Internal problem ID [5690]
Internal file name [OUTPUT/4938_Sunday_June_05_2022_03_11_04_PM_64268551/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: 12.
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]]
\[ \boxed {y^{\prime \prime }-2 y^{\prime }-3 y=0} \] With initial conditions \begin {align*} [y \left (4\right ) = -3, y^{\prime }\left (4\right ) = -17] \end {align*}
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) &=-2\\ q(t) &=-3\\ F &=0 \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }-2 y^{\prime }-3 y = 0 \end {align*}
The domain of \(p(t)=-2\) is \[
\{-\infty Since both initial conditions are not at zero, then let \begin {align*} y(0) &= c_{1}\\ 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 )-2 s Y \left (s \right )+2 y \left (0\right )-3 Y \left (s \right ) = 0\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=c_{1}\\ 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} -s c_{1} -2 s Y \left (s \right )+2 c_{1} -3 Y \left (s \right ) = 0 \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {s c_{1} -2 c_{1} +c_{2}}{s^{2}-2 s -3} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= \frac {\frac {3 c_{1}}{4}-\frac {c_{2}}{4}}{s +1}+\frac {\frac {c_{1}}{4}+\frac {c_{2}}{4}}{s -3} \] The inverse Laplace of each term above
is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {\frac {3 c_{1}}{4}-\frac {c_{2}}{4}}{s +1}\right ) &= \frac {\left (3 c_{1} -c_{2} \right ) {\mathrm e}^{-t}}{4}\\ \mathcal {L}^{-1}\left (\frac {\frac {c_{1}}{4}+\frac {c_{2}}{4}}{s -3}\right ) &= \frac {\left (c_{1} +c_{2} \right ) {\mathrm e}^{3 t}}{4} \end {align*}
Adding the above results and simplifying gives \[ y=\frac {{\mathrm e}^{t} \left (2 c_{1} \cosh \left (2 t \right )+\sinh \left (2 t \right ) \left (-c_{1} +c_{2} \right )\right )}{2} \] Since both initial conditions given are not at
zero, then we need to setup two equations to solve for \(c_{1},c_{1}\). At \(t=4\) the first equation becomes, using
the above solution \begin {align*} -3 &= \frac {{\mathrm e}^{4} \left (2 c_{1} \cosh \left (8\right )+\sinh \left (8\right ) \left (-c_{1} +c_{2} \right )\right )}{2} \end {align*}
And taking derivative of the solution and evaluating at \(t=4\) gives the second equation as
\begin {align*} -17 &= \frac {{\mathrm e}^{4} \left (2 c_{1} \cosh \left (8\right )+\sinh \left (8\right ) \left (-c_{1} +c_{2} \right )\right )}{2}+\frac {{\mathrm e}^{4} \left (4 c_{1} \sinh \left (8\right )+2 \cosh \left (8\right ) \left (-c_{1} +c_{2} \right )\right )}{2} \end {align*}
Solving gives \begin {align*} c_{1} &= -\frac {{\mathrm e}^{-4} \left (-7 \sinh \left (8\right )+3 \cosh \left (8\right )\right )}{\cosh \left (8\right )^{2}-\sinh \left (8\right )^{2}}\\ c_{2} &= -\frac {\left (17 \cosh \left (8\right )-13 \sinh \left (8\right )\right ) {\mathrm e}^{-4}}{\cosh \left (8\right )^{2}-\sinh \left (8\right )^{2}} \end {align*}
Subtituting these in the solution obtained above gives \begin {align*} y &= \frac {{\mathrm e}^{t} \left (-\frac {2 \,{\mathrm e}^{-4} \left (-7 \sinh \left (8\right )+3 \cosh \left (8\right )\right ) \cosh \left (2 t \right )}{\cosh \left (8\right )^{2}-\sinh \left (8\right )^{2}}+\sinh \left (2 t \right ) \left (\frac {{\mathrm e}^{-4} \left (-7 \sinh \left (8\right )+3 \cosh \left (8\right )\right )}{\cosh \left (8\right )^{2}-\sinh \left (8\right )^{2}}-\frac {\left (17 \cosh \left (8\right )-13 \sinh \left (8\right )\right ) {\mathrm e}^{-4}}{\cosh \left (8\right )^{2}-\sinh \left (8\right )^{2}}\right )\right )}{2}\\ &= -3 \left (\left (\cosh \left (8\right )-\frac {7 \sinh \left (8\right )}{3}\right ) \cosh \left (2 t \right )+\frac {7 \left (\cosh \left (8\right )-\frac {3 \sinh \left (8\right )}{7}\right ) \sinh \left (2 t \right )}{3}\right ) {\mathrm e}^{t -4} \end {align*}
