Internal problem ID [5212]
Internal file name [OUTPUT/4705_Sunday_June_05_2022_03_03_44_PM_45275139/index.tex]
Book: Schaums Outline Differential Equations, 4th edition. Bronson and Costa. McGraw Hill
2014
Section: Chapter 24. Solutions of linear DE by Laplace transforms. Supplementary Problems.
page 248
Problem number: Problem 24.36.
ODE order: 4.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_laplace"
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 ) = 1, y^{\prime }\left (0\right ) = 0, y^{\prime \prime }\left (0\right ) = 0, y^{\prime \prime \prime }\left (0\right ) = 0] \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 )\\ \mathcal {L}\left (y^{\prime \prime \prime \prime }\right ) &= s^4 Y(s) - y'''(0) - s y''(0) - s^2 y'(0)- s^3 y \left (0\right ) \end {align*}
The given ode becomes an algebraic equation in the Laplace domain \[ s^{4} Y \left (s \right )-y^{\prime \prime \prime }\left (0\right )-s y^{\prime \prime }\left (0\right )-s^{2} y^{\prime }\left (0\right )-s^{3} y \left (0\right )-Y \left (s \right ) = 0\tag {1} \] But the initial conditions are \begin {align*} y \left (0\right )&=1\\ y^{\prime }\left (0\right )&=0\\ y^{\prime \prime }\left (0\right )&=0\\ y^{\prime \prime \prime }\left (0\right )&=0 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \[ s^{4} Y \left (s \right )-s^{3}-Y \left (s \right ) = 0 \] Solving the above equation for \(Y(s)\) results in \[ Y(s) = \frac {s^{3}}{s^{4}-1} \] Applying partial fractions decomposition results in \[ Y(s)= \frac {1}{4 s -4}+\frac {1}{4 s -4 i}+\frac {1}{4 s +4 i}+\frac {1}{4 s +4} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {1}{4 s -4}\right ) &= \frac {{\mathrm e}^{x}}{4}\\ \mathcal {L}^{-1}\left (\frac {1}{4 s -4 i}\right ) &= \frac {{\mathrm e}^{i x}}{4}\\ \mathcal {L}^{-1}\left (\frac {1}{4 s +4 i}\right ) &= \frac {{\mathrm e}^{-i x}}{4}\\ \mathcal {L}^{-1}\left (\frac {1}{4 s +4}\right ) &= \frac {{\mathrm e}^{-x}}{4} \end {align*}
Adding the above results and simplifying gives \[ y=\frac {\cos \left (x \right )}{2}+\frac {\cosh \left (x \right )}{2} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\cos \left (x \right )}{2}+\frac {\cosh \left (x \right )}{2} \\ \end{align*}
Verification of solutions
\[ y = \frac {\cos \left (x \right )}{2}+\frac {\cosh \left (x \right )}{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime \prime }-y=0, y \left (0\right )=1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0, 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 (\cos \left (x \right )-\mathrm {I} \sin \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 (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right ) \\ -\cos \left (x \right )+\mathrm {I} \sin \left (x \right ) \\ \mathrm {I} \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right ) \\ \cos \left (x \right )-\mathrm {I} \sin \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}}=c_{1} {\mathrm e}^{-x}\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=-c_{1} {\mathrm e}^{-x}+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 )=1 \\ {} & {} & 1=-c_{1} +c_{2} -c_{4} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=c_{1} {\mathrm e}^{-x}+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 \}}}}=0 \\ {} & {} & 0=c_{1} +c_{2} -c_{3} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-c_{1} {\mathrm e}^{-x}+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 }=c_{1} {\mathrm e}^{-x}+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} =-\frac {1}{4}, c_{2} =\frac {1}{4}, c_{3} =0, c_{4} =-\frac {1}{2}\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\cos \left (x \right )}{2}+\frac {{\mathrm e}^{-x}}{4}+\frac {{\mathrm e}^{x}}{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: 0.422 (sec). Leaf size: 13
dsolve([diff(y(x),x$4)-y(x)=0,y(0) = 1, D(y)(0) = 0, (D@@2)(y)(0) = 0, (D@@3)(y)(0) = 0],y(x), singsol=all)
\[ y \left (x \right ) = \frac {\cos \left (x \right )}{2}+\frac {\cosh \left (x \right )}{2} \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 22
DSolve[{y''''[x]-y[x]==0,{y[0]==1,y'[0]==0,y''[0]==0,y'''[0]==0}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{4} \left (e^{-x}+e^x+2 \cos (x)\right ) \]