Internal problem ID [1488]
Internal file name [OUTPUT/1489_Sunday_June_05_2022_02_19_29_AM_18528297/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 9 Introduction to Linear Higher Order Equations. Section 9.2. constant
coefficient. Page 483
Problem number: section 9.2, problem 24.
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]]
\[ \boxed {y^{\prime \prime \prime \prime }-6 y^{\prime \prime \prime }+7 y^{\prime \prime }+6 y^{\prime }-8 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = -2, y^{\prime }\left (0\right ) = -8, y^{\prime \prime }\left (0\right ) = -14, y^{\prime \prime \prime }\left (0\right ) = -62] \end {align*}
The characteristic equation is \[ \lambda ^{4}-6 \lambda ^{3}+7 \lambda ^{2}+6 \lambda -8 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= 2\\ \lambda _3 &= 4\\ \lambda _4 &= -1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x}+c_{3} {\mathrm e}^{2 x}+{\mathrm e}^{4 x} c_{4} \] 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}^{2 x}\\ y_4 &= {\mathrm e}^{4 x} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x}+c_{3} {\mathrm e}^{2 x}+{\mathrm e}^{4 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 \(y = -2\) and \(x = 0\) in the above gives \begin {align*} -2 = c_{1} +c_{2} +c_{3} +c_{4}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x}+2 c_{3} {\mathrm e}^{2 x}+4 \,{\mathrm e}^{4 x} c_{4} \end {align*}
substituting \(y^{\prime } = -8\) and \(x = 0\) in the above gives \begin {align*} -8 = -c_{1} +c_{2} +2 c_{3} +4 c_{4}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x}+4 c_{3} {\mathrm e}^{2 x}+16 \,{\mathrm e}^{4 x} c_{4} \end {align*}
substituting \(y^{\prime \prime } = -14\) and \(x = 0\) in the above gives \begin {align*} -14 = c_{1} +c_{2} +4 c_{3} +16 c_{4}\tag {3A} \end {align*}
Taking three derivatives of the solution gives \begin {align*} y^{\prime \prime \prime } = -c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x}+8 c_{3} {\mathrm e}^{2 x}+64 \,{\mathrm e}^{4 x} c_{4} \end {align*}
substituting \(y^{\prime \prime \prime } = -62\) and \(x = 0\) in the above gives \begin {align*} -62 = -c_{1} +c_{2} +8 c_{3} +64 c_{4}\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}&=2\\ c_{2}&=-4\\ c_{3}&=1\\ c_{4}&=-1 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = 2 \,{\mathrm e}^{-x}-4 \,{\mathrm e}^{x}+{\mathrm e}^{2 x}-{\mathrm e}^{4 x} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= 2 \,{\mathrm e}^{-x}-4 \,{\mathrm e}^{x}+{\mathrm e}^{2 x}-{\mathrm e}^{4 x} \\ \end{align*}
Verification of solutions
\[ y = 2 \,{\mathrm e}^{-x}-4 \,{\mathrm e}^{x}+{\mathrm e}^{2 x}-{\mathrm e}^{4 x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime \prime }-6 y^{\prime \prime \prime }+7 y^{\prime \prime }+6 y^{\prime }-8 y=0, y \left (0\right )=-2, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-8, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-14, y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-62\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 )=6 y_{4}\left (x \right )-7 y_{3}\left (x \right )-6 y_{2}\left (x \right )+8 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 )=6 y_{4}\left (x \right )-7 y_{3}\left (x \right )-6 y_{2}\left (x \right )+8 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 \\ 8 & -6 & -7 & 6 \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 \\ 8 & -6 & -7 & 6 \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 [2, \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [4, \left [\begin {array}{c} \frac {1}{64} \\ \frac {1}{16} \\ \frac {1}{4} \\ 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 eigenpair}\hspace {3pt} \\ {} & {} & \left [2, \left [\begin {array}{c} \frac {1}{8} \\ \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 x}\cdot \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [4, \left [\begin {array}{c} \frac {1}{64} \\ \frac {1}{16} \\ \frac {1}{4} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{4}={\mathrm e}^{4 x}\cdot \left [\begin {array}{c} \frac {1}{64} \\ \frac {1}{16} \\ \frac {1}{4} \\ 1 \end {array}\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}+c_{4} {\moverset {\rightarrow }{y}}_{4} \\ \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 ]+c_{3} {\mathrm e}^{2 x}\cdot \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]+{\mathrm e}^{4 x} c_{4} \cdot \left [\begin {array}{c} \frac {1}{64} \\ \frac {1}{16} \\ \frac {1}{4} \\ 1 \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}+\frac {c_{3} {\mathrm e}^{2 x}}{8}+\frac {{\mathrm e}^{4 x} c_{4}}{64} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=-2 \\ {} & {} & -2=-c_{1} +c_{2} +\frac {c_{3}}{8}+\frac {c_{4}}{64} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x}+\frac {c_{3} {\mathrm e}^{2 x}}{4}+\frac {{\mathrm e}^{4 x} c_{4}}{16} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-8 \\ {} & {} & -8=c_{1} +c_{2} +\frac {c_{3}}{4}+\frac {c_{4}}{16} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x}+\frac {c_{3} {\mathrm e}^{2 x}}{2}+\frac {{\mathrm e}^{4 x} c_{4}}{4} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-14 \\ {} & {} & -14=-c_{1} +c_{2} +\frac {c_{3}}{2}+\frac {c_{4}}{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_{3} {\mathrm e}^{2 x}+{\mathrm e}^{4 x} c_{4} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-62 \\ {} & {} & -62=c_{1} +c_{2} +c_{3} +c_{4} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =-2, c_{2} =-4, c_{3} =8, c_{4} =-64\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=2 \,{\mathrm e}^{-x}-4 \,{\mathrm e}^{x}+{\mathrm e}^{2 x}-{\mathrm e}^{4 x} \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: 25
dsolve([diff(y(x),x$4)-6*diff(y(x),x$3)+7*diff(y(x),x$2)+6*diff(y(x),x)-8*y(x)=0,y(0) = -2, D(y)(0) = -8, (D@@2)(y)(0) = -14, (D@@3)(y)(0) = -62],y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{2 x}-{\mathrm e}^{4 x}+2 \,{\mathrm e}^{-x}-4 \,{\mathrm e}^{x} \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 30
DSolve[{y''''[x]-6*y'''[x]+7*y''[x]+6*y'[x]-8*y[x]==0,{y[0]==-2,y'[0]==-8,y''[0]==-14,y'''[0]==-62}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to 2 e^{-x}-4 e^x+e^{2 x}-e^{4 x} \]