problem Problem 13(b)

Internal problem ID [12345]
Internal ﬁle name [OUTPUT/10998_Monday_October_02_2023_02_47_47_AM_18318295/index.tex]

Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A. Dobrushkin. CRC Press 2015
Section: Chapter 5.6 Laplace transform. Nonhomogeneous equations. Problems page 368
Problem number: Problem 13(b).
ODE order: 3.
ODE degree: 1.

\[ \boxed {y^{\prime \prime \prime }-2 y^{\prime \prime }-y^{\prime }+2 y=4 t} \] With initial conditions \begin {align*} [y \left (0\right ) = 2, y^{\prime }\left (0\right ) = -2, y^{\prime \prime }\left (0\right ) = 4] \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 \[ s^{3} Y \left (s \right )-y^{\prime \prime }\left (0\right )-s y^{\prime }\left (0\right )-s^{2} y \left (0\right )-2 s^{2} Y \left (s \right )+2 y^{\prime }\left (0\right )+2 s y \left (0\right )-s Y \left (s \right )+y \left (0\right )+2 Y \left (s \right ) = \frac {4}{s^{2}}\tag {1} \] But the initial conditions are \begin {align*} y \left (0\right )&=2\\ y^{\prime }\left (0\right )&=-2\\ y^{\prime \prime }\left (0\right )&=4 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \[ s^{3} Y \left (s \right )-6+6 s -2 s^{2}-2 s^{2} Y \left (s \right )-s Y \left (s \right )+2 Y \left (s \right ) = \frac {4}{s^{2}} \] Solving the above equation for \(Y(s)\) results in \[ Y(s) = \frac {2 s^{4}-6 s^{3}+6 s^{2}+4}{s^{2} \left (s^{3}-2 s^{2}-s +2\right )} \] Applying partial fractions decomposition results in \[ Y(s)= \frac {2}{s^{2}}-\frac {3}{s -1}+\frac {1}{s}+\frac {3}{s +1}+\frac {1}{s -2} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {2}{s^{2}}\right ) &= 2 t\\ \mathcal {L}^{-1}\left (-\frac {3}{s -1}\right ) &= -3 \,{\mathrm e}^{t}\\ \mathcal {L}^{-1}\left (\frac {1}{s}\right ) &= 1\\ \mathcal {L}^{-1}\left (\frac {3}{s +1}\right ) &= 3 \,{\mathrm e}^{-t}\\ \mathcal {L}^{-1}\left (\frac {1}{s -2}\right ) &= {\mathrm e}^{2 t} \end {align*}

Adding the above results and simplifying gives \[ y=1-6 \sinh \left (t \right )+{\mathrm e}^{2 t}+2 t \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= 1-6 \sinh \left (t \right )+{\mathrm e}^{2 t}+2 t \\ \end{align*}

Veriﬁcation of solutions

\[ y = 1-6 \sinh \left (t \right )+{\mathrm e}^{2 t}+2 t \] Veriﬁed OK.

