7.5 problem 13

Internal problem ID [6692]
Internal file name [OUTPUT/5940_Sunday_June_05_2022_04_02_56_PM_13728986/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. 7.4.1 DERIVATIVES OF A TRANSFORM. Page 309
Problem number: 13.
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]]

\[ \boxed {y^{\prime \prime }+16 y=\left \{\begin {array}{cc} \cos \left (4 t \right ) & 0\le t <\pi \\ 0 & \pi \le t \end {array}\right .} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 1] \end {align*}

7.5.1 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) &=0\\ q(t) &=16\\ F &=\left \{\begin {array}{cc} 0 & t <0 \\ \cos \left (4 t \right ) & t <\pi \\ 0 & \pi \le t \end {array}\right . \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+16 y = \left \{\begin {array}{cc} 0 & t <0 \\ \cos \left (4 t \right ) & t <\pi \\ 0 & \pi \le t \end {array}\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 )+16 Y \left (s \right ) = \frac {\left (-{\mathrm e}^{-\pi s}+1\right ) s}{s^{2}+16}\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+16 Y \left (s \right ) = \frac {\left (-{\mathrm e}^{-\pi s}+1\right ) s}{s^{2}+16} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = -\frac {s \,{\mathrm e}^{-\pi s}-s^{2}-s -16}{\left (s^{2}+16\right )^{2}} \end {align*}

Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (-\frac {s \,{\mathrm e}^{-\pi s}-s^{2}-s -16}{\left (s^{2}+16\right )^{2}}\right )\\ &= \frac {\sin \left (4 t \right ) \left (2+t -\operatorname {Heaviside}\left (t -\pi \right ) \left (t -\pi \right )\right )}{8} \end {align*}

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

Summary

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

Verification of solutions

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

7.5.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+16 y=\left \{\begin {array}{cc} 0 & t <0 \\ \cos \left (4 t \right ) & t <\pi \\ 0 & \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}+16=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {-64}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-4 \,\mathrm {I}, 4 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )=\cos \left (4 t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=\sin \left (4 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 (4 t \right )+c_{2} \sin \left (4 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 <0 \\ \cos \left (4 t \right ) & t <\pi \\ 0 & \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 (4 t \right ) & \sin \left (4 t \right ) \\ -4 \sin \left (4 t \right ) & 4 \cos \left (4 t \right ) \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=4 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\cos \left (4 t \right ) \left (\int \left (\left \{\begin {array}{cc} 0 & t <0 \\ \frac {\sin \left (8 t \right )}{8} & t <\pi \\ 0 & \pi \le t \end {array}\right .\right )d t \right )+\sin \left (4 t \right ) \left (\int \left (\left \{\begin {array}{cc} 0 & t <0 \\ \frac {\cos \left (4 t \right )^{2}}{4} & t <\pi \\ 0 & \pi \le t \end {array}\right .\right )d t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\frac {\sin \left (4 t \right ) \left (\left \{\begin {array}{cc} 0 & t \le 0 \\ t & t \le \pi \\ \pi & \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`
 

✓ Solution by Maple

Time used: 2.703 (sec). Leaf size: 29

dsolve([diff(y(t),t$2)+16*y(t)=piecewise(0<=t and t<Pi,cos(4*t),t>= Pi,0),y(0) = 0, D(y)(0) = 1],y(t), singsol=all)
 

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

✓ Solution by Mathematica

Time used: 0.135 (sec). Leaf size: 60

DSolve[{y''[t]+16*y[t]==Piecewise[{{Cos[4*t],0<=t<Pi},{0,t>=Pi}}],{y[0]==1,y'[0]==1}},y[t],t,IncludeSingularSolutions -> True]
 

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