Internal problem ID [13630]
Internal file name [OUTPUT/12802_Saturday_February_17_2024_08_44_53_AM_78520361/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 19. Arbitrary Homogeneous linear equations with constant coefficients.
Additional exercises page 369
Problem number: 19.3 (c).
ODE order: 4.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_linear_constant_coefficients_ODE"
Maple gives the following as the ode type
[[_high_order, _missing_x]]
\[ \boxed {y^{\prime \prime \prime \prime }+26 y^{\prime \prime }+25 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 6, y^{\prime }\left (0\right ) = -28, y^{\prime \prime }\left (0\right ) = -102, y^{\prime \prime \prime }\left (0\right ) = 622] \end {align*}
The characteristic equation is \[ \lambda ^{4}+26 \lambda ^{2}+25 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 5 i\\ \lambda _2 &= -5 i\\ \lambda _3 &= i\\ \lambda _4 &= -i \end {align*}
Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{i x} c_{1} +{\mathrm e}^{-5 i x} c_{2} +{\mathrm e}^{5 i x} c_{3} +{\mathrm e}^{-i x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= {\mathrm e}^{i x}\\ y_2 &= {\mathrm e}^{-5 i x}\\ y_3 &= {\mathrm e}^{5 i x}\\ y_4 &= {\mathrm e}^{-i x} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = {\mathrm e}^{i x} c_{1} +{\mathrm e}^{-5 i x} c_{2} +{\mathrm e}^{5 i x} c_{3} +{\mathrm e}^{-i x} c_{4} \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 6\) and \(x = 0\) in the above gives \begin {align*} 6 = c_{1} +c_{2} +c_{3} +c_{4}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = i {\mathrm e}^{i x} c_{1} -5 i {\mathrm e}^{-5 i x} c_{2} +5 i {\mathrm e}^{5 i x} c_{3} -i {\mathrm e}^{-i x} c_{4} \end {align*}
substituting \(y^{\prime } = -28\) and \(x = 0\) in the above gives \begin {align*} -28 = i \left (c_{1} -5 c_{2} +5 c_{3} -c_{4} \right )\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = -{\mathrm e}^{i x} c_{1} -25 \,{\mathrm e}^{-5 i x} c_{2} -25 \,{\mathrm e}^{5 i x} c_{3} -{\mathrm e}^{-i x} c_{4} \end {align*}
substituting \(y^{\prime \prime } = -102\) and \(x = 0\) in the above gives \begin {align*} -102 = -c_{1} -25 c_{2} -25 c_{3} -c_{4}\tag {3A} \end {align*}
Taking three derivatives of the solution gives \begin {align*} y^{\prime \prime \prime } = -i {\mathrm e}^{i x} c_{1} +125 i {\mathrm e}^{-5 i x} c_{2} -125 i {\mathrm e}^{5 i x} c_{3} +i {\mathrm e}^{-i x} c_{4} \end {align*}
substituting \(y^{\prime \prime \prime } = 622\) and \(x = 0\) in the above gives \begin {align*} 622 = -i \left (c_{1} -125 c_{2} +125 c_{3} -c_{4} \right )\tag {4A} \end {align*}
Equations {1A,2A,3A,4A} are now solved for \(\{c_{1}, c_{2}, c_{3}, c_{4}\}\). Solving for the constants gives \begin {align*} c_{1}&=1+\frac {13 i}{8}\\ c_{2}&=2-\frac {99 i}{40}\\ c_{3}&=2+\frac {99 i}{40}\\ c_{4}&=1-\frac {13 i}{8} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = 4 \cos \left (5 x \right )-\frac {99 \sin \left (5 x \right )}{20}+2 \cos \left (x \right )-\frac {13 \sin \left (x \right )}{4} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= 4 \cos \left (5 x \right )-\frac {99 \sin \left (5 x \right )}{20}+2 \cos \left (x \right )-\frac {13 \sin \left (x \right )}{4} \\ \end{align*}
Verification of solutions
