problem 28

Internal problem ID [6675]
Internal ﬁle name [OUTPUT/5923_Sunday_June_05_2022_04_01_45_PM_85455363/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: 28.
ODE order: 2.
ODE degree: 1.

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

\[ \boxed {2 y^{\prime \prime }+20 y^{\prime }+51 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 2, y^{\prime }\left (0\right ) = 0] \end {align*}

6.8.1 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) &=10\\ q(t) &={\frac {51}{2}}\\ F &=0 \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+10 y^{\prime }+\frac {51 y}{2} = 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*} 2 s^{2} Y \left (s \right )-2 y^{\prime }\left (0\right )-2 s y \left (0\right )+20 s Y \left (s \right )-20 y \left (0\right )+51 Y \left (s \right ) = 0\tag {1} \end {align*}

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

Substituting these initial conditions in above in Eq (1) gives \begin {align*} 2 s^{2} Y \left (s \right )-40-4 s +20 s Y \left (s \right )+51 Y \left (s \right ) = 0 \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {4 s +40}{2 s^{2}+20 s +51} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {1-5 i \sqrt {2}}{s +5-\frac {i \sqrt {2}}{2}}+\frac {1+5 i \sqrt {2}}{s +5+\frac {i \sqrt {2}}{2}} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {1-5 i \sqrt {2}}{s +5-\frac {i \sqrt {2}}{2}}\right ) &= \left (1-5 i \sqrt {2}\right ) {\mathrm e}^{-\frac {\left (-i \sqrt {2}+10\right ) t}{2}}\\ \mathcal {L}^{-1}\left (\frac {1+5 i \sqrt {2}}{s +5+\frac {i \sqrt {2}}{2}}\right ) &= \left (1+5 i \sqrt {2}\right ) {\mathrm e}^{-\frac {\left (i \sqrt {2}+10\right ) t}{2}} \end {align*}

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

Summary

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

Veriﬁcation of solutions

\[ y = 2 \,{\mathrm e}^{-5 t} \left (\cos \left (\frac {\sqrt {2}\, t}{2}\right )+5 \sqrt {2}\, \sin \left (\frac {\sqrt {2}\, t}{2}\right )\right ) \] Veriﬁed OK.

6.8.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [2 \frac {d}{d t}y^{\prime }+20 y^{\prime }+51 y=0, y \left (0\right )=2, 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 \\ {} & {} & \frac {d}{d t}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y^{\prime }=-10 y^{\prime }-\frac {51 y}{2} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y^{\prime }+10 y^{\prime }+\frac {51 y}{2}=0 \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}+10 r +\frac {51}{2}=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-10\right )\pm \left (\sqrt {-2}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-5-\frac {\mathrm {I} \sqrt {2}}{2}, -5+\frac {\mathrm {I} \sqrt {2}}{2}\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-5 t} \cos \left (\frac {\sqrt {2}\, t}{2}\right ) \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-5 t} \sin \left (\frac {\sqrt {2}\, t}{2}\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}^{-5 t} \cos \left (\frac {\sqrt {2}\, t}{2}\right )+c_{2} {\mathrm e}^{-5 t} \sin \left (\frac {\sqrt {2}\, t}{2}\right ) \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-5 t} \cos \left (\frac {\sqrt {2}\, t}{2}\right )+c_{2} {\mathrm e}^{-5 t} \sin \left (\frac {\sqrt {2}\, t}{2}\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 }=-5 c_{1} {\mathrm e}^{-5 t} \cos \left (\frac {\sqrt {2}\, t}{2}\right )-\frac {c_{1} {\mathrm e}^{-5 t} \sqrt {2}\, \sin \left (\frac {\sqrt {2}\, t}{2}\right )}{2}-5 c_{2} {\mathrm e}^{-5 t} \sin \left (\frac {\sqrt {2}\, t}{2}\right )+\frac {c_{2} {\mathrm e}^{-5 t} \sqrt {2}\, \cos \left (\frac {\sqrt {2}\, t}{2}\right )}{2} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-5 c_{1} +\frac {c_{2} \sqrt {2}}{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =2, c_{2} =10 \sqrt {2}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=2 \,{\mathrm e}^{-5 t} \left (\cos \left (\frac {\sqrt {2}\, t}{2}\right )+5 \sqrt {2}\, \sin \left (\frac {\sqrt {2}\, t}{2}\right )\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=2 \,{\mathrm e}^{-5 t} \left (\cos \left (\frac {\sqrt {2}\, t}{2}\right )+5 \sqrt {2}\, \sin \left (\frac {\sqrt {2}\, t}{2}\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`

\[ y \left (t \right ) = 2 \,{\mathrm e}^{-5 t} \left (\cos \left (\frac {\sqrt {2}\, t}{2}\right )+5 \sqrt {2}\, \sin \left (\frac {\sqrt {2}\, t}{2}\right )\right ) \]

\[ y(t)\to 2 e^{-5 t} \left (5 \sqrt {2} \sin \left (\frac {t}{\sqrt {2}}\right )+\cos \left (\frac {t}{\sqrt {2}}\right )\right ) \]

6.8 problem 28

6.8.1 Existence and uniqueness analysis

6.8.2 Maple step by step solution