Internal problem ID [13571]
Internal file name [OUTPUT/12743_Saturday_February_17_2024_08_44_26_AM_69163266/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 15. General solutions to Homogeneous linear differential equations. Additional
exercises page 294
Problem number: 15.5 (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]]
\[ \boxed {y^{\prime \prime \prime \prime }-y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 4, y^{\prime \prime }\left (0\right ) = 0, y^{\prime \prime \prime }\left (0\right ) = 0] \end {align*}
The characteristic equation is \[ \lambda ^{4}-1 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= -1\\ \lambda _3 &= i\\ \lambda _4 &= -i \end {align*}
Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{-x} c_{1} +c_{2} {\mathrm e}^{x}+{\mathrm e}^{i x} c_{3} +{\mathrm e}^{-i x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= {\mathrm e}^{-x}\\ y_2 &= {\mathrm e}^{x}\\ y_3 &= {\mathrm e}^{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}^{-x} c_{1} +c_{2} {\mathrm e}^{x}+{\mathrm e}^{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 \(y = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = c_{1} +c_{2} +c_{3} +c_{4}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -{\mathrm e}^{-x} c_{1} +c_{2} {\mathrm e}^{x}+i {\mathrm e}^{i x} c_{3} -i {\mathrm e}^{-i x} c_{4} \end {align*}
substituting \(y^{\prime } = 4\) and \(x = 0\) in the above gives \begin {align*} 4 = c_{3} i-c_{4} i-c_{1} +c_{2}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = {\mathrm e}^{-x} c_{1} +c_{2} {\mathrm e}^{x}-{\mathrm e}^{i x} c_{3} -{\mathrm e}^{-i x} c_{4} \end {align*}
substituting \(y^{\prime \prime } = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = c_{1} +c_{2} -c_{3} -c_{4}\tag {3A} \end {align*}
Taking three derivatives of the solution gives \begin {align*} y^{\prime \prime \prime } = -{\mathrm e}^{-x} c_{1} +c_{2} {\mathrm e}^{x}-i {\mathrm e}^{i x} c_{3} +i {\mathrm e}^{-i x} c_{4} \end {align*}
substituting \(y^{\prime \prime \prime } = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = -c_{3} i+c_{4} i-c_{1} +c_{2}\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\\ c_{2}&=1\\ c_{3}&=-i\\ c_{4}&=i \end {align*}
Substituting these values back in above solution results in \begin {align*} y = -{\mathrm e}^{-x}+{\mathrm e}^{x}+2 \sin \left (x \right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -{\mathrm e}^{-x}+{\mathrm e}^{x}+2 \sin \left (x \right ) \\ \end{align*}
Verification of solutions
\[ y = -{\mathrm e}^{-x}+{\mathrm e}^{x}+2 \sin \left (x \right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime \prime }-y=0, y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=4, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0, y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0\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 )=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 )=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 \\ 1 & 0 & 0 & 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 \\ 1 & 0 & 0 & 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 [-1, \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]\right ], \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 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 ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-1, \left [\begin {array}{c} -1 \\ 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 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{x}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \end {array}\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}+c_{2} {\moverset {\rightarrow }{y}}_{2}+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}}={\mathrm e}^{-x} c_{1} \cdot \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]+c_{2} {\mathrm e}^{x}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]+\left [\begin {array}{c} -c_{3} \sin \left (x \right )-c_{4} \cos \left (x \right ) \\ -c_{3} \cos \left (x \right )+c_{4} \sin \left (x \right ) \\ c_{3} \sin \left (x \right )+c_{4} \cos \left (x \right ) \\ c_{3} \cos \left (x \right )-c_{4} \sin \left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=-{\mathrm e}^{-x} c_{1} +c_{2} {\mathrm e}^{x}-c_{4} \cos \left (x \right )-c_{3} \sin \left (x \right ) \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=-c_{1} +c_{2} -c_{4} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }={\mathrm e}^{-x} c_{1} +c_{2} {\mathrm e}^{x}+c_{4} \sin \left (x \right )-c_{3} \cos \left (x \right ) \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=4 \\ {} & {} & 4=c_{1} +c_{2} -c_{3} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-{\mathrm e}^{-x} c_{1} +c_{2} {\mathrm e}^{x}+c_{4} \cos \left (x \right )+c_{3} \sin \left (x \right ) \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-c_{1} +c_{2} +c_{4} \\ \bullet & {} & \textrm {Calculate the 3rd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }={\mathrm e}^{-x} c_{1} +c_{2} {\mathrm e}^{x}-c_{4} \sin \left (x \right )+c_{3} \cos \left (x \right ) \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=c_{1} +c_{2} +c_{3} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =1, c_{2} =1, c_{3} =-2, c_{4} =0\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-{\mathrm e}^{-x}+{\mathrm e}^{x}+2 \sin \left (x \right ) \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: 0.031 (sec). Leaf size: 17
dsolve([diff(y(x),x$4)-y(x)=0,y(0) = 0, D(y)(0) = 4, (D@@2)(y)(0) = 0, (D@@3)(y)(0) = 0],y(x), singsol=all)
\[ y \left (x \right ) = -{\mathrm e}^{-x}+{\mathrm e}^{x}+2 \sin \left (x \right ) \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 20
DSolve[{y''''[x]-y[x]==0,{y[0]==0,y'[0]==4,y''[0]==0,y'''[0]==0}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -e^{-x}+e^x+2 \sin (x) \]