Internal problem ID [4605]
Internal file name [OUTPUT/4098_Sunday_June_05_2022_12_22_02_PM_55432396/index.tex]
Book: Ordinary Differential Equations, By Tenenbaum and Pollard. Dover, NY 1963
Section: Chapter 4. Higher order linear differential equations. Lesson 20. Constant
coefficients
Problem number: Exercise 20, problem 35, page 220.
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]]
\[ \boxed {3 y^{\prime \prime \prime }+5 y^{\prime \prime }+y^{\prime }-y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 1, y^{\prime \prime }\left (0\right ) = -1] \end {align*}
The characteristic equation is \[ 3 \lambda ^{3}+5 \lambda ^{2}+\lambda -1 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= {\frac {1}{3}}\\ \lambda _2 &= -1\\ \lambda _3 &= -1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{-x} x +{\mathrm e}^{\frac {x}{3}} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= {\mathrm e}^{-x}\\ y_2 &= x \,{\mathrm e}^{-x}\\ y_3 &= {\mathrm e}^{\frac {x}{3}} \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} x +{\mathrm e}^{\frac {x}{3}} c_{3} \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 \(y = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = c_{1} +c_{3}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -c_{1} {\mathrm e}^{-x}-c_{2} {\mathrm e}^{-x} x +c_{2} {\mathrm e}^{-x}+\frac {{\mathrm e}^{\frac {x}{3}} c_{3}}{3} \end {align*}
substituting \(y^{\prime } = 1\) and \(x = 0\) in the above gives \begin {align*} 1 = -c_{1} +c_{2} +\frac {c_{3}}{3}\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} x -2 c_{2} {\mathrm e}^{-x}+\frac {{\mathrm e}^{\frac {x}{3}} c_{3}}{9} \end {align*}
substituting \(y^{\prime \prime } = -1\) and \(x = 0\) in the above gives \begin {align*} -1 = c_{1} -2 c_{2} +\frac {c_{3}}{9}\tag {3A} \end {align*}
Equations {1A,2A,3A} are now solved for \(\{c_{1}, c_{2}, c_{3}\}\). Solving for the constants gives \begin {align*} c_{1}&=-{\frac {9}{16}}\\ c_{2}&={\frac {1}{4}}\\ c_{3}&={\frac {9}{16}} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = -\frac {9 \,{\mathrm e}^{-x}}{16}+\frac {x \,{\mathrm e}^{-x}}{4}+\frac {9 \,{\mathrm e}^{\frac {x}{3}}}{16} \end {align*}
Which simplifies to \[ y = \frac {\left (9 \,{\mathrm e}^{\frac {4 x}{3}}+4 x -9\right ) {\mathrm e}^{-x}}{16} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (9 \,{\mathrm e}^{\frac {4 x}{3}}+4 x -9\right ) {\mathrm e}^{-x}}{16} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (9 \,{\mathrm e}^{\frac {4 x}{3}}+4 x -9\right ) {\mathrm e}^{-x}}{16} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [3 y^{\prime \prime \prime }+5 y^{\prime \prime }+y^{\prime }-y=0, y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=1, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-1\right ] \\ \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 {5 y^{\prime \prime }}{3}-\frac {y^{\prime }}{3}+\frac {y}{3} \\ \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 {5 y^{\prime \prime }}{3}+\frac {y^{\prime }}{3}-\frac {y}{3}=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 {5 y_{3}\left (x \right )}{3}-\frac {y_{2}\left (x \right )}{3}+\frac {y_{1}\left (x \right )}{3} \\ & {} & \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 {5 y_{3}\left (x \right )}{3}-\frac {y_{2}\left (x \right )}{3}+\frac {y_{1}\left (x \right )}{3}\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}{3} & -\frac {1}{3} & -\frac {5}{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 \\ \frac {1}{3} & -\frac {1}{3} & -\frac {5}{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 [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ], \left [-1, \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [\frac {1}{3}, \left [\begin {array}{c} 9 \\ 3 \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair, with eigenvalue of algebraic multiplicity 2}\hspace {3pt} \\ {} & {} & \left [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {First solution from eigenvalue}\hspace {3pt} -1 \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}\left (x \right )={\mathrm e}^{-x}\cdot \left [\begin {array}{c} 1 \\ -1 \\ 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 =-1\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 =-1 \\ \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} -1 \\ {} & {} & \left (\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ \frac {1}{3} & -\frac {1}{3} & -\frac {5}{3} \end {array}\right ]-\left (-1\right )\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} 1 \\ -1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Choice of}\hspace {3pt} {\moverset {\rightarrow }{p}} \\ {} & {} & {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Second solution from eigenvalue}\hspace {3pt} -1 \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{-x}\cdot \left (x \cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]+\left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ) \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [\frac {1}{3}, \left [\begin {array}{c} 9 \\ 3 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{\frac {x}{3}}\cdot \left [\begin {array}{c} 9 \\ 3 \\ 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}^{-x}\cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]+c_{2} {\mathrm e}^{-x}\cdot \left (x \cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]+\left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right )+{\mathrm e}^{\frac {x}{3}} c_{3} \cdot \left [\begin {array}{c} 9 \\ 3 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\left (9 \,{\mathrm e}^{\frac {4 x}{3}} c_{3} +c_{2} x +c_{1} +c_{2} \right ) {\mathrm e}^{-x} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=9 c_{3} +c_{1} +c_{2} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\left (12 \,{\mathrm e}^{\frac {4 x}{3}} c_{3} +c_{2} \right ) {\mathrm e}^{-x}-\left (9 \,{\mathrm e}^{\frac {4 x}{3}} c_{3} +c_{2} x +c_{1} +c_{2} \right ) {\mathrm e}^{-x} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=3 c_{3} -c_{1} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=16 \,{\mathrm e}^{\frac {4 x}{3}} c_{3} {\mathrm e}^{-x}-2 \left (12 \,{\mathrm e}^{\frac {4 x}{3}} c_{3} +c_{2} \right ) {\mathrm e}^{-x}+\left (9 \,{\mathrm e}^{\frac {4 x}{3}} c_{3} +c_{2} x +c_{1} +c_{2} \right ) {\mathrm e}^{-x} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-1 \\ {} & {} & -1=c_{3} -c_{2} +c_{1} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =-\frac {13}{16}, c_{2} =\frac {1}{4}, c_{3} =\frac {1}{16}\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\left (9 \,{\mathrm e}^{\frac {4 x}{3}}+4 x -9\right ) {\mathrm e}^{-x}}{16} \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: 21
dsolve([3*diff(y(x),x$3)+5*diff(y(x),x$2)+diff(y(x),x)-y(x)=0,y(0) = 0, D(y)(0) = 1, (D@@2)(y)(0) = -1],y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (9 \,{\mathrm e}^{\frac {4 x}{3}}+4 x -9\right ) {\mathrm e}^{-x}}{16} \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 28
DSolve[{3*y'''[x]+5*y''[x]+y'[x]-y[x]==0,{y[0]==0,y'[0]==1,y''[0]==-1}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{16} e^{-x} \left (4 x+9 e^{4 x/3}-9\right ) \]