {8}{5}+\pi \right ) \sin \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}}}{4}+\left (4-5 t \right ) \mathit {Heaviside}\left (t -\frac {\pi }{2}\right )+5 \left (-\frac {4}{5}+\frac {\left (-3 \sin \left (t \right )+4 \cos \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}}}{5}+t \right ) \mathit {Heaviside}\left (t \right ) \\ \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 {5 \mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) \left (\left (\frac {12}{5}+\pi \right ) \cos \left (t \right )-2 \left (-\frac {8}{5}+\pi \right ) \sin \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}}}{4}+\left (4-5 t \right ) \mathit {Heaviside}\left (t -\frac {\pi }{2}\right )+5 \left (-\frac {4}{5}+\frac {\left (-3 \sin \left (t \right )+4 \cos \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}}}{5}+t \right ) \mathit {Heaviside}\left (t \right ) \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=2 \\ {} & {} & 2=c_{1} \\ {} & \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 {5 \mathit {Dirac}\left (t -\frac {\pi }{2}\right ) \left (\left (\frac {12}{5}+\pi \right ) \cos \left (t \right )-2 \left (-\frac {8}{5}+\pi \right ) \sin \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}}}{4}-\frac {5 \mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) \left (-\left (\frac {12}{5}+\pi \right ) \sin \left (t \right )-2 \left (-\frac {8}{5}+\pi \right ) \cos \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}}}{4}+\frac {5 \mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) \left (\left (\frac {12}{5}+\pi \right ) \cos \left (t \right )-2 \left (-\frac {8}{5}+\pi \right ) \sin \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}}}{8}-5 \mathit {Heaviside}\left (t -\frac {\pi }{2}\right )+\left (4-5 t \right ) \mathit {Dirac}\left (t -\frac {\pi }{2}\right )+5 \left (\frac {\left (-3 \cos \left (t \right )-4 \sin \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}}}{5}-\frac {\left (-3 \sin \left (t \right )+4 \cos \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}}}{10}+1\right ) \mathit {Heaviside}\left (t \right )+5 \left (-\frac {4}{5}+\frac {\left (-3 \sin \left (t \right )+4 \cos \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}}}{5}+t \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 \}}}}=2 \\ {} & {} & 2=-\frac {c_{1}}{2}+c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =2, c_{2} =3\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {5 \mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) \left (\left (\frac {12}{5}+\pi \right ) \cos \left (t \right )-2 \left (-\frac {8}{5}+\pi \right ) \sin \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}}}{4}+\left (4-5 t \right ) \mathit {Heaviside}\left (t -\frac {\pi }{2}\right )+\left (\left (-3 \sin \left (t \right )+4 \cos \left (t \right )\right ) \mathit {Heaviside}\left (t \right )+2 \cos \left (t \right )+3 \sin \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}}+\mathit {Heaviside}\left (t \right ) \left (5 t -4\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {5 \mathit {Heaviside}\left (t -\frac {\pi }{2}\right ) \left (\left (\frac {12}{5}+\pi \right ) \cos \left (t \right )-2 \left (-\frac {8}{5}+\pi \right ) \sin \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}}}{4}+\left (4-5 t \right ) \mathit {Heaviside}\left (t -\frac {\pi }{2}\right )+\left (\left (-3 \sin \left (t \right )+4 \cos \left (t \right )\right ) \mathit {Heaviside}\left (t \right )+2 \cos \left (t \right )+3 \sin \left (t \right )\right ) {\mathrm e}^{-\frac {t}{2}}+\mathit {Heaviside}\left (t \right ) \left (5 t -4\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`

\[ y \left (t \right ) = -4+\left (\frac {5}{4}-\frac {5 i}{8}\right ) \pi \operatorname {Heaviside}\left (t -\frac {\pi }{2}\right ) {\mathrm e}^{\left (\frac {1}{4}-\frac {i}{2}\right ) \left (-2 t +\pi \right )}+\left (\frac {5}{4}+\frac {5 i}{8}\right ) \pi \operatorname {Heaviside}\left (t -\frac {\pi }{2}\right ) {\mathrm e}^{\left (\frac {1}{4}+\frac {i}{2}\right ) \left (-2 t +\pi \right )}-3 \left (\cos \left (t \right )+\frac {4 \sin \left (t \right )}{3}\right ) \operatorname {Heaviside}\left (t -\frac {\pi }{2}\right ) {\mathrm e}^{-\frac {t}{2}+\frac {\pi }{4}}+\left (4-5 t \right ) \operatorname {Heaviside}\left (t -\frac {\pi }{2}\right )+6 \cos \left (t \right ) {\mathrm e}^{-\frac {t}{2}}+5 t \]

\[ y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} 5 t+6 e^{-t/2} \cos (t)-4 & t\geq 0\land 2 t\leq \pi \\ e^{-t/2} (2 \cos (t)+3 \sin (t)) & t<0 \\ \frac {1}{4} e^{-t/2} \left (\left (24-e^{\pi /4} (12+5 \pi )\right ) \cos (t)+2 e^{\pi /4} (-8+5 \pi ) \sin (t)\right ) & \text {True} \\ \end {array} \\ \end {array} \]

4.23 problem Problem 3(i)

4.23.1 Existence and uniqueness analysis

4.23.2 Maple step by step solution