problem 27

Internal problem ID [13222]
Internal ﬁle name [OUTPUT/11878_Sunday_December_03_2023_07_22_50_PM_75951210/index.tex]

Book: DIFFERENTIAL EQUATIONS by Paul Blanchard, Robert L. Devaney, Glen R. Hall. 4th edition. Brooks/Cole. Boston, USA. 2012
Section: Chapter 6. Laplace transform. Section 6.3 page 600
Problem number: 27.
ODE order: 2.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime }+4 y=8} \] With initial conditions \begin {align*} [y \left (0\right ) = 11, y^{\prime }\left (0\right ) = 5] \end {align*}

19.1.1 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*}

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 )+4 Y \left (s \right ) = \frac {8}{s}\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&=11\\ y'(0) &=5 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-5-11 s +4 Y \left (s \right ) = \frac {8}{s} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {11 s^{2}+5 s +8}{s \left (s^{2}+4\right )} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {\frac {9}{2}-\frac {5 i}{4}}{s -2 i}+\frac {\frac {9}{2}+\frac {5 i}{4}}{s +2 i}+\frac {2}{s} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {\frac {9}{2}-\frac {5 i}{4}}{s -2 i}\right ) &= \left (\frac {9}{2}-\frac {5 i}{4}\right ) {\mathrm e}^{2 i t}\\ \mathcal {L}^{-1}\left (\frac {\frac {9}{2}+\frac {5 i}{4}}{s +2 i}\right ) &= \left (\frac {9}{2}+\frac {5 i}{4}\right ) {\mathrm e}^{-2 i t}\\ \mathcal {L}^{-1}\left (\frac {2}{s}\right ) &= 2 \end {align*}

Adding the above results and simplifying gives \[ y=2+9 \cos \left (2 t \right )+\frac {5 \sin \left (2 t \right )}{2} \] Simplifying the solution gives \[ y = 2+9 \cos \left (2 t \right )+\frac {5 \sin \left (2 t \right )}{2} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= 2+9 \cos \left (2 t \right )+\frac {5 \sin \left (2 t \right )}{2} \\ \end{align*}

Veriﬁcation of solutions

\[ y = 2+9 \cos \left (2 t \right )+\frac {5 \sin \left (2 t \right )}{2} \] Veriﬁed OK.

19.1.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+4 y=8, y \left (0\right )=11, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=5\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}+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} \\ {} & {} & y_{1}\left (t \right )=\cos \left (2 t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=\sin \left (2 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 (2 t \right )+c_{2} \sin \left (2 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 )=8\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 (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 (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=2 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-4 \cos \left (2 t \right ) \left (\int \sin \left (2 t \right )d t \right )+4 \sin \left (2 t \right ) \left (\int \cos \left (2 t \right )d t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=2 \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} \cos \left (2 t \right )+c_{2} \sin \left (2 t \right )+2 \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} \cos \left (2 t \right )+c_{2} \sin \left (2 t \right )+2 \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=11 \\ {} & {} & 11=c_{1} +2 \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-2 c_{1} \sin \left (2 t \right )+2 c_{2} \cos \left (2 t \right ) \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=5 \\ {} & {} & 5=2 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =9, c_{2} =\frac {5}{2}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=2+9 \cos \left (2 t \right )+\frac {5 \sin \left (2 t \right )}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=2+9 \cos \left (2 t \right )+\frac {5 \sin \left (2 t \right )}{2} \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`

\[ y \left (t \right ) = 9 \cos \left (2 t \right )+\frac {5 \sin \left (2 t \right )}{2}+2 \]

19.1 problem 27

19.1.1 Existence and uniqueness analysis

19.1.2 Maple step by step solution