Internal problem ID [12306]
Internal file name [OUTPUT/10959_Saturday_September_30_2023_08_26_37_PM_89131153/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 26.
ODE order: 3.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_laplace"
Maple gives the following as the ode type
[[_3rd_order, _missing_x]]
\[ \boxed {y^{\prime \prime \prime }-6 y^{\prime \prime }+10 y^{\prime }=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = 3, y^{\prime \prime }\left (0\right ) = 8] \end {align*}
Solving using the Laplace transform method. Let \[ \mathcal {L}\left (y\right ) =Y(s) \] 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 )\\ \mathcal {L}\left (y^{\prime \prime \prime }\right ) &= s^3 Y(s) - y''(0) - s y'(0) - s^2 y \left (0\right ) \end {align*}
The given ode becomes an algebraic equation in the Laplace domain \[ s^{3} Y \left (s \right )-y^{\prime \prime }\left (0\right )-s y^{\prime }\left (0\right )-s^{2} y \left (0\right )-6 s^{2} Y \left (s \right )+6 y^{\prime }\left (0\right )+6 s y \left (0\right )+10 s Y \left (s \right )-10 y \left (0\right ) = 0\tag {1} \] But the initial conditions are \begin {align*} y \left (0\right )&=1\\ y^{\prime }\left (0\right )&=3\\ y^{\prime \prime }\left (0\right )&=8 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \[ s^{3} Y \left (s \right )+3 s -s^{2}-6 s^{2} Y \left (s \right )+10 s Y \left (s \right ) = 0 \] Solving the above equation for \(Y(s)\) results in \[ Y(s) = \frac {s -3}{s^{2}-6 s +10} \] Applying partial fractions decomposition results in \[ Y(s)= \frac {1}{2 s -6-2 i}+\frac {1}{2 s -6+2 i} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {1}{2 s -6-2 i}\right ) &= \frac {{\mathrm e}^{\left (3+i\right ) t}}{2}\\ \mathcal {L}^{-1}\left (\frac {1}{2 s -6+2 i}\right ) &= \frac {{\mathrm e}^{\left (3-i\right ) t}}{2} \end {align*}
Adding the above results and simplifying gives \[ y={\mathrm e}^{3 t} \cos \left (t \right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{3 t} \cos \left (t \right ) \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{3 t} \cos \left (t \right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime }-6 y^{\prime \prime }+10 y^{\prime }=0, y \left (0\right )=1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=3, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=8\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\prime \prime \prime } \\ \square & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{1}\left (t \right ) \\ {} & {} & y_{1}\left (t \right )=y \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (t \right ) \\ {} & {} & y_{2}\left (t \right )=y^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (t \right ) \\ {} & {} & y_{3}\left (t \right )=y^{\prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{3}^{\prime }\left (t \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{3}^{\prime }\left (t \right )=6 y_{3}\left (t \right )-10 y_{2}\left (t \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (t \right )=y_{1}^{\prime }\left (t \right ), y_{3}\left (t \right )=y_{2}^{\prime }\left (t \right ), y_{3}^{\prime }\left (t \right )=6 y_{3}\left (t \right )-10 y_{2}\left (t \right )\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (t \right )=\left [\begin {array}{c} y_{1}\left (t \right ) \\ y_{2}\left (t \right ) \\ y_{3}\left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (t \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & -10 & 6 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \right ) \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & -10 & 6 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (t \right )=A \cdot {\moverset {\rightarrow }{y}}\left (t \right ) \\ \bullet & {} & \textrm {To solve the system, find the eigenvalues and eigenvectors of}\hspace {3pt} A \\ \bullet & {} & \textrm {Eigenpairs of}\hspace {3pt} A \\ {} & {} & \left [\left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ], \left [3-\mathrm {I}, \left [\begin {array}{c} \frac {2}{25}+\frac {3 \,\mathrm {I}}{50} \\ \frac {3}{10}+\frac {\mathrm {I}}{10} \\ 1 \end {array}\right ]\right ], \left [3+\mathrm {I}, \left [\begin {array}{c} \frac {2}{25}-\frac {3 \,\mathrm {I}}{50} \\ \frac {3}{10}-\frac {\mathrm {I}}{10} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}=\left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [3-\mathrm {I}, \left [\begin {array}{c} \frac {2}{25}+\frac {3 \,\mathrm {I}}{50} \\ \frac {3}{10}+\frac {\mathrm {I}}{10} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (3-\mathrm {I}\right ) t}\cdot \left [\begin {array}{c} \frac {2}{25}+\frac {3 \,\mathrm {I}}{50} \\ \frac {3}{10}+\frac {\mathrm {I}}{10} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Use Euler identity to write solution in terms of}\hspace {3pt} \sin \hspace {3pt}\textrm {and}\hspace {3pt} \cos \\ {} & {} & {\mathrm e}^{3 t}\cdot \left (\cos \left (t \right )-\mathrm {I} \sin \left (t \right )\right )\cdot \left [\begin {array}{c} \frac {2}{25}+\frac {3 \,\mathrm {I}}{50} \\ \frac {3}{10}+\frac {\mathrm {I}}{10} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{3 t}\cdot \left [\begin {array}{c} \left (\frac {2}{25}+\frac {3 \,\mathrm {I}}{50}\right ) \left (\cos \left (t \right )-\mathrm {I} \sin \left (t \right )\right ) \\ \left (\frac {3}{10}+\frac {\mathrm {I}}{10}\right ) \left (\cos \left (t \right )-\mathrm {I} \sin \left (t \right )\right ) \\ \cos \left (t \right )-\mathrm {I} \sin \left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{y}}_{2}\left (t \right )={\mathrm e}^{3 t}\cdot \left [\begin {array}{c} \frac {2 \cos \left (t \right )}{25}+\frac {3 \sin \left (t \right )}{50} \\ \frac {3 \cos \left (t \right )}{10}+\frac {\sin \left (t \right )}{10} \\ \cos \left (t \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (t \right )={\mathrm e}^{3 t}\cdot \left [\begin {array}{c} -\frac {2 \sin \left (t \right )}{25}+\frac {3 \cos \left (t \right )}{50} \\ -\frac {3 \sin \left (t \right )}{10}+\frac {\cos \left (t \right )}{10} \\ -\sin \left (t \right ) \end {array}\right ]\right ] \\ \bullet & {} & \textrm {General solution to the system of ODEs}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (t \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (t \right ) \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{2} {\mathrm e}^{3 t}\cdot \left [\begin {array}{c} \frac {2 \cos \left (t \right )}{25}+\frac {3 \sin \left (t \right )}{50} \\ \frac {3 \cos \left (t \right )}{10}+\frac {\sin \left (t \right )}{10} \\ \cos \left (t \right ) \end {array}\right ]+c_{3} {\mathrm e}^{3 t}\cdot \left [\begin {array}{c} -\frac {2 \sin \left (t \right )}{25}+\frac {3 \cos \left (t \right )}{50} \\ -\frac {3 \sin \left (t \right )}{10}+\frac {\cos \left (t \right )}{10} \\ -\sin \left (t \right ) \end {array}\right ]+\left [\begin {array}{c} c_{1} \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (\left (4 c_{2} +3 c_{3} \right ) \cos \left (t \right )+3 \sin \left (t \right ) \left (c_{2} -\frac {4 c_{3}}{3}\right )\right ) {\mathrm e}^{3 t}}{50}+c_{1} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=\frac {2 c_{2}}{25}+\frac {3 c_{3}}{50}+c_{1} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\left (-\left (4 c_{2} +3 c_{3} \right ) \sin \left (t \right )+3 \cos \left (t \right ) \left (c_{2} -\frac {4 c_{3}}{3}\right )\right ) {\mathrm e}^{3 t}}{50}+\frac {3 \left (\left (4 c_{2} +3 c_{3} \right ) \cos \left (t \right )+3 \sin \left (t \right ) \left (c_{2} -\frac {4 c_{3}}{3}\right )\right ) {\mathrm e}^{3 t}}{50} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=3 \\ {} & {} & 3=\frac {3 c_{2}}{10}+\frac {c_{3}}{10} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {\left (-\left (4 c_{2} +3 c_{3} \right ) \cos \left (t \right )-3 \sin \left (t \right ) \left (c_{2} -\frac {4 c_{3}}{3}\right )\right ) {\mathrm e}^{3 t}}{50}+\frac {3 \left (-\left (4 c_{2} +3 c_{3} \right ) \sin \left (t \right )+3 \cos \left (t \right ) \left (c_{2} -\frac {4 c_{3}}{3}\right )\right ) {\mathrm e}^{3 t}}{25}+\frac {9 \left (\left (4 c_{2} +3 c_{3} \right ) \cos \left (t \right )+3 \sin \left (t \right ) \left (c_{2} -\frac {4 c_{3}}{3}\right )\right ) {\mathrm e}^{3 t}}{50} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=8 \\ {} & {} & 8=c_{2} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =0, c_{2} =8, c_{3} =6\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{3 t} \cos \left (t \right ) \end {array} \]
Maple trace
`Methods for third order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 4.531 (sec). Leaf size: 11
dsolve([diff(y(t),t$3)-6*diff(y(t),t$2)+10*diff(y(t),t)=0,y(0) = 1, D(y)(0) = 3, (D@@2)(y)(0) = 8],y(t), singsol=all)
\[ y \left (t \right ) = {\mathrm e}^{3 t} \cos \left (t \right ) \]
✓ Solution by Mathematica
Time used: 0.212 (sec). Leaf size: 13
DSolve[{y'''[t]-6*y''[t]+10*y'[t]==0,{y[0]==1,y'[0]==3,y''[0]==8}},y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to e^{3 t} \cos (t) \]