Internal problem ID [833]
Internal file name [OUTPUT/833_Sunday_June_05_2022_01_50_45_AM_6567445/index.tex]
Book: Elementary differential equations and boundary value problems, 11th ed., Boyce, DiPrima,
Meade
Section: Chapter 6.2, The Laplace Transform. Solution of Initial Value Problems. page
255
Problem number: 8.
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]]
\[ \boxed {y^{\prime \prime }-y^{\prime }-6 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = -1] \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) &=-1\\ q(t) &=-6\\ F &=0 \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }-y^{\prime }-6 y = 0 \end {align*}
The domain of \(p(t)=-1\) 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 )-s Y \left (s \right )+y \left (0\right )-6 Y \left (s \right ) = 0\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=1\\ y'(0) &=-1 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )+2-s -s Y \left (s \right )-6 Y \left (s \right ) = 0 \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {s -2}{s^{2}-s -6} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= \frac {1}{5 s -15}+\frac {4}{5 \left (s +2\right )} \] The inverse Laplace of each term above
is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {1}{5 s -15}\right ) &= \frac {{\mathrm e}^{3 t}}{5}\\ \mathcal {L}^{-1}\left (\frac {4}{5 \left (s +2\right )}\right ) &= \frac {4 \,{\mathrm e}^{-2 t}}{5} \end {align*}
Adding the above results and simplifying gives \[ y=\frac {{\mathrm e}^{3 t}}{5}+\frac {4 \,{\mathrm e}^{-2 t}}{5} \] Simplifying the solution gives \[
y = \frac {\left ({\mathrm e}^{5 t}+4\right ) {\mathrm e}^{-2 t}}{5}
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {\left ({\mathrm e}^{5 t}+4\right ) {\mathrm e}^{-2 t}}{5} \\
\end{align*} Verification of solutions
\[
y = \frac {\left ({\mathrm e}^{5 t}+4\right ) {\mathrm e}^{-2 t}}{5}
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-y^{\prime }-6 y=0, y \left (0\right )=1, 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 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}-r -6=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +2\right ) \left (r -3\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-2, 3\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-2 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=c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{3 t} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{3 t} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=c_{1} +c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-2 c_{1} {\mathrm e}^{-2 t}+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 {=}0\right \}}}}=-1 \\ {} & {} & -1=-2 c_{1} +3 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =\frac {4}{5}, c_{2} =\frac {1}{5}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {\left ({\mathrm e}^{5 t}+4\right ) {\mathrm e}^{-2 t}}{5} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\left ({\mathrm e}^{5 t}+4\right ) {\mathrm e}^{-2 t}}{5} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.609 (sec). Leaf size: 17
\[
y \left (t \right ) = \frac {\left ({\mathrm e}^{5 t}+4\right ) {\mathrm e}^{-2 t}}{5}
\]
✓ Solution by Mathematica
Time used: 0.013 (sec). Leaf size: 21
\[
y(t)\to \frac {1}{5} e^{-2 t} \left (e^{5 t}+4\right )
\]
3.1.2 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)-diff(y(t),t)-6*y(t)=0,y(0) = 1, D(y)(0) = -1],y(t), singsol=all)
DSolve[{y''[t]-y'[t]-6*y[t]==0,{y[0]==1,y'[0]==-1}},y[t],t,IncludeSingularSolutions -> True]