19.6 problem 28.8 (c)

Internal problem ID [13867]
Internal file name [OUTPUT/13039_Friday_February_23_2024_06_54_18_AM_97496273/index.tex]

Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell. second edition. CRC Press. FL, USA. 2020
Section: Chapter 28. The inverse Laplace transform. Additional Exercises. page 509
Problem number: 28.8 (c).
ODE order: 2.
ODE degree: 1.

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

Maple gives the following as the ode type

[[_2nd_order, _missing_x]]

\[ \boxed {y^{\prime \prime }+6 y^{\prime }+13 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 2, y^{\prime }\left (0\right ) = 8] \end {align*}

19.6.1 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) &=6\\ q(t) &=13\\ F &=0 \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+6 y^{\prime }+13 y = 0 \end {align*}

The domain of \(p(t)=6\) is \[ \{-\infty

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 )+6 s Y \left (s \right )-6 y \left (0\right )+13 Y \left (s \right ) = 0\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&=2\\ y'(0) &=8 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-20-2 s +6 s Y \left (s \right )+13 Y \left (s \right ) = 0 \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {2 s +20}{s^{2}+6 s +13} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {1-\frac {7 i}{2}}{s +3-2 i}+\frac {1+\frac {7 i}{2}}{s +3+2 i} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {1-\frac {7 i}{2}}{s +3-2 i}\right ) &= \left (1-\frac {7 i}{2}\right ) {\mathrm e}^{\left (-3+2 i\right ) t}\\ \mathcal {L}^{-1}\left (\frac {1+\frac {7 i}{2}}{s +3+2 i}\right ) &= \left (1+\frac {7 i}{2}\right ) {\mathrm e}^{\left (-3-2 i\right ) t} \end {align*}

Adding the above results and simplifying gives \[ y={\mathrm e}^{-3 t} \left (2 \cos \left (2 t \right )+7 \sin \left (2 t \right )\right ) \] Simplifying the solution gives \[ y = {\mathrm e}^{-3 t} \left (2 \cos \left (2 t \right )+7 \sin \left (2 t \right )\right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{-3 t} \left (2 \cos \left (2 t \right )+7 \sin \left (2 t \right )\right ) \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = {\mathrm e}^{-3 t} \left (2 \cos \left (2 t \right )+7 \sin \left (2 t \right )\right ) \] Verified OK.

19.6.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+6 y^{\prime }+13 y=0, y \left (0\right )=2, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=8\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 ODE}\hspace {3pt} \\ {} & {} & r^{2}+6 r +13=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-6\right )\pm \left (\sqrt {-16}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-3-2 \,\mathrm {I}, -3+2 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-3 t} \cos \left (2 t \right ) \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-3 t} \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 ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-3 t} \cos \left (2 t \right )+c_{2} {\mathrm e}^{-3 t} \sin \left (2 t \right ) \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-3 t} \cos \left (2 t \right )+c_{2} {\mathrm e}^{-3 t} \sin \left (2 t \right ) \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=2 \\ {} & {} & 2=c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-3 c_{1} {\mathrm e}^{-3 t} \cos \left (2 t \right )-2 c_{1} {\mathrm e}^{-3 t} \sin \left (2 t \right )-3 c_{2} {\mathrm e}^{-3 t} \sin \left (2 t \right )+2 c_{2} {\mathrm e}^{-3 t} \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 \}}}}=8 \\ {} & {} & 8=-3 c_{1} +2 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =2, c_{2} =7\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-3 t} \left (2 \cos \left (2 t \right )+7 \sin \left (2 t \right )\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-3 t} \left (2 \cos \left (2 t \right )+7 \sin \left (2 t \right )\right ) \end {array} \]

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
<- constant coefficients successful`
 

✓ Solution by Maple

Time used: 4.75 (sec). Leaf size: 22

dsolve([diff(y(t),t$2)+6*diff(y(t),t)+13*y(t)=0,y(0) = 2, D(y)(0) = 8],y(t), singsol=all)
 

\[ y \left (t \right ) = {\mathrm e}^{-3 t} \left (2 \cos \left (2 t \right )+7 \sin \left (2 t \right )\right ) \]

✓ Solution by Mathematica

Time used: 0.017 (sec). Leaf size: 24

DSolve[{y''[t]+6*y'[t]+13*y[t]==0,{y[0]==2,y'[0]==8}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to e^{-3 t} (7 \sin (2 t)+2 \cos (2 t)) \]