2.3 problem Problem 3

2.3.1 Maple step by step solution

Internal problem ID [12165]
Internal file name [OUTPUT/10818_Thursday_September_21_2023_05_47_23_AM_40160976/index.tex]

Book: Differential equations and the calculus of variations by L. ElSGOLTS. MIR PUBLISHERS, MOSCOW, Third printing 1977.
Section: Chapter 2, DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER. Problems page 172
Problem number: Problem 3.
ODE order: 3.
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

[[_3rd_order, _missing_x]]

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

Therefore the homogeneous solution is \[ y_h(x)=c_{1} +{\mathrm e}^{\left (\frac {3}{2}+\frac {\sqrt {5}}{2}\right ) x} c_{2} +{\mathrm e}^{\left (\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= 1\\ y_2 &= {\mathrm e}^{\left (\frac {3}{2}+\frac {\sqrt {5}}{2}\right ) x}\\ y_3 &= {\mathrm e}^{\left (\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) x} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} +{\mathrm e}^{\left (\frac {3}{2}+\frac {\sqrt {5}}{2}\right ) x} c_{2} +{\mathrm e}^{\left (\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) x} c_{3} \\ \end{align*}

Verification of solutions

\[ y = c_{1} +{\mathrm e}^{\left (\frac {3}{2}+\frac {\sqrt {5}}{2}\right ) x} c_{2} +{\mathrm e}^{\left (\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) x} c_{3} \] Verified OK.

2.3.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }+\frac {d}{d x}y^{\prime \prime }-3 \frac {d}{d x}y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & \frac {d}{d x}y^{\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 )=\frac {d}{d x}y^{\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 )=-y_{2}\left (x \right )+3 y_{3}\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_{3}^{\prime }\left (x \right )=-y_{2}\left (x \right )+3 y_{3}\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 ) \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 \\ 0 & -1 & 3 \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 \\ 0 & -1 & 3 \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 [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ], \left [\frac {3}{2}-\frac {\sqrt {5}}{2}, \left [\begin {array}{c} \frac {1}{\left (\frac {3}{2}-\frac {\sqrt {5}}{2}\right )^{2}} \\ \frac {1}{\frac {3}{2}-\frac {\sqrt {5}}{2}} \\ 1 \end {array}\right ]\right ], \left [\frac {3}{2}+\frac {\sqrt {5}}{2}, \left [\begin {array}{c} \frac {1}{\left (\frac {3}{2}+\frac {\sqrt {5}}{2}\right )^{2}} \\ \frac {1}{\frac {3}{2}+\frac {\sqrt {5}}{2}} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}=\left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [\frac {3}{2}-\frac {\sqrt {5}}{2}, \left [\begin {array}{c} \frac {1}{\left (\frac {3}{2}-\frac {\sqrt {5}}{2}\right )^{2}} \\ \frac {1}{\frac {3}{2}-\frac {\sqrt {5}}{2}} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{\left (\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) x}\cdot \left [\begin {array}{c} \frac {1}{\left (\frac {3}{2}-\frac {\sqrt {5}}{2}\right )^{2}} \\ \frac {1}{\frac {3}{2}-\frac {\sqrt {5}}{2}} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [\frac {3}{2}+\frac {\sqrt {5}}{2}, \left [\begin {array}{c} \frac {1}{\left (\frac {3}{2}+\frac {\sqrt {5}}{2}\right )^{2}} \\ \frac {1}{\frac {3}{2}+\frac {\sqrt {5}}{2}} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{\left (\frac {3}{2}+\frac {\sqrt {5}}{2}\right ) x}\cdot \left [\begin {array}{c} \frac {1}{\left (\frac {3}{2}+\frac {\sqrt {5}}{2}\right )^{2}} \\ \frac {1}{\frac {3}{2}+\frac {\sqrt {5}}{2}} \\ 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} \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{2} {\mathrm e}^{\left (\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) x}\cdot \left [\begin {array}{c} \frac {1}{\left (\frac {3}{2}-\frac {\sqrt {5}}{2}\right )^{2}} \\ \frac {1}{\frac {3}{2}-\frac {\sqrt {5}}{2}} \\ 1 \end {array}\right ]+c_{3} {\mathrm e}^{\left (\frac {3}{2}+\frac {\sqrt {5}}{2}\right ) x}\cdot \left [\begin {array}{c} \frac {1}{\left (\frac {3}{2}+\frac {\sqrt {5}}{2}\right )^{2}} \\ \frac {1}{\frac {3}{2}+\frac {\sqrt {5}}{2}} \\ 1 \end {array}\right ]+\left [\begin {array}{c} c_{1} \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {c_{3} \left (7-3 \sqrt {5}\right ) {\mathrm e}^{\frac {\left (3+\sqrt {5}\right ) x}{2}}}{2}+\frac {c_{2} \left (3 \sqrt {5}+7\right ) {\mathrm e}^{-\frac {\left (\sqrt {5}-3\right ) x}{2}}}{2}+c_{1} \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`
 

Solution by Maple

Time used: 0.0 (sec). Leaf size: 28

dsolve(diff(y(x),x)+diff(y(x),x$3)-3*diff(y(x),x$2)=0,y(x), singsol=all)
 

\[ y \left (x \right ) = c_{1} +c_{2} {\mathrm e}^{\frac {\left (3+\sqrt {5}\right ) x}{2}}+c_{3} {\mathrm e}^{-\frac {\left (\sqrt {5}-3\right ) x}{2}} \]

Solution by Mathematica

Time used: 0.384 (sec). Leaf size: 57

DSolve[y'[x]+y'''[x]-3*y''[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {1}{2} e^{-\frac {1}{2} \left (\sqrt {5}-3\right ) x} \left (\left (3+\sqrt {5}\right ) c_1-\left (\sqrt {5}-3\right ) c_2 e^{\sqrt {5} x}\right )+c_3 \]