Internal problem ID [12297]
Internal file name [OUTPUT/10950_Saturday_September_30_2023_08_26_35_PM_64669449/index.tex]
Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A.
Dobrushkin. CRC Press 2015
Section: Chapter 5.5 Laplace transform. Homogeneous equations. Problems page 357
Problem number: Problem 17.
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]]
\[ \boxed {3 y^{\prime \prime }+8 y^{\prime }-3 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 3, y^{\prime }\left (0\right ) = -4] \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) &={\frac {8}{3}}\\ q(t) &=-1\\ F &=0 \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+\frac {8 y^{\prime }}{3}-y = 0 \end {align*}
The domain of \(p(t)={\frac {8}{3}}\) 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*} 3 s^{2} Y \left (s \right )-3 y^{\prime }\left (0\right )-3 s y \left (0\right )+8 s Y \left (s \right )-8 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 )&=3\\ y'(0) &=-4 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} 3 s^{2} Y \left (s \right )-12-9 s +8 s Y \left (s \right )-3 Y \left (s \right ) = 0 \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {9 s +12}{3 s^{2}+8 s -3} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= \frac {3}{2 \left (s -\frac {1}{3}\right )}+\frac {3}{2 \left (s +3\right )} \] The inverse Laplace of each term above
is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {3}{2 \left (s -\frac {1}{3}\right )}\right ) &= \frac {3 \,{\mathrm e}^{\frac {t}{3}}}{2}\\ \mathcal {L}^{-1}\left (\frac {3}{2 \left (s +3\right )}\right ) &= \frac {3 \,{\mathrm e}^{-3 t}}{2} \end {align*}
Adding the above results and simplifying gives \[ y=3 \,{\mathrm e}^{-\frac {4 t}{3}} \cosh \left (\frac {5 t}{3}\right ) \] Simplifying the solution gives \[
y = 3 \,{\mathrm e}^{-\frac {4 t}{3}} \cosh \left (\frac {5 t}{3}\right )
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= 3 \,{\mathrm e}^{-\frac {4 t}{3}} \cosh \left (\frac {5 t}{3}\right ) \\
\end{align*} Verification of solutions
\[
y = 3 \,{\mathrm e}^{-\frac {4 t}{3}} \cosh \left (\frac {5 t}{3}\right )
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [3 y^{\prime \prime }+8 y^{\prime }-3 y=0, y \left (0\right )=3, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-4\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {8 y^{\prime }}{3}+y \\ \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} \\ {} & {} & y^{\prime \prime }+\frac {8 y^{\prime }}{3}-y=0 \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}+\frac {8}{3} r -1=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \frac {\left (r +3\right ) \left (3 r -1\right )}{3}=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-3, \frac {1}{3}\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-3 t} \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{\frac {t}{3}} \\ \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}^{-3 t}+c_{2} {\mathrm e}^{\frac {t}{3}} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-3 t}+c_{2} {\mathrm e}^{\frac {t}{3}} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=3 \\ {} & {} & 3=c_{1} +c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-3 c_{1} {\mathrm e}^{-3 t}+\frac {c_{2} {\mathrm e}^{\frac {t}{3}}}{3} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-4 \\ {} & {} & -4=-3 c_{1} +\frac {c_{2}}{3} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =\frac {3}{2}, c_{2} =\frac {3}{2}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {3 \left ({\mathrm e}^{\frac {10 t}{3}}+1\right ) {\mathrm e}^{-3 t}}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {3 \left ({\mathrm e}^{\frac {10 t}{3}}+1\right ) {\mathrm e}^{-3 t}}{2} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 4.625 (sec). Leaf size: 14
\[
y \left (t \right ) = 3 \,{\mathrm e}^{-\frac {4 t}{3}} \cosh \left (\frac {5 t}{3}\right )
\]
✓ Solution by Mathematica
Time used: 0.023 (sec). Leaf size: 23
\[
y(t)\to \frac {3}{2} e^{-3 t} \left (e^{10 t/3}+1\right )
\]
3.16.2 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([3*diff(y(t),t$2)+8*diff(y(t),t)-3*y(t)=0,y(0) = 3, D(y)(0) = -4],y(t), singsol=all)
DSolve[{3*y''[t]+8*y'[t]-3*y[t]==0,{y[0]==3,y'[0]==-4}},y[t],t,IncludeSingularSolutions -> True]