problem 15(a)

Internal problem ID [6709]
Internal ﬁle name [OUTPUT/5957_Sunday_June_05_2022_04_04_22_PM_51342204/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. EXERCISES 7.5. Page 315
Problem number: 15(a).
ODE order: 2.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime }+2 y^{\prime }+10 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 1] \end {align*}

8.13.1 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*}

Hence the ode is \begin {align*} y^{\prime \prime }+2 y^{\prime }+10 y = 0 \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 )+2 s Y \left (s \right )-2 y \left (0\right )+10 Y \left (s \right ) = 0\tag {1} \end {align*}

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

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-1+2 s Y \left (s \right )+10 Y \left (s \right ) = 0 \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {1}{s^{2}+2 s +10} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= -\frac {i}{6 \left (s +1-3 i\right )}+\frac {i}{6 s +6+18 i} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {i}{6 \left (s +1-3 i\right )}\right ) &= -\frac {i {\mathrm e}^{\left (-1+3 i\right ) t}}{6}\\ \mathcal {L}^{-1}\left (\frac {i}{6 s +6+18 i}\right ) &= \frac {i {\mathrm e}^{\left (-1-3 i\right ) t}}{6} \end {align*}

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

Summary

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

Veriﬁcation of solutions

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

8.13.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+2 y^{\prime }+10 y=0, y \left (0\right )=0, y^{\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 \\ {} & {} & \frac {d}{d t}y^{\prime } \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}+2 r +10=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-2\right )\pm \left (\sqrt {-36}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-1-3 \,\mathrm {I}, -1+3 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-t} \cos \left (3 t \right ) \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-t} \sin \left (3 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}^{-t} \cos \left (3 t \right )+c_{2} {\mathrm e}^{-t} \sin \left (3 t \right ) \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-t} \cos \left (3 t \right )+c_{2} {\mathrm e}^{-t} \sin \left (3 t \right ) \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-c_{1} {\mathrm e}^{-t} \cos \left (3 t \right )-3 c_{1} {\mathrm e}^{-t} \sin \left (3 t \right )-c_{2} {\mathrm e}^{-t} \sin \left (3 t \right )+3 c_{2} {\mathrm e}^{-t} \cos \left (3 t \right ) \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=-c_{1} +3 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =0, c_{2} =\frac {1}{3}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {{\mathrm e}^{-t} \sin \left (3 t \right )}{3} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {{\mathrm e}^{-t} \sin \left (3 t \right )}{3} \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`

\[ y \left (t \right ) = \frac {{\mathrm e}^{-t} \sin \left (3 t \right )}{3} \]

8.13 problem 15(a)

8.13.1 Existence and uniqueness analysis

8.13.2 Maple step by step solution