problem Problem 24.46

Internal problem ID [5215]
Internal ﬁle name [OUTPUT/4708_Sunday_June_05_2022_03_03_47_PM_35789450/index.tex]

Book: Schaums Outline Diﬀerential Equations, 4th edition. Bronson and Costa. McGraw Hill 2014
Section: Chapter 24. Solutions of linear DE by Laplace transforms. Supplementary Problems. page 248
Problem number: Problem 24.46.
ODE order: 2.
ODE degree: 1.

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

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

5.16.1 Existence and uniqueness analysis

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

Where here \begin {align*} p(t) &=9\\ q(t) &=14\\ F &=\frac {\sin \left (t \right )}{2} \end {align*}

Hence the ode is \begin {align*} q^{\prime \prime }+9 q^{\prime }+14 q = \frac {\sin \left (t \right )}{2} \end {align*}

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

Taking the Laplace transform of the ode and using the relations that \begin {align*} \mathcal {L}\left (q^{\prime }\right ) &= s Y(s) - q \left (0\right )\\ \mathcal {L}\left (q^{\prime \prime }\right ) &= s^2 Y(s) - q'(0) - s q \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 )-q^{\prime }\left (0\right )-s q \left (0\right )+9 s Y \left (s \right )-9 q \left (0\right )+14 Y \left (s \right ) = \frac {1}{2 s^{2}+2}\tag {1} \end {align*}

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

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

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {2 s^{2}+3}{2 \left (s^{2}+1\right ) \left (s^{2}+9 s +14\right )} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {-\frac {9}{1000}-\frac {13 i}{1000}}{s -i}+\frac {-\frac {9}{1000}+\frac {13 i}{1000}}{s +i}+\frac {11}{50 \left (s +2\right )}-\frac {101}{500 \left (s +7\right )} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {-\frac {9}{1000}-\frac {13 i}{1000}}{s -i}\right ) &= \left (-\frac {9}{1000}-\frac {13 i}{1000}\right ) {\mathrm e}^{i t}\\ \mathcal {L}^{-1}\left (\frac {-\frac {9}{1000}+\frac {13 i}{1000}}{s +i}\right ) &= \left (-\frac {9}{1000}+\frac {13 i}{1000}\right ) {\mathrm e}^{-i t}\\ \mathcal {L}^{-1}\left (\frac {11}{50 \left (s +2\right )}\right ) &= \frac {11 \,{\mathrm e}^{-2 t}}{50}\\ \mathcal {L}^{-1}\left (-\frac {101}{500 \left (s +7\right )}\right ) &= -\frac {101 \,{\mathrm e}^{-7 t}}{500} \end {align*}

Adding the above results and simplifying gives \[ q=-\frac {101 \,{\mathrm e}^{-7 t}}{500}+\frac {11 \,{\mathrm e}^{-2 t}}{50}-\frac {9 \cos \left (t \right )}{500}+\frac {13 \sin \left (t \right )}{500} \] Simplifying the solution gives \[ q = -\frac {101 \,{\mathrm e}^{-7 t}}{500}+\frac {11 \,{\mathrm e}^{-2 t}}{50}-\frac {9 \cos \left (t \right )}{500}+\frac {13 \sin \left (t \right )}{500} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} q &= -\frac {101 \,{\mathrm e}^{-7 t}}{500}+\frac {11 \,{\mathrm e}^{-2 t}}{50}-\frac {9 \cos \left (t \right )}{500}+\frac {13 \sin \left (t \right )}{500} \\ \end{align*}

Veriﬁcation of solutions

\[ q = -\frac {101 \,{\mathrm e}^{-7 t}}{500}+\frac {11 \,{\mathrm e}^{-2 t}}{50}-\frac {9 \cos \left (t \right )}{500}+\frac {13 \sin \left (t \right )}{500} \] Veriﬁed OK.

