problem 69

Internal problem ID [6686]
Internal ﬁle name [OUTPUT/5934_Sunday_June_05_2022_04_02_24_PM_76411984/index.tex]

Book: DIFFERENTIAL EQUATIONS with Boundary Value Problems. DENNIS G. ZILL, WARREN S. WRIGHT, MICHAEL R. CULLEN. Brooks/Cole. Boston, MA. 2013. 8th edition.
Section: CHAPTER 7 THE LAPLACE TRANSFORM. 7.3.1 TRANSLATION ON THE s-AXIS. Page 297
Problem number: 69.
ODE order: 2.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime }+y=\left \{\begin {array}{cc} 0 & 0\le t <\pi \\ 1 & \pi \le t <2 \pi \\ 0 & 2 \pi \le t \end {array}\right .} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 1] \end {align*}

6.19.1 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) &=1\\ F &=\left \{\begin {array}{cc} 0 & t <\pi \\ 1 & t <2 \pi \\ 0 & 2 \pi \le t \end {array}\right . \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+y = \left \{\begin {array}{cc} 0 & t <\pi \\ 1 & t <2 \pi \\ 0 & 2 \pi \le t \end {array}\right . \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*} s^{2} Y \left (s \right )-y^{\prime }\left (0\right )-s y \left (0\right )+Y \left (s \right ) = \frac {{\mathrm e}^{-s \pi }-{\mathrm e}^{-2 s \pi }}{s}\tag {1} \end {align*}

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

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

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

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

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} \sin \left (t \right ) & t <\pi \\ \sin \left (t \right )+2 \cos \left (\frac {t}{2}\right )^{2} & t <2 \pi \\ \sin \left (t \right )+2 \cos \left (\frac {t}{2}\right )^{2}-2 \sin \left (\frac {t}{2}\right )^{2} & 2 \pi \le t \end {array}\right . \] Simplifying the solution gives \[ y = \sin \left (t \right )+\left (\left \{\begin {array}{cc} 0 & t <\pi \\ 1+\cos \left (t \right ) & t <2 \pi \\ 2 \cos \left (t \right ) & 2 \pi \le t \end {array}\right .\right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \sin \left (t \right )+\left (\left \{\begin {array}{cc} 0 & t <\pi \\ 1+\cos \left (t \right ) & t <2 \pi \\ 2 \cos \left (t \right ) & 2 \pi \le t \end {array}\right .\right ) \\ \end{align*}

Veriﬁcation of solutions

\[ y = \sin \left (t \right )+\left (\left \{\begin {array}{cc} 0 & t <\pi \\ 1+\cos \left (t \right ) & t <2 \pi \\ 2 \cos \left (t \right ) & 2 \pi \le t \end {array}\right .\right ) \] Veriﬁed OK.

6.19.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+y=\left \{\begin {array}{cc} 0 & t <\pi \\ 1 & t <2 \pi \\ 0 & 2 \pi \le t \end {array}\right ., y \left (0\right )=0, 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 {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+1=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {-4}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (\mathrm {-I}, \mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )=\cos \left (t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=\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} \cos \left (t \right )+c_{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 )=\left \{\begin {array}{cc} 0 & t <\pi \\ 1 & t <2 \pi \\ 0 & 2 \pi \le t \end {array}\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 (t \right ) & \sin \left (t \right ) \\ -\sin \left (t \right ) & \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 )=1 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\cos \left (t \right ) \left (\int \left (\left \{\begin {array}{cc} 0 & t <\pi \\ \sin \left (t \right ) & t <2 \pi \\ 0 & 2 \pi \le t \end {array}\right .\right )d t \right )+\sin \left (t \right ) \left (\int \left (\left \{\begin {array}{cc} 0 & t <\pi \\ \cos \left (t \right ) & t <2 \pi \\ 0 & 2 \pi \le t \end {array}\right .\right )d t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\left \{\begin {array}{cc} 0 & t \le \pi \\ 1+\cos \left (t \right ) & t \le 2 \pi \\ 2 \cos \left (t \right ) & 2 \pi

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`

\[ y \left (t \right ) = \sin \left (t \right )+\left (\left \{\begin {array}{cc} 0 & t <\pi \\ \cos \left (t \right )+1 & t <2 \pi \\ 2 \cos \left (t \right ) & 2 \pi \le t \end {array}\right .\right ) \]

\[ y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} \sin (t) & t\leq \pi \\ \cos (t)+\sin (t)+1 & \pi

6.19 problem 69

6.19.1 Existence and uniqueness analysis

6.19.2 Maple step by step solution