6.5 problem 5

Internal problem ID [5683]
Internal file name [OUTPUT/4931_Sunday_June_05_2022_03_10_55_PM_72200753/index.tex]

Book: ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section: Chapter 6. Laplace Transforms. Problem set 6.2, page 216
Problem number: 5.
ODE order: 2.
ODE degree: 1.

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

Maple gives the following as the ode type

[[_2nd_order, _missing_x]]

\[ \boxed {y^{\prime \prime }-\frac {y}{4}=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 12, y^{\prime }\left (0\right ) = 0] \end {align*}

6.5.1 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) &=0\\ q(t) &=-{\frac {1}{4}}\\ F &=0 \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }-\frac {y}{4} = 0 \end {align*}

The domain of \(p(t)=0\) 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 )-\frac {Y \left (s \right )}{4} = 0\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&=12\\ y'(0) &=0 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-12 s -\frac {Y \left (s \right )}{4} = 0 \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {48 s}{4 s^{2}-1} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {6}{s -\frac {1}{2}}+\frac {6}{s +\frac {1}{2}} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {6}{s -\frac {1}{2}}\right ) &= 6 \,{\mathrm e}^{\frac {t}{2}}\\ \mathcal {L}^{-1}\left (\frac {6}{s +\frac {1}{2}}\right ) &= 6 \,{\mathrm e}^{-\frac {t}{2}} \end {align*}

Adding the above results and simplifying gives \[ y=12 \cosh \left (\frac {t}{2}\right ) \] Simplifying the solution gives \[ y = 12 \cosh \left (\frac {t}{2}\right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= 12 \cosh \left (\frac {t}{2}\right ) \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = 12 \cosh \left (\frac {t}{2}\right ) \] Verified OK.

6.5.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-\frac {y}{4}=0, y \left (0\right )=12, 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 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}-\frac {1}{4}=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \frac {\left (2 r -1\right ) \left (2 r +1\right )}{4}=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-\frac {1}{2}, \frac {1}{2}\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-\frac {t}{2}} \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{\frac {t}{2}} \\ \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}^{-\frac {t}{2}}+c_{2} {\mathrm e}^{\frac {t}{2}} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-\frac {t}{2}}+c_{2} {\mathrm e}^{\frac {t}{2}} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=12 \\ {} & {} & 12=c_{1} +c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {c_{1} {\mathrm e}^{-\frac {t}{2}}}{2}+\frac {c_{2} {\mathrm e}^{\frac {t}{2}}}{2} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-\frac {c_{1}}{2}+\frac {c_{2}}{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =6, c_{2} =6\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=6 \,{\mathrm e}^{-\frac {t}{2}}+6 \,{\mathrm e}^{\frac {t}{2}} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=6 \,{\mathrm e}^{-\frac {t}{2}}+6 \,{\mathrm e}^{\frac {t}{2}} \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: 0.781 (sec). Leaf size: 10

dsolve([diff(y(t),t$2)-1/4*y(t)=0,y(0) = 12, D(y)(0) = 0],y(t), singsol=all)
 

\[ y \left (t \right ) = 12 \cosh \left (\frac {t}{2}\right ) \]

✓ Solution by Mathematica

Time used: 0.013 (sec). Leaf size: 19

DSolve[{y''[t]-1/4*y[t]==0,{y[0]==12,y'[0]==0}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to 6 e^{-t/2} \left (e^t+1\right ) \]