Simplifying the solution gives \[
y = -3 \left (\left (\cosh \left (8\right )-\frac {7 \sinh \left (8\right )}{3}\right ) \cosh \left (2 t \right )+\frac {7 \left (\cosh \left (8\right )-\frac {3 \sinh \left (8\right )}{7}\right ) \sinh \left (2 t \right )}{3}\right ) {\mathrm e}^{t -4}
\] The solution(s) found are the following \begin{align*}
\tag{1} y &= -3 \left (\left (\cosh \left (8\right )-\frac {7 \sinh \left (8\right )}{3}\right ) \cosh \left (2 t \right )+\frac {7 \left (\cosh \left (8\right )-\frac {3 \sinh \left (8\right )}{7}\right ) \sinh \left (2 t \right )}{3}\right ) {\mathrm e}^{t -4} \\
\end{align*} Verification of solutions
\[
y = -3 \left (\left (\cosh \left (8\right )-\frac {7 \sinh \left (8\right )}{3}\right ) \cosh \left (2 t \right )+\frac {7 \left (\cosh \left (8\right )-\frac {3 \sinh \left (8\right )}{7}\right ) \sinh \left (2 t \right )}{3}\right ) {\mathrm e}^{t -4}
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-2 y^{\prime }-3 y=0, y \left (4\right )=-3, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}4\right \}}}}=-17\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}-2 r -3=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +1\right ) \left (r -3\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-1, 3\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-t} \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{3 t} \\ \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={\mathrm e}^{-t} c_{1} +c_{2} {\mathrm e}^{3 t} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y={\mathrm e}^{-t} c_{1} +c_{2} {\mathrm e}^{3 t} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (4\right )=-3 \\ {} & {} & -3={\mathrm e}^{-4} c_{1} +c_{2} {\mathrm e}^{12} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-{\mathrm e}^{-t} c_{1} +3 c_{2} {\mathrm e}^{3 t} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}4\right \}}}}=-17 \\ {} & {} & -17=-{\mathrm e}^{-4} c_{1} +3 c_{2} {\mathrm e}^{12} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =\frac {2}{{\mathrm e}^{-4}}, c_{2} =-\frac {5}{{\mathrm e}^{12}}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=2 \,{\mathrm e}^{-t +4}-5 \,{\mathrm e}^{-12+3 t} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=2 \,{\mathrm e}^{-t +4}-5 \,{\mathrm e}^{-12+3 t} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.672 (sec). Leaf size: 21
\[
y \left (t \right ) = -5 \,{\mathrm e}^{3 t -12}+2 \,{\mathrm e}^{-t +4}
\]
✓ Solution by Mathematica
Time used: 0.013 (sec). Leaf size: 24
\[
y(t)\to 2 e^{4-t}-5 e^{3 (t-4)}
\]
6.12.2 Maple step by step solution
`Methods for second order ODEs:
--- Trying classification methods ---
trying a quadrature
checking if the LODE has constant coefficients
<- constant coefficients successful`
dsolve([diff(y(t),t$2)-2*diff(y(t),t)-3*y(t)=0,y(4) = -3, D(y)(4) = -17],y(t), singsol=all)
DSolve[{y''[t]-2*y'[t]-3*y[t]==0,{y[4]==-3,y'[4]==-17}},y[t],t,IncludeSingularSolutions -> True]