15.11 problem 11

Internal problem ID [11517]
Internal file name [OUTPUT/10500_Thursday_May_18_2023_04_21_14_AM_69301582/index.tex]

Book: A First Course in Differential Equations by J. David Logan. Third Edition. Springer-Verlag, NY. 2015.
Section: Chapter 3, Laplace transform. Section 3.2.1 Initial value problems. Exercises page 156
Problem number: 11.
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

[[_linear, `class A`]]

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

15.11.1 Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as \begin {align*} x^{\prime \prime } + p(t)x^{\prime } + q(t) x &= F \end {align*}

Where here \begin {align*} p(t) &=0\\ q(t) &=4\\ F &=\cos \left (2 t \right ) \left (1-\operatorname {Heaviside}\left (-2 \pi +t \right )\right ) \end {align*}

Hence the ode is \begin {align*} x^{\prime \prime }+4 x = \cos \left (2 t \right ) \left (1-\operatorname {Heaviside}\left (-2 \pi +t \right )\right ) \end {align*}

The domain of \(p(t)=0\) is \[ \{-\infty

Solving using the Laplace transform method. Let \begin {align*} \mathcal {L}\left (x\right ) =Y(s) \end {align*}

Taking the Laplace transform of the ode and using the relations that \begin {align*} \mathcal {L}\left (x^{\prime }\right ) &= s Y(s) - x \left (0\right )\\ \mathcal {L}\left (x^{\prime \prime }\right ) &= s^2 Y(s) - x'(0) - s x \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 )-x^{\prime }\left (0\right )-s x \left (0\right )+4 Y \left (s \right ) = \frac {s \left (-{\mathrm e}^{-2 s \pi }+1\right )}{s^{2}+4}\tag {1} \end {align*}

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

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

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

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

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

Summary

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

Verification of solutions

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

15.11.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{\prime \prime }+4 x=\cos \left (2 t \right ) \left (1-\mathit {Heaviside}\left (-2 \pi +t \right )\right ), x \left (0\right )=0, x^{\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 \\ {} & {} & x^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & x^{\prime \prime }=-4 x-\cos \left (2 t \right ) \left (-1+\mathit {Heaviside}\left (-2 \pi +t \right )\right ) \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} x\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & x^{\prime \prime }+4 x=-\cos \left (2 t \right ) \left (-1+\mathit {Heaviside}\left (-2 \pi +t \right )\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} \\ {} & {} & x_{1}\left (t \right )=\cos \left (2 t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & x_{2}\left (t \right )=\sin \left (2 t \right ) \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & x=c_{1} x_{1}\left (t \right )+c_{2} x_{2}\left (t \right )+x_{p}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & x=c_{1} \cos \left (2 t \right )+c_{2} \sin \left (2 t \right )+x_{p}\left (t \right ) \\ \square & {} & \textrm {Find a particular solution}\hspace {3pt} x_{p}\left (t \right )\hspace {3pt}\textrm {of the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Use variation of parameters to find}\hspace {3pt} x_{p}\hspace {3pt}\textrm {here}\hspace {3pt} f \left (t \right )\hspace {3pt}\textrm {is the forcing function}\hspace {3pt} \\ {} & {} & \left [x_{p}\left (t \right )=-x_{1}\left (t \right ) \left (\int \frac {x_{2}\left (t \right ) f \left (t \right )}{W \left (x_{1}\left (t \right ), x_{2}\left (t \right )\right )}d t \right )+x_{2}\left (t \right ) \left (\int \frac {x_{1}\left (t \right ) f \left (t \right )}{W \left (x_{1}\left (t \right ), x_{2}\left (t \right )\right )}d t \right ), f \left (t \right )=-\cos \left (2 t \right ) \left (-1+\mathit {Heaviside}\left (-2 \pi +t \right )\right )\right ] \\ {} & \circ & \textrm {Wronskian of solutions of the homogeneous equation}\hspace {3pt} \\ {} & {} & W \left (x_{1}\left (t \right ), x_{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 (x_{1}\left (t \right ), x_{2}\left (t \right )\right )=2 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} x_{p}\left (t \right ) \\ {} & {} & x_{p}\left (t \right )=\frac {\cos \left (2 t \right ) \left (\int \sin \left (4 t \right ) \left (-1+\mathit {Heaviside}\left (-2 \pi +t \right )\right )d t \right )}{4}-\frac {\sin \left (2 t \right ) \left (\int \cos \left (2 t \right )^{2} \left (-1+\mathit {Heaviside}\left (-2 \pi +t \right )\right )d t \right )}{2} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & x_{p}\left (t \right )=-\frac {\sin \left (2 t \right ) \left (-2 \pi +t \right ) \mathit {Heaviside}\left (-2 \pi +t \right )}{4}+\frac {\sin \left (2 t \right ) t}{4}+\frac {\cos \left (2 t \right )}{16} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & x=c_{1} \cos \left (2 t \right )+c_{2} \sin \left (2 t \right )-\frac {\sin \left (2 t \right ) \left (-2 \pi +t \right ) \mathit {Heaviside}\left (-2 \pi +t \right )}{4}+\frac {\sin \left (2 t \right ) t}{4}+\frac {\cos \left (2 t \right )}{16} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} x=c_{1} \cos \left (2 t \right )+c_{2} \sin \left (2 t \right )-\frac {\sin \left (2 t \right ) \left (-2 \pi +t \right ) \mathit {Heaviside}\left (-2 \pi +t \right )}{4}+\frac {\sin \left (2 t \right ) t}{4}+\frac {\cos \left (2 t \right )}{16} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} x \left (0\right )=0 \\ {} & {} & 0=c_{1} +\frac {1}{16} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & x^{\prime }=-2 c_{1} \sin \left (2 t \right )+2 c_{2} \cos \left (2 t \right )-\frac {\cos \left (2 t \right ) \left (-2 \pi +t \right ) \mathit {Heaviside}\left (-2 \pi +t \right )}{2}-\frac {\sin \left (2 t \right ) \mathit {Heaviside}\left (-2 \pi +t \right )}{4}-\frac {\sin \left (2 t \right ) \left (-2 \pi +t \right ) \mathit {Dirac}\left (-2 \pi +t \right )}{4}+\frac {\cos \left (2 t \right ) t}{2}+\frac {\sin \left (2 t \right )}{8} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} x^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=2 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\frac {1}{16}, c_{2} =0\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & x=-\frac {\sin \left (2 t \right ) \left (\mathit {Heaviside}\left (-2 \pi +t \right ) \left (-2 \pi +t \right )-t \right )}{4} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & x=-\frac {\sin \left (2 t \right ) \left (\mathit {Heaviside}\left (-2 \pi +t \right ) \left (-2 \pi +t \right )-t \right )}{4} \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
<- 1st order linear successful`
 

✗ Solution by Maple

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

\[ \text {No solution found} \]

✓ Solution by Mathematica

Time used: 0.168 (sec). Leaf size: 28

DSolve[{x''[t]+4*x[t]==Cos[2*t]*UnitStep[2*Pi-t],{x[0]==0,x'[0]==0}},x[t],t,IncludeSingularSolutions -> True]
 

\[ x(t)\to \begin {array}{cc} \{ & \begin {array}{cc} \pi \cos (t) \sin (t) & t>2 \pi \\ \frac {1}{2} t \cos (t) \sin (t) & \text {True} \\ \end {array} \\ \end {array} \]