4.38.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime }-2 y^{\prime \prime }-y^{\prime }+2 y=4 t , y \left (0\right )=2, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-2, y^{\prime \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} 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 )=4 t +2 y_{3}\left (t \right )+y_{2}\left (t \right )-2 y_{1}\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 )=4 t +2 y_{3}\left (t \right )+y_{2}\left (t \right )-2 y_{1}\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 \\ -2 & 1 & 2 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \right )+\left [\begin {array}{c} 0 \\ 0 \\ 4 t \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (t \right )=\left [\begin {array}{c} 0 \\ 0 \\ 4 t \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -2 & 1 & 2 \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 )+{\moverset {\rightarrow }{f}} \\ \bullet & {} & \textrm {To solve the system, find the eigenvalues and eigenvectors of}\hspace {3pt} A \\ \bullet & {} & \textrm {Eigenpairs of}\hspace {3pt} A \\ {} & {} & \left [\left [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ], \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]\right ], \left [2, \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-t}\cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{t}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [2, \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{2 t}\cdot \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {General solution of the system of ODEs can be written in terms of the particular solution}\hspace {3pt} {\moverset {\rightarrow }{y}}_{p}\left (t \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}\left (t \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3}+{\moverset {\rightarrow }{y}}_{p}\left (t \right ) \\ \square & {} & \textrm {Fundamental matrix}\hspace {3pt} \\ {} & \circ & \textrm {Let}\hspace {3pt} \phi \left (t \right )\hspace {3pt}\textrm {be the matrix whose columns are the independent solutions of the homogeneous system.}\hspace {3pt} \\ {} & {} & \phi \left (t \right )=\left [\begin {array}{ccc} {\mathrm e}^{-t} & {\mathrm e}^{t} & \frac {{\mathrm e}^{2 t}}{4} \\ -{\mathrm e}^{-t} & {\mathrm e}^{t} & \frac {{\mathrm e}^{2 t}}{2} \\ {\mathrm e}^{-t} & {\mathrm e}^{t} & {\mathrm e}^{2 t} \end {array}\right ] \\ {} & \circ & \textrm {The fundamental matrix,}\hspace {3pt} \Phi \left (t \right )\hspace {3pt}\textrm {is a normalized version of}\hspace {3pt} \phi \left (t \right )\hspace {3pt}\textrm {satisfying}\hspace {3pt} \Phi \left (0\right )=I \hspace {3pt}\textrm {where}\hspace {3pt} I \hspace {3pt}\textrm {is the identity matrix}\hspace {3pt} \\ {} & {} & \Phi \left (t \right )=\phi \left (t \right )\cdot \frac {1}{\phi \left (0\right )} \\ {} & \circ & \textrm {Substitute the value of}\hspace {3pt} \phi \left (t \right )\hspace {3pt}\textrm {and}\hspace {3pt} \phi \left (0\right ) \\ {} & {} & \Phi \left (t \right )=\left [\begin {array}{ccc} {\mathrm e}^{-t} & {\mathrm e}^{t} & \frac {{\mathrm e}^{2 t}}{4} \\ -{\mathrm e}^{-t} & {\mathrm e}^{t} & \frac {{\mathrm e}^{2 t}}{2} \\ {\mathrm e}^{-t} & {\mathrm e}^{t} & {\mathrm e}^{2 t} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} 1 & 1 & \frac {1}{4} \\ -1 & 1 & \frac {1}{2} \\ 1 & 1 & 1 \end {array}\right ]} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (t \right )=\left [\begin {array}{ccc} \frac {{\mathrm e}^{-t}}{3}+{\mathrm e}^{t}-\frac {{\mathrm e}^{2 t}}{3} & -\frac {{\mathrm e}^{-t}}{2}+\frac {{\mathrm e}^{t}}{2} & \frac {{\mathrm e}^{-t}}{6}-\frac {{\mathrm e}^{t}}{2}+\frac {{\mathrm e}^{2 t}}{3} \\ -\frac {{\mathrm e}^{-t}}{3}+{\mathrm e}^{t}-\frac {2 \,{\mathrm e}^{2 t}}{3} & \frac {{\mathrm e}^{-t}}{2}+\frac {{\mathrm e}^{t}}{2} & -\frac {{\mathrm e}^{-t}}{6}-\frac {{\mathrm e}^{t}}{2}+\frac {2 \,{\mathrm e}^{2 t}}{3} \\ \frac {{\mathrm e}^{-t}}{3}+{\mathrm e}^{t}-\frac {4 \,{\mathrm e}^{2 t}}{3} & -\frac {{\mathrm e}^{-t}}{2}+\frac {{\mathrm e}^{t}}{2} & \frac {{\mathrm e}^{-t}}{6}-\frac {{\mathrm e}^{t}}{2}+\frac {4 \,{\mathrm e}^{2 t}}{3} \end {array}\right ] \\ \square & {} & \textrm {Find a particular solution of the system of ODEs using variation of parameters}\hspace {3pt} \\ {} & \circ & \textrm {Let the particular solution be the fundamental matrix multiplied by}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right )\hspace {3pt}\textrm {and solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (t \right )=\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right ) \\ {} & \circ & \textrm {Take the derivative of the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}^{\prime }\left (t \right )=\Phi ^{\prime }\left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right ) \\ {} & \circ & \textrm {Substitute particular solution and its derivative into the system of ODEs}\hspace {3pt} \\ {} & {} & \Phi ^{\prime }\left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )=A \cdot \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+{\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {The fundamental matrix has columns that are solutions to the homogeneous system so its derivative follows that of the homogeneous system}\hspace {3pt} \\ {} & {} & A \cdot \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )=A \cdot \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+{\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {Cancel like terms}\hspace {3pt} \\ {} & {} & \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )={\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {Multiply by the inverse of the fundamental matrix}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )=\frac {1}{\Phi \left (t \right )}\cdot {\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {Integrate to solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right ) \\ {} & {} & {\moverset {\rightarrow }{v}}\left (t \right )=\int _{0}^{t}\frac {1}{\Phi \left (s \right )}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \\ {} & \circ & \textrm {Plug}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right )\hspace {3pt}\textrm {into the equation for the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (t \right )=\Phi \left (t \right )\cdot \left (\int _{0}^{t}\frac {1}{\Phi \left (s \right )}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \right ) \\ {} & \circ & \textrm {Plug in the fundamental matrix and the forcing function and compute}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (t \right )=\left [\begin {array}{c} 1+\frac {{\mathrm e}^{2 t}}{3}+2 t +\frac {2 \,{\mathrm e}^{-t}}{3}-2 \,{\mathrm e}^{t} \\ 2+\frac {2 \,{\mathrm e}^{2 t}}{3}-\frac {2 \,{\mathrm e}^{-t}}{3}-2 \,{\mathrm e}^{t} \\ \frac {4 \,{\mathrm e}^{2 t}}{3}+\frac {2 \,{\mathrm e}^{-t}}{3}-2 \,{\mathrm e}^{t} \end {array}\right ] \\ \bullet & {} & \textrm {Plug particular solution back into general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (t \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3}+\left [\begin {array}{c} 1+\frac {{\mathrm e}^{2 t}}{3}+2 t +\frac {2 \,{\mathrm e}^{-t}}{3}-2 \,{\mathrm e}^{t} \\ 2+\frac {2 \,{\mathrm e}^{2 t}}{3}-\frac {2 \,{\mathrm e}^{-t}}{3}-2 \,{\mathrm e}^{t} \\ \frac {4 \,{\mathrm e}^{2 t}}{3}+\frac {2 \,{\mathrm e}^{-t}}{3}-2 \,{\mathrm e}^{t} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=1+\frac {\left (2+3 c_{1} \right ) {\mathrm e}^{-t}}{3}+\frac {\left (3 c_{3} +4\right ) {\mathrm e}^{2 t}}{12}+\left (-2+c_{2} \right ) {\mathrm e}^{t}+2 t \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=2 \\ {} & {} & 2=c_{1} +\frac {c_{3}}{4}+c_{2} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {\left (2+3 c_{1} \right ) {\mathrm e}^{-t}}{3}+\frac {\left (3 c_{3} +4\right ) {\mathrm e}^{2 t}}{6}+\left (-2+c_{2} \right ) {\mathrm e}^{t}+2 \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-2 \\ {} & {} & -2=-c_{1} +\frac {c_{3}}{2}+c_{2} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {\left (2+3 c_{1} \right ) {\mathrm e}^{-t}}{3}+\frac {\left (3 c_{3} +4\right ) {\mathrm e}^{2 t}}{3}+\left (-2+c_{2} \right ) {\mathrm e}^{t} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=4 \\ {} & {} & 4=c_{1} +c_{3} +c_{2} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =\frac {7}{3}, c_{2} =-1, c_{3} =\frac {8}{3}\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=1+3 \,{\mathrm e}^{-t}+{\mathrm e}^{2 t}-3 \,{\mathrm e}^{t}+2 t \end {array} \]

Maple trace

`Methods for third order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 3; linear nonhomogeneous with symmetry [0,1] 
trying high order linear exact nonhomogeneous 
trying differential order: 3; missing the dependent variable 
checking if the LODE has constant coefficients 
<- constant coefficients successful`

4.38 problem Problem 13(b)

4.38.1 Maple step by step solution