Internal problem ID [12805]
Internal file name [OUTPUT/11458_Saturday_November_04_2023_08_47_24_AM_22492681/index.tex]
Book: Ordinary Differential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 5. The Laplace Transform Method. Exercises 5.3, page 255
Problem number: 14.
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]]
\[ \boxed {y^{\prime \prime \prime }-y^{\prime \prime }+4 y^{\prime }-4 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 5, y^{\prime \prime }\left (0\right ) = 5] \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 \[ s^{3} Y \left (s \right )-y^{\prime \prime }\left (0\right )-s y^{\prime }\left (0\right )-s^{2} y \left (0\right )-s^{2} Y \left (s \right )+y^{\prime }\left (0\right )+s y \left (0\right )+4 s Y \left (s \right )-4 y \left (0\right )-4 Y \left (s \right ) = 0\tag {1} \] But the initial conditions are \begin {align*} y \left (0\right )&=0\\ y^{\prime }\left (0\right )&=5\\ y^{\prime \prime }\left (0\right )&=5 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \[ s^{3} Y \left (s \right )-5 s -s^{2} Y \left (s \right )+4 s Y \left (s \right )-4 Y \left (s \right ) = 0 \] Solving the above equation for \(Y(s)\) results in \[ Y(s) = \frac {5 s}{s^{3}-s^{2}+4 s -4} \] Applying partial fractions decomposition results in \[ Y(s)= \frac {-\frac {1}{2}-i}{s -2 i}+\frac {-\frac {1}{2}+i}{s +2 i}+\frac {1}{s -1} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {-\frac {1}{2}-i}{s -2 i}\right ) &= \left (-\frac {1}{2}-i\right ) {\mathrm e}^{2 i x}\\ \mathcal {L}^{-1}\left (\frac {-\frac {1}{2}+i}{s +2 i}\right ) &= \left (-\frac {1}{2}+i\right ) {\mathrm e}^{-2 i x}\\ \mathcal {L}^{-1}\left (\frac {1}{s -1}\right ) &= {\mathrm e}^{x} \end {align*}
Adding the above results and simplifying gives \[ y={\mathrm e}^{x}-\cos \left (2 x \right )+2 \sin \left (2 x \right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{x}-\cos \left (2 x \right )+2 \sin \left (2 x \right ) \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{x}-\cos \left (2 x \right )+2 \sin \left (2 x \right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime }-y^{\prime \prime }+4 y^{\prime }-4 y=0, y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=5, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=5\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 (x \right ) \\ {} & {} & y_{1}\left (x \right )=y \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (x \right ) \\ {} & {} & y_{2}\left (x \right )=y^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (x \right ) \\ {} & {} & y_{3}\left (x \right )=y^{\prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{3}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{3}^{\prime }\left (x \right )=y_{3}\left (x \right )-4 y_{2}\left (x \right )+4 y_{1}\left (x \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (x \right )=y_{1}^{\prime }\left (x \right ), y_{3}\left (x \right )=y_{2}^{\prime }\left (x \right ), y_{3}^{\prime }\left (x \right )=y_{3}\left (x \right )-4 y_{2}\left (x \right )+4 y_{1}\left (x \right )\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=\left [\begin {array}{c} y_{1}\left (x \right ) \\ y_{2}\left (x \right ) \\ y_{3}\left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 4 & -4 & 1 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right ) \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 4 & -4 & 1 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=A \cdot {\moverset {\rightarrow }{y}}\left (x \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 [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]\right ], \left [-2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ], \left [2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {1}{4} \\ -\frac {\mathrm {I}}{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}^{x}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-2 \,\mathrm {I} x}\cdot \left [\begin {array}{c} -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 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 \\ {} & {} & \left (\cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right )\right )\cdot \left [\begin {array}{c} -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} -\frac {\cos \left (2 x \right )}{4}+\frac {\mathrm {I} \sin \left (2 x \right )}{4} \\ \frac {\mathrm {I}}{2} \left (\cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right )\right ) \\ \cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \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 (x \right )=\left [\begin {array}{c} -\frac {\cos \left (2 x \right )}{4} \\ \frac {\sin \left (2 x \right )}{2} \\ \cos \left (2 x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (x \right )=\left [\begin {array}{c} \frac {\sin \left (2 x \right )}{4} \\ \frac {\cos \left (2 x \right )}{2} \\ -\sin \left (2 x \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 (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right ) \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\mathrm e}^{x}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]+\left [\begin {array}{c} -\frac {c_{2} \cos \left (2 x \right )}{4}+\frac {c_{3} \sin \left (2 x \right )}{4} \\ \frac {\sin \left (2 x \right ) c_{2}}{2}+\frac {c_{3} \cos \left (2 x \right )}{2} \\ c_{2} \cos \left (2 x \right )-c_{3} \sin \left (2 x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{x}+\frac {c_{3} \sin \left (2 x \right )}{4}-\frac {c_{2} \cos \left (2 x \right )}{4} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} -\frac {c_{2}}{4} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=c_{1} {\mathrm e}^{x}+\frac {c_{3} \cos \left (2 x \right )}{2}+\frac {\sin \left (2 x \right ) c_{2}}{2} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=5 \\ {} & {} & 5=c_{1} +\frac {c_{3}}{2} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=c_{1} {\mathrm e}^{x}-c_{3} \sin \left (2 x \right )+c_{2} \cos \left (2 x \right ) \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=5 \\ {} & {} & 5=c_{1} +c_{2} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =1, c_{2} =4, c_{3} =8\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{x}-\cos \left (2 x \right )+2 \sin \left (2 x \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: 5.813 (sec). Leaf size: 19
dsolve([diff(y(x),x$3)-diff(y(x),x$2)+4*diff(y(x),x)-4*y(x)=0,y(0) = 0, D(y)(0) = 5, (D@@2)(y)(0) = 5],y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{x}-\cos \left (2 x \right )+2 \sin \left (2 x \right ) \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 21
DSolve[{y'''[x]-y''[x]+4*y'[x]-4*y[x]==0,{y[0]==0,y'[0]==5,y''[0]==5}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^x+2 \sin (2 x)-\cos (2 x) \]