\[ y = 4 \cos \left (5 x \right )-\frac {99 \sin \left (5 x \right )}{20}+2 \cos \left (x \right )-\frac {13 \sin \left (x \right )}{4} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime \prime }+26 y^{\prime \prime }+25 y=0, y \left (0\right )=6, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-28, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-102, y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=622\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & y^{\prime \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 {Define new variable}\hspace {3pt} y_{4}\left (x \right ) \\ {} & {} & y_{4}\left (x \right )=y^{\prime \prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{4}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{4}^{\prime }\left (x \right )=-26 y_{3}\left (x \right )-25 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_{4}\left (x \right )=y_{3}^{\prime }\left (x \right ), y_{4}^{\prime }\left (x \right )=-26 y_{3}\left (x \right )-25 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 ) \\ y_{4}\left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -25 & 0 & -26 & 0 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right ) \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -25 & 0 & -26 & 0 \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 [-5 \,\mathrm {I}, \left [\begin {array}{c} -\frac {\mathrm {I}}{125} \\ -\frac {1}{25} \\ \frac {\mathrm {I}}{5} \\ 1 \end {array}\right ]\right ], \left [\mathrm {-I}, \left [\begin {array}{c} \mathrm {-I} \\ -1 \\ \mathrm {I} \\ 1 \end {array}\right ]\right ], \left [\mathrm {I}, \left [\begin {array}{c} \mathrm {I} \\ -1 \\ \mathrm {-I} \\ 1 \end {array}\right ]\right ], \left [5 \,\mathrm {I}, \left [\begin {array}{c} \frac {\mathrm {I}}{125} \\ -\frac {1}{25} \\ -\frac {\mathrm {I}}{5} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-5 \,\mathrm {I}, \left [\begin {array}{c} -\frac {\mathrm {I}}{125} \\ -\frac {1}{25} \\ \frac {\mathrm {I}}{5} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-5 \,\mathrm {I} x}\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{125} \\ -\frac {1}{25} \\ \frac {\mathrm {I}}{5} \\ 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 (5 x \right )-\mathrm {I} \sin \left (5 x \right )\right )\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{125} \\ -\frac {1}{25} \\ \frac {\mathrm {I}}{5} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} -\frac {\mathrm {I}}{125} \left (\cos \left (5 x \right )-\mathrm {I} \sin \left (5 x \right )\right ) \\ -\frac {\cos \left (5 x \right )}{25}+\frac {\mathrm {I} \sin \left (5 x \right )}{25} \\ \frac {\mathrm {I}}{5} \left (\cos \left (5 x \right )-\mathrm {I} \sin \left (5 x \right )\right ) \\ \cos \left (5 x \right )-\mathrm {I} \sin \left (5 x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{y}}_{1}\left (x \right )=\left [\begin {array}{c} -\frac {\sin \left (5 x \right )}{125} \\ -\frac {\cos \left (5 x \right )}{25} \\ \frac {\sin \left (5 x \right )}{5} \\ \cos \left (5 x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{2}\left (x \right )=\left [\begin {array}{c} -\frac {\cos \left (5 x \right )}{125} \\ \frac {\sin \left (5 x \right )}{25} \\ \frac {\cos \left (5 x \right )}{5} \\ -\sin \left (5 x \right ) \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [\mathrm {-I}, \left [\begin {array}{c} \mathrm {-I} \\ -1 \\ \mathrm {I} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\mathrm {-I} x}\cdot \left [\begin {array}{c} \mathrm {-I} \\ -1 \\ \mathrm {I} \\ 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 (-\mathrm {I} \sin \left (x \right )+\cos \left (x \right )\right )\cdot \left [\begin {array}{c} \mathrm {-I} \\ -1 \\ \mathrm {I} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} \mathrm {-I} \left (-\mathrm {I} \sin \left (x \right )+\cos \left (x \right )\right ) \\ -\cos \left (x \right )+\mathrm {I} \sin \left (x \right ) \\ \mathrm {I} \left (-\mathrm {I} \sin \left (x \right )+\cos \left (x \right )\right ) \\ -\mathrm {I} \sin \left (x \right )+\cos \left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{y}}_{3}\left (x \right )=\left [\begin {array}{c} -\sin \left (x \right ) \\ -\cos \left (x \right ) \\ \sin \left (x \right ) \\ \cos \left (x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{4}\left (x \right )=\left [\begin {array}{c} -\cos \left (x \right ) \\ \sin \left (x \right ) \\ \cos \left (x \right ) \\ -\sin \left (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}\left (x \right )+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+c_{4} {\moverset {\rightarrow }{y}}_{4}\left (x \right ) \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=\left [\begin {array}{c} -c_{4} \cos \left (x \right )-c_{3} \sin \left (x \right )-\frac {c_{2} \cos \left (5 x \right )}{125}-\frac {c_{1} \sin \left (5 x \right )}{125} \\ c_{4} \sin \left (x \right )-c_{3} \cos \left (x \right )+\frac {c_{2} \sin \left (5 x \right )}{25}-\frac {c_{1} \cos \left (5 x \right )}{25} \\ c_{4} \cos \left (x \right )+c_{3} \sin \left (x \right )+\frac {c_{2} \cos \left (5 x \right )}{5}+\frac {c_{1} \sin \left (5 x \right )}{5} \\ -c_{4} \sin \left (x \right )+c_{3} \cos \left (x \right )-c_{2} \sin \left (5 x \right )+c_{1} \cos \left (5 x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=-c_{4} \cos \left (x \right )-c_{3} \sin \left (x \right )-\frac {c_{2} \cos \left (5 x \right )}{125}-\frac {c_{1} \sin \left (5 x \right )}{125} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=6 \\ {} & {} & 6=-c_{4} -\frac {c_{2}}{125} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=c_{4} \sin \left (x \right )-c_{3} \cos \left (x \right )+\frac {c_{2} \sin \left (5 x \right )}{25}-\frac {c_{1} \cos \left (5 x \right )}{25} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-28 \\ {} & {} & -28=-c_{3} -\frac {c_{1}}{25} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=c_{4} \cos \left (x \right )+c_{3} \sin \left (x \right )+\frac {c_{2} \cos \left (5 x \right )}{5}+\frac {c_{1} \sin \left (5 x \right )}{5} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-102 \\ {} & {} & -102=c_{4} +\frac {c_{2}}{5} \\ \bullet & {} & \textrm {Calculate the 3rd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=-c_{4} \sin \left (x \right )+c_{3} \cos \left (x \right )-c_{2} \sin \left (5 x \right )+c_{1} \cos \left (5 x \right ) \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=622 \\ {} & {} & 622=c_{1} +c_{3} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =\frac {2475}{4}, c_{2} =-500, c_{3} =\frac {13}{4}, c_{4} =-2\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=4 \cos \left (5 x \right )-\frac {99 \sin \left (5 x \right )}{20}+2 \cos \left (x \right )-\frac {13 \sin \left (x \right )}{4} \end {array} \]
Maple trace
`Methods for high order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 25
dsolve([diff(y(x),x$4)+26*diff(y(x),x$2)+25*y(x)=0,y(0) = 6, D(y)(0) = -28, (D@@2)(y)(0) = -102, (D@@3)(y)(0) = 622],y(x), singsol=all)
\[ y \left (x \right ) = -\frac {13 \sin \left (x \right )}{4}+2 \cos \left (x \right )-\frac {99 \sin \left (5 x \right )}{20}+4 \cos \left (5 x \right ) \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 30
DSolve[{y''''[x]+26*y''[x]+25*y[x]==0,{y[0]==6,y'[0]==-28,y''[0]==-102,y'''[0]==622}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -\frac {13 \sin (x)}{4}-\frac {99}{20} \sin (5 x)+2 \cos (x)+4 \cos (5 x) \]