problem Problem 3(c)

Internal problem ID [12324]
Internal ﬁle name [OUTPUT/10977_Monday_October_02_2023_02_47_40_AM_57838993/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(c).
ODE order: 2.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime }+9 y=24 \sin \left (t \right ) \left (\operatorname {Heaviside}\left (t \right )+\operatorname {Heaviside}\left (t -\pi \right )\right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 0] \end {align*}

4.17.1 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) &=9\\ F &=24 \sin \left (t \right ) \left (\operatorname {Heaviside}\left (t \right )+\operatorname {Heaviside}\left (t -\pi \right )\right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+9 y = 24 \sin \left (t \right ) \left (\operatorname {Heaviside}\left (t \right )+\operatorname {Heaviside}\left (t -\pi \right )\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 )+9 Y \left (s \right ) = \frac {24-24 \,{\mathrm e}^{-s \pi }}{s^{2}+1}\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 )+9 Y \left (s \right ) = \frac {24-24 \,{\mathrm e}^{-s \pi }}{s^{2}+1} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = -\frac {24 \left (-1+{\mathrm e}^{-s \pi }\right )}{\left (s^{2}+1\right ) \left (s^{2}+9\right )} \end {align*}

Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (-\frac {24 \left (-1+{\mathrm e}^{-s \pi }\right )}{\left (s^{2}+1\right ) \left (s^{2}+9\right )}\right )\\ &= 4 \left (1+\operatorname {Heaviside}\left (t -\pi \right )\right ) \sin \left (t \right )^{3} \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} 4 \sin \left (t \right )^{3} & t <\pi \\ 8 \sin \left (t \right )^{3} & \pi \le t \end {array}\right . \] Simplifying the solution gives \[ y = \left (\left \{\begin {array}{cc} 4 & t <\pi \\ 8 & \pi \le t \end {array}\right .\right ) \sin \left (t \right )^{3} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left (\left \{\begin {array}{cc} 4 & t <\pi \\ 8 & \pi \le t \end {array}\right .\right ) \sin \left (t \right )^{3} \\ \end{align*}

Veriﬁcation of solutions

\[ y = \left (\left \{\begin {array}{cc} 4 & t <\pi \\ 8 & \pi \le t \end {array}\right .\right ) \sin \left (t \right )^{3} \] Veriﬁed OK.

4.17.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+9 y=\left (24 \mathit {Heaviside}\left (t \right )+24 \mathit {Heaviside}\left (t -\pi \right )\right ) \sin \left (t \right ), 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 }=-9 y+24 \sin \left (t \right ) \left (\mathit {Heaviside}\left (t \right )+\mathit {Heaviside}\left (t -\pi \right )\right ) \\ \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 }+9 y=24 \sin \left (t \right ) \left (\mathit {Heaviside}\left (t \right )+\mathit {Heaviside}\left (t -\pi \right )\right ) \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+9=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {-36}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-3 \,\mathrm {I}, 3 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )=\cos \left (3 t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=\sin \left (3 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 (3 t \right )+c_{2} \sin \left (3 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 )=24 \sin \left (t \right ) \left (\mathit {Heaviside}\left (t \right )+\mathit {Heaviside}\left (t -\pi \right )\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 (3 t \right ) & \sin \left (3 t \right ) \\ -3 \sin \left (3 t \right ) & 3 \cos \left (3 t \right ) \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=3 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-8 \cos \left (3 t \right ) \left (\int \sin \left (3 t \right ) \sin \left (t \right ) \left (\mathit {Heaviside}\left (t \right )+\mathit {Heaviside}\left (t -\pi \right )\right )d t \right )+8 \sin \left (3 t \right ) \left (\int \cos \left (3 t \right ) \sin \left (t \right ) \left (\mathit {Heaviside}\left (t \right )+\mathit {Heaviside}\left (t -\pi \right )\right )d t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=4 \sin \left (t \right )^{3} \left (\mathit {Heaviside}\left (t \right )+\mathit {Heaviside}\left (t -\pi \right )\right ) \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} \cos \left (3 t \right )+c_{2} \sin \left (3 t \right )+4 \sin \left (t \right )^{3} \left (\mathit {Heaviside}\left (t \right )+\mathit {Heaviside}\left (t -\pi \right )\right ) \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} \cos \left (3 t \right )+c_{2} \sin \left (3 t \right )+4 \sin \left (t \right )^{3} \left (\mathit {Heaviside}\left (t \right )+\mathit {Heaviside}\left (t -\pi \right )\right ) \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-3 c_{1} \sin \left (3 t \right )+3 c_{2} \cos \left (3 t \right )+12 \sin \left (t \right )^{2} \left (\mathit {Heaviside}\left (t \right )+\mathit {Heaviside}\left (t -\pi \right )\right ) \cos \left (t \right )+4 \sin \left (t \right )^{3} \left (\mathit {Dirac}\left (t \right )+\mathit {Dirac}\left (t -\pi \right )\right ) \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=3 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =0, c_{2} =0\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=4 \sin \left (t \right )^{3} \left (\mathit {Heaviside}\left (t \right )+\mathit {Heaviside}\left (t -\pi \right )\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=4 \sin \left (t \right )^{3} \left (\mathit {Heaviside}\left (t \right )+\mathit {Heaviside}\left (t -\pi \right )\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`

\[ y \left (t \right ) = 4 \left (1+\operatorname {Heaviside}\left (t -\pi \right )\right ) \sin \left (t \right )^{3} \]

4.17 problem Problem 3(c)

4.17.1 Existence and uniqueness analysis

4.17.2 Maple step by step solution