problem 10

Internal problem ID [2126]
Internal ﬁle name [OUTPUT/2126_Monday_February_26_2024_09_17_45_AM_53084472/index.tex]

Book: Diﬀerential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath. Boston. 1964
Section: Exercise 18, page 82
Problem number: 10.
ODE order: 3.
ODE degree: 1.

The type(s) of ODE detected by this program : "higher_order_linear_constant_coeﬃcients_ODE"

\[ \boxed {4 y^{\prime \prime \prime }-8 y^{\prime \prime }+5 y^{\prime }-y=0} \] The characteristic equation is \[ 4 \lambda ^{3}-8 \lambda ^{2}+5 \lambda -1 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= {\frac {1}{2}}\\ \lambda _3 &= {\frac {1}{2}} \end {align*}

Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{x}+c_{2} {\mathrm e}^{\frac {x}{2}}+x \,{\mathrm e}^{\frac {x}{2}} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= {\mathrm e}^{x}\\ y_2 &= {\mathrm e}^{\frac {x}{2}}\\ y_3 &= {\mathrm e}^{\frac {x}{2}} x \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{x}+c_{2} {\mathrm e}^{\frac {x}{2}}+x \,{\mathrm e}^{\frac {x}{2}} c_{3} \\ \end{align*}

Veriﬁcation of solutions

\[ y = c_{1} {\mathrm e}^{x}+c_{2} {\mathrm e}^{\frac {x}{2}}+x \,{\mathrm e}^{\frac {x}{2}} c_{3} \] Veriﬁed OK.

10.10.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 y^{\prime \prime \prime }-8 y^{\prime \prime }+5 y^{\prime }-y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\prime \prime \prime } \\ \bullet & {} & \textrm {Isolate 3rd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=2 y^{\prime \prime }-\frac {5 y^{\prime }}{4}+\frac {y}{4} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }-2 y^{\prime \prime }+\frac {5 y^{\prime }}{4}-\frac {y}{4}=0 \\ \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 )=2 y_{3}\left (x \right )-\frac {5 y_{2}\left (x \right )}{4}+\frac {y_{1}\left (x \right )}{4} \\ & {} & \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 )=2 y_{3}\left (x \right )-\frac {5 y_{2}\left (x \right )}{4}+\frac {y_{1}\left (x \right )}{4}\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 \\ \frac {1}{4} & -\frac {5}{4} & 2 \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 \\ \frac {1}{4} & -\frac {5}{4} & 2 \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 [\frac {1}{2}, \left [\begin {array}{c} 4 \\ 2 \\ 1 \end {array}\right ]\right ], \left [\frac {1}{2}, \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair, with eigenvalue of algebraic multiplicity 2}\hspace {3pt} \\ {} & {} & \left [\frac {1}{2}, \left [\begin {array}{c} 4 \\ 2 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {First solution from eigenvalue}\hspace {3pt} \frac {1}{2} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}\left (x \right )={\mathrm e}^{\frac {x}{2}}\cdot \left [\begin {array}{c} 4 \\ 2 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Form of the 2nd homogeneous solution where}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {is to be solved for,}\hspace {3pt} \lambda =\frac {1}{2}\hspace {3pt}\textrm {is the eigenvalue, and}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is the eigenvector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Note that the}\hspace {3pt} x \hspace {3pt}\textrm {multiplying}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {makes this solution linearly independent to the 1st solution obtained from}\hspace {3pt} \lambda =\frac {1}{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} {\moverset {\rightarrow }{y}}_{2}\left (x \right )\hspace {3pt}\textrm {into the homogeneous system}\hspace {3pt} \\ {} & {} & \lambda \,{\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda x} {\moverset {\rightarrow }{v}}=\left ({\mathrm e}^{\lambda x} A \right )\cdot \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Use the fact that}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is an eigenvector of}\hspace {3pt} A \\ {} & {} & \lambda \,{\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda x} {\moverset {\rightarrow }{v}}={\mathrm e}^{\lambda x} \left (\lambda x {\moverset {\rightarrow }{v}}+A \cdot {\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Simplify equation}\hspace {3pt} \\ {} & {} & \lambda {\moverset {\rightarrow }{p}}+{\moverset {\rightarrow }{v}}=A \cdot {\moverset {\rightarrow }{p}} \\ \bullet & {} & \textrm {Make use of the identity matrix}\hspace {3pt} \mathrm {I} \\ {} & {} & \left (\lambda \cdot I \right )\cdot {\moverset {\rightarrow }{p}}+{\moverset {\rightarrow }{v}}=A \cdot {\moverset {\rightarrow }{p}} \\ \bullet & {} & \textrm {Condition}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {must meet for}\hspace {3pt} {\moverset {\rightarrow }{y}}_{2}\left (x \right )\hspace {3pt}\textrm {to be a solution to the homogeneous system}\hspace {3pt} \\ {} & {} & \left (A -\lambda \cdot I \right )\cdot {\moverset {\rightarrow }{p}}={\moverset {\rightarrow }{v}} \\ \bullet & {} & \textrm {Choose}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {to use in the second solution to the homogeneous system from eigenvalue}\hspace {3pt} \frac {1}{2} \\ {} & {} & \left (\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ \frac {1}{4} & -\frac {5}{4} & 2 \end {array}\right ]-\frac {1}{2}\cdot \left [\begin {array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {array}\right ]\right )\cdot {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} 4 \\ 2 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Choice of}\hspace {3pt} {\moverset {\rightarrow }{p}} \\ {} & {} & {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} -8 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Second solution from eigenvalue}\hspace {3pt} \frac {1}{2} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{\frac {x}{2}}\cdot \left (x \cdot \left [\begin {array}{c} 4 \\ 2 \\ 1 \end {array}\right ]+\left [\begin {array}{c} -8 \\ 0 \\ 0 \end {array}\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}}_{3}={\mathrm e}^{x}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\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} \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\mathrm e}^{\frac {x}{2}}\cdot \left [\begin {array}{c} 4 \\ 2 \\ 1 \end {array}\right ]+c_{2} {\mathrm e}^{\frac {x}{2}}\cdot \left (x \cdot \left [\begin {array}{c} 4 \\ 2 \\ 1 \end {array}\right ]+\left [\begin {array}{c} -8 \\ 0 \\ 0 \end {array}\right ]\right )+{\mathrm e}^{x} c_{3} \cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\left (\left (4 x -8\right ) c_{2} +4 c_{1} \right ) {\mathrm e}^{\frac {x}{2}}+{\mathrm e}^{x} c_{3} \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`

\[ y \left (x \right ) = \left (c_{3} x +c_{2} \right ) {\mathrm e}^{\frac {x}{2}}+{\mathrm e}^{x} c_{1} \]

10.10 problem 10

10.10.1 Maple step by step solution