problem 32

Internal problem ID [6679]
Internal ﬁle name [OUTPUT/5927_Sunday_June_05_2022_04_01_54_PM_48314026/index.tex]

Book: DIFFERENTIAL EQUATIONS with Boundary Value Problems. DENNIS G. ZILL, WARREN S. WRIGHT, MICHAEL R. CULLEN. Brooks/Cole. Boston, MA. 2013. 8th edition.
Section: CHAPTER 7 THE LAPLACE TRANSFORM. 7.3.1 TRANSLATION ON THE s-AXIS. Page 297
Problem number: 32.
ODE order: 2.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime }+8 y^{\prime }+20 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (\pi \right ) = 0] \end {align*}

Since initial condition \(y'(0)\) is not at zero, then let \begin {align*} y'(0) &= c_{2} \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 )+8 s Y \left (s \right )-8 y \left (0\right )+20 Y \left (s \right ) = 0\tag {1} \end {align*}

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

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-c_{2} +8 s Y \left (s \right )+20 Y \left (s \right ) = 0 \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {c_{2}}{s^{2}+8 s +20} \end {align*}

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

Adding the above results and simplifying gives \[ y=\frac {c_{2} {\mathrm e}^{-4 t} \sin \left (2 t \right )}{2} \] Since one initial condition \(y'(\pi )\) is not at zero, then we need to setup an equation to solve for \(c_{2}\). At \(t=\pi \). Taking derivative of the solution and evaluating at \(t=\pi \) gives \begin {align*} 0 &= c_{2} {\mathrm e}^{-4 \pi } \end {align*}

Subtituting this in the solution obtained above gives \begin {align*} y &= 0\\ &= 0 \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= 0 \\ \end{align*}

Veriﬁcation of solutions

\[ y = 0 \] Veriﬁed OK.

6.12.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+8 y^{\prime }+20 y=0, y \left (0\right )=0, y^{\prime }\left (\pi \right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d t}y^{\prime } \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}+8 r +20=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-8\right )\pm \left (\sqrt {-16}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-4-2 \,\mathrm {I}, -4+2 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-4 t} \cos \left (2 t \right ) \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-4 t} \sin \left (2 t \right ) \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y=c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-4 t} \cos \left (2 t \right )+c_{2} {\mathrm e}^{-4 t} \sin \left (2 t \right ) \end {array} \]

Maple trace

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

6.12 problem 32

6.12.1 Maple step by step solution