5.16.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [q^{\prime \prime }+9 q^{\prime }+14 q=\frac {\sin \left (t \right )}{2}, q \left (0\right )=0, q^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & q^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+9 r +14=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +7\right ) \left (r +2\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-7, -2\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & q_{1}\left (t \right )={\mathrm e}^{-7 t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & q_{2}\left (t \right )={\mathrm e}^{-2 t} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & q=c_{1} q_{1}\left (t \right )+c_{2} q_{2}\left (t \right )+q_{p}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & q=c_{1} {\mathrm e}^{-7 t}+c_{2} {\mathrm e}^{-2 t}+q_{p}\left (t \right ) \\ \square & {} & \textrm {Find a particular solution}\hspace {3pt} q_{p}\left (t \right )\hspace {3pt}\textrm {of the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Use variation of parameters to find}\hspace {3pt} q_{p}\hspace {3pt}\textrm {here}\hspace {3pt} f \left (t \right )\hspace {3pt}\textrm {is the forcing function}\hspace {3pt} \\ {} & {} & \left [q_{p}\left (t \right )=-q_{1}\left (t \right ) \left (\int \frac {q_{2}\left (t \right ) f \left (t \right )}{W \left (q_{1}\left (t \right ), q_{2}\left (t \right )\right )}d t \right )+q_{2}\left (t \right ) \left (\int \frac {q_{1}\left (t \right ) f \left (t \right )}{W \left (q_{1}\left (t \right ), q_{2}\left (t \right )\right )}d t \right ), f \left (t \right )=\frac {\sin \left (t \right )}{2}\right ] \\ {} & \circ & \textrm {Wronskian of solutions of the homogeneous equation}\hspace {3pt} \\ {} & {} & W \left (q_{1}\left (t \right ), q_{2}\left (t \right )\right )=\left [\begin {array}{cc} {\mathrm e}^{-7 t} & {\mathrm e}^{-2 t} \\ -7 \,{\mathrm e}^{-7 t} & -2 \,{\mathrm e}^{-2 t} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (q_{1}\left (t \right ), q_{2}\left (t \right )\right )=5 \,{\mathrm e}^{-9 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} q_{p}\left (t \right ) \\ {} & {} & q_{p}\left (t \right )=-\frac {{\mathrm e}^{-7 t} \left (\int \sin \left (t \right ) {\mathrm e}^{7 t}d t \right )}{10}+\frac {{\mathrm e}^{-2 t} \left (\int \sin \left (t \right ) {\mathrm e}^{2 t}d t \right )}{10} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & q_{p}\left (t \right )=-\frac {9 \cos \left (t \right )}{500}+\frac {13 \sin \left (t \right )}{500} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & q=c_{1} {\mathrm e}^{-7 t}+c_{2} {\mathrm e}^{-2 t}-\frac {9 \cos \left (t \right )}{500}+\frac {13 \sin \left (t \right )}{500} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} q=c_{1} {\mathrm e}^{-7 t}+c_{2} {\mathrm e}^{-2 t}-\frac {9 \cos \left (t \right )}{500}+\frac {13 \sin \left (t \right )}{500} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} q \left (0\right )=0 \\ {} & {} & 0=c_{1} +c_{2} -\frac {9}{500} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & q^{\prime }=-7 c_{1} {\mathrm e}^{-7 t}-2 c_{2} {\mathrm e}^{-2 t}+\frac {9 \sin \left (t \right )}{500}+\frac {13 \cos \left (t \right )}{500} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} q^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=-7 c_{1} -2 c_{2} +\frac {13}{500} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\frac {101}{500}, c_{2} =\frac {11}{50}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & q=-\frac {101 \,{\mathrm e}^{-7 t}}{500}+\frac {11 \,{\mathrm e}^{-2 t}}{50}-\frac {9 \cos \left (t \right )}{500}+\frac {13 \sin \left (t \right )}{500} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & q=-\frac {101 \,{\mathrm e}^{-7 t}}{500}+\frac {11 \,{\mathrm e}^{-2 t}}{50}-\frac {9 \cos \left (t \right )}{500}+\frac {13 \sin \left (t \right )}{500} \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`

\[ q \left (t \right ) = -\frac {9 \cos \left (t \right )}{500}+\frac {13 \sin \left (t \right )}{500}-\frac {101 \,{\mathrm e}^{-7 t}}{500}+\frac {11 \,{\mathrm e}^{-2 t}}{50} \]

\[ q(t)\to \frac {1}{500} \left (-101 e^{-7 t}+110 e^{-2 t}+13 \sin (t)-9 \cos (t)\right ) \]

5.16 problem Problem 24.46

5.16.1 Existence and uniqueness analysis

5.16.2 Maple step by step solution