Internal problem ID [2866]
Internal file name [OUTPUT/2358_Sunday_June_05_2022_03_00_26_AM_76402982/index.tex]
Book: Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth
edition, 2015
Section: Chapter 10, The Laplace Transform and Some Elementary Applications. Exercises for
10.4. page 689
Problem number: Problem 28.
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]]
\[ \boxed {y^{\prime \prime }-y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = A, y^{\prime }\left (0\right ) = B] \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) &=0\\ q(t) &=-1\\ F &=0 \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }-y = 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 )-Y \left (s \right ) = 0\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=A\\ y'(0) &=B \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-B -s A -Y \left (s \right ) = 0 \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {s A +B}{s^{2}-1} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= \frac {\frac {A}{2}+\frac {B}{2}}{s -1}+\frac {\frac {A}{2}-\frac {B}{2}}{s +1} \] The inverse Laplace of each term above
is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {\frac {A}{2}+\frac {B}{2}}{s -1}\right ) &= \frac {\left (A +B \right ) {\mathrm e}^{t}}{2}\\ \mathcal {L}^{-1}\left (\frac {\frac {A}{2}-\frac {B}{2}}{s +1}\right ) &= \frac {\left (A -B \right ) {\mathrm e}^{-t}}{2} \end {align*}
Adding the above results and simplifying gives \[ y=A \cosh \left (t \right )+B \sinh \left (t \right ) \] Simplifying the solution gives \[
y = A \cosh \left (t \right )+B \sinh \left (t \right )
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= A \cosh \left (t \right )+B \sinh \left (t \right ) \\
\end{align*} Verification of solutions
\[
y = A \cosh \left (t \right )+B \sinh \left (t \right )
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }-y=0, y \left (0\right )=A , y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=B \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}-1=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -1\right ) \left (r +1\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-1, 1\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}^{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=c_{1} {\mathrm e}^{-t}+c_{2} {\mathrm e}^{t} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-t}+c_{2} {\mathrm e}^{t} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=A \\ {} & {} & A =c_{1} +c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-c_{1} {\mathrm e}^{-t}+c_{2} {\mathrm e}^{t} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=B \\ {} & {} & B =-c_{1} +c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =\frac {A}{2}-\frac {B}{2}, c_{2} =\frac {A}{2}+\frac {B}{2}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {\left (A -B \right ) {\mathrm e}^{-t}}{2}+\frac {\left (A +B \right ) {\mathrm e}^{t}}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\left (A -B \right ) {\mathrm e}^{-t}}{2}+\frac {\left (A +B \right ) {\mathrm e}^{t}}{2} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 1.703 (sec). Leaf size: 13
\[
y \left (t \right ) = A \cosh \left (t \right )+B \sinh \left (t \right )
\]
✓ Solution by Mathematica
Time used: 0.012 (sec). Leaf size: 33
\[
y(t)\to \frac {1}{2} e^{-t} \left (a \left (e^{2 t}+1\right )+b \left (e^{2 t}-1\right )\right )
\]
13.28.2 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)-y(t)=0,y(0) = A, D(y)(0) = B],y(t), singsol=all)
DSolve[{y''[t]-y[t]==0,{y[0]==a,y'[0]==b}},y[t],t,IncludeSingularSolutions -> True]