problem 15

Internal problem ID [840]
Internal ﬁle name [OUTPUT/840_Sunday_June_05_2022_01_50_53_AM_32515336/index.tex]

Book: Elementary diﬀerential equations and boundary value problems, 11th ed., Boyce, DiPrima, Meade
Section: Chapter 6.2, The Laplace Transform. Solution of Initial Value Problems. page 255
Problem number: 15.
ODE order: 2.
ODE degree: 1.

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

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

3.8.1 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) &=\omega ^{2}\\ F &=\cos \left (2 t \right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+\omega ^{2} y = \cos \left (2 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 )+\omega ^{2} Y \left (s \right ) = \frac {s}{s^{2}+4}\tag {1} \end {align*}

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

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

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

Applying partial fractions decomposition results in \[ Y(s)= \frac {1}{2 \left (\omega ^{2}-4\right ) \left (s -2 i\right )}+\frac {1}{2 \left (\omega ^{2}-4\right ) \left (s +2 i\right )}+\frac {\omega ^{2}-5}{\left (2 \omega ^{2}-8\right ) \left (s -\sqrt {-\omega ^{2}}\right )}+\frac {\omega ^{2}-5}{\left (2 \omega ^{2}-8\right ) \left (s +\sqrt {-\omega ^{2}}\right )} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {1}{2 \left (\omega ^{2}-4\right ) \left (s -2 i\right )}\right ) &= \frac {{\mathrm e}^{2 i t}}{2 \omega ^{2}-8}\\ \mathcal {L}^{-1}\left (\frac {1}{2 \left (\omega ^{2}-4\right ) \left (s +2 i\right )}\right ) &= \frac {{\mathrm e}^{-2 i t}}{2 \omega ^{2}-8}\\ \mathcal {L}^{-1}\left (\frac {\omega ^{2}-5}{\left (2 \omega ^{2}-8\right ) \left (s -\sqrt {-\omega ^{2}}\right )}\right ) &= \frac {\left (\omega ^{2}-5\right ) {\mathrm e}^{i \operatorname {csgn}\left (i \omega \right ) \omega t}}{2 \omega ^{2}-8}\\ \mathcal {L}^{-1}\left (\frac {\omega ^{2}-5}{\left (2 \omega ^{2}-8\right ) \left (s +\sqrt {-\omega ^{2}}\right )}\right ) &= \frac {\left (\omega ^{2}-5\right ) {\mathrm e}^{-i \operatorname {csgn}\left (i \omega \right ) \omega t}}{2 \omega ^{2}-8} \end {align*}

Adding the above results and simplifying gives \[ y=\frac {\cos \left (2 t \right )+\cos \left (\omega t \right ) \left (\omega ^{2}-5\right )}{\omega ^{2}-4} \] Simplifying the solution gives \[ y = \frac {\cos \left (2 t \right )+\cos \left (\omega t \right ) \left (\omega ^{2}-5\right )}{\omega ^{2}-4} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\cos \left (2 t \right )+\cos \left (\omega t \right ) \left (\omega ^{2}-5\right )}{\omega ^{2}-4} \\ \end{align*}

Veriﬁcation of solutions

\[ y = \frac {\cos \left (2 t \right )+\cos \left (\omega t \right ) \left (\omega ^{2}-5\right )}{\omega ^{2}-4} \] Veriﬁed OK.

3.8.2 Maple step by step solution

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

\[ y \left (t \right ) = \frac {\cos \left (2 t \right )+\cos \left (\omega t \right ) \left (\omega ^{2}-5\right )}{\omega ^{2}-4} \]

3.8 problem 15

3.8.1 Existence and uniqueness analysis

3.8.2 Maple step by step solution