Internal problem ID [300]
Internal file name [OUTPUT/300_Sunday_June_05_2022_01_38_24_AM_29981424/index.tex
]
Book: Differential equations and linear algebra, 4th ed., Edwards and Penney
Section: Section 5.3, Higher-Order Linear Differential Equations. Homogeneous Equations with
Constant Coefficients. Page 300
Problem number: problem 28.
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]]
\[ \boxed {2 y^{\prime \prime \prime }-y^{\prime \prime }-5 y^{\prime }-2 y=0} \] The characteristic equation is \[ 2 \lambda ^{3}-\lambda ^{2}-5 \lambda -2 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 2\\ \lambda _2 &= -{\frac {1}{2}}\\ \lambda _3 &= -1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{2 x}+{\mathrm e}^{-\frac {x}{2}} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= {\mathrm e}^{-x}\\ y_2 &= {\mathrm e}^{2 x}\\ y_3 &= {\mathrm e}^{-\frac {x}{2}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{2 x}+{\mathrm e}^{-\frac {x}{2}} c_{3} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{2 x}+{\mathrm e}^{-\frac {x}{2}} c_{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 y^{\prime \prime \prime }-y^{\prime \prime }-5 y^{\prime }-2 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 }=\frac {y^{\prime \prime }}{2}+\frac {5 y^{\prime }}{2}+y \\ \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 }-\frac {y^{\prime \prime }}{2}-\frac {5 y^{\prime }}{2}-y=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 )=\frac {y_{3}\left (x \right )}{2}+\frac {5 y_{2}\left (x \right )}{2}+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_{3}^{\prime }\left (x \right )=\frac {y_{3}\left (x \right )}{2}+\frac {5 y_{2}\left (x \right )}{2}+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 ) \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 \\ 1 & \frac {5}{2} & \frac {1}{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 \\ 1 & \frac {5}{2} & \frac {1}{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 [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ], \left [-\frac {1}{2}, \left [\begin {array}{c} 4 \\ -2 \\ 1 \end {array}\right ]\right ], \left [2, \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\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}}_{1}={\mathrm e}^{-x}\cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-\frac {1}{2}, \left [\begin {array}{c} 4 \\ -2 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{-\frac {x}{2}}\cdot \left [\begin {array}{c} 4 \\ -2 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [2, \left [\begin {array}{c} \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}{4} \\ \frac {1}{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_{1} {\mathrm e}^{-x}\cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]+c_{2} {\mathrm e}^{-\frac {x}{2}}\cdot \left [\begin {array}{c} 4 \\ -2 \\ 1 \end {array}\right ]+c_{3} {\mathrm e}^{2 x}\cdot \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left ({\mathrm e}^{3 x} c_{3} +16 c_{2} {\mathrm e}^{\frac {x}{2}}+4 c_{1} \right ) {\mathrm e}^{-x}}{4} \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: 0.016 (sec). Leaf size: 23
dsolve(2*diff(y(x),x$3)-diff(y(x),x$2)-5*diff(y(x),x)-2*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \left (c_{2} {\mathrm e}^{3 x}+c_{1} {\mathrm e}^{\frac {x}{2}}+c_{3} \right ) {\mathrm e}^{-x} \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 32
DSolve[2*y'''[x]-y''[x]-5*y'[x]-2*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{-x} \left (c_1 e^{x/2}+c_3 e^{3 x}+c_2\right ) \]