problem Problem 27

Internal problem ID [2865]
Internal ﬁle name [OUTPUT/2357_Sunday_June_05_2022_03_00_23_AM_86244952/index.tex]

Book: Diﬀerential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth edition, 2015
Section: Chapter 10, The Laplace Transform and Some Elementary Applications. Exercises for 10.4. page 689
Problem number: Problem 27.
ODE order: 2.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime }+9 y=7 \sin \left (4 t \right )+14 \cos \left (4 t \right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = 2] \end {align*}

13.27.1 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 &=7 \sin \left (4 t \right )+14 \cos \left (4 t \right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+9 y = 7 \sin \left (4 t \right )+14 \cos \left (4 t \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 {28+14 s}{s^{2}+16}\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&=1\\ y'(0) &=2 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-2-s +9 Y \left (s \right ) = \frac {28+14 s}{s^{2}+16} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {s^{3}+2 s^{2}+30 s +60}{\left (s^{2}+16\right ) \left (s^{2}+9\right )} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {-1+\frac {i}{2}}{s -4 i}+\frac {-1-\frac {i}{2}}{s +4 i}+\frac {\frac {3}{2}-i}{s -3 i}+\frac {\frac {3}{2}+i}{s +3 i} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {-1+\frac {i}{2}}{s -4 i}\right ) &= \left (-1+\frac {i}{2}\right ) {\mathrm e}^{4 i t}\\ \mathcal {L}^{-1}\left (\frac {-1-\frac {i}{2}}{s +4 i}\right ) &= \left (-1-\frac {i}{2}\right ) {\mathrm e}^{-4 i t}\\ \mathcal {L}^{-1}\left (\frac {\frac {3}{2}-i}{s -3 i}\right ) &= \left (\frac {3}{2}-i\right ) {\mathrm e}^{3 i t}\\ \mathcal {L}^{-1}\left (\frac {\frac {3}{2}+i}{s +3 i}\right ) &= \left (\frac {3}{2}+i\right ) {\mathrm e}^{-3 i t} \end {align*}

Adding the above results and simplifying gives \[ y=-2 \cos \left (4 t \right )-\sin \left (4 t \right )+3 \cos \left (3 t \right )+2 \sin \left (3 t \right ) \] Simplifying the solution gives \[ y = -2 \cos \left (4 t \right )-\sin \left (4 t \right )+3 \cos \left (3 t \right )+2 \sin \left (3 t \right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -2 \cos \left (4 t \right )-\sin \left (4 t \right )+3 \cos \left (3 t \right )+2 \sin \left (3 t \right ) \\ \end{align*}

Veriﬁcation of solutions

\[ y = -2 \cos \left (4 t \right )-\sin \left (4 t \right )+3 \cos \left (3 t \right )+2 \sin \left (3 t \right ) \] Veriﬁed OK.

13.27.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+9 y=7 \sin \left (4 t \right )+14 \cos \left (4 t \right ), y \left (0\right )=1, 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 {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 )=7 \sin \left (4 t \right )+14 \cos \left (4 t \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 )=-\frac {7 \cos \left (3 t \right ) \left (\int \sin \left (3 t \right ) \left (\sin \left (4 t \right )+2 \cos \left (4 t \right )\right )d t \right )}{3}+\frac {7 \sin \left (3 t \right ) \left (\int \cos \left (3 t \right ) \left (\sin \left (4 t \right )+2 \cos \left (4 t \right )\right )d t \right )}{3} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\sin \left (4 t \right )-2 \cos \left (4 t \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 )-\sin \left (4 t \right )-2 \cos \left (4 t \right ) \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} \cos \left (3 t \right )+c_{2} \sin \left (3 t \right )-\sin \left (4 t \right )-2 \cos \left (4 t \right ) \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=c_{1} -2 \\ {} & \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 )-4 \cos \left (4 t \right )+8 \sin \left (4 t \right ) \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=2 \\ {} & {} & 2=-4+3 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =3, c_{2} =2\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-2 \cos \left (4 t \right )-\sin \left (4 t \right )+3 \cos \left (3 t \right )+2 \sin \left (3 t \right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-2 \cos \left (4 t \right )-\sin \left (4 t \right )+3 \cos \left (3 t \right )+2 \sin \left (3 t \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`

\[ y \left (t \right ) = -2 \cos \left (4 t \right )-\sin \left (4 t \right )+3 \cos \left (3 t \right )+2 \sin \left (3 t \right ) \]

\[ y(t)\to \frac {1}{8} \left (-7 \sin (4 t)+11 \sqrt {2} \sin \left (2 \sqrt {2} t\right )-14 \cos (4 t)+22 \cos \left (2 \sqrt {2} t\right )\right ) \]

13.27 problem Problem 27

13.27.1 Existence and uniqueness analysis

13.27.2 Maple step by step solution