Internal problem ID [1459]
Internal file name [OUTPUT/1460_Sunday_June_05_2022_02_18_47_AM_32992798/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 9.1. Page
471
Problem number: section 9.1, problem 5(b) 1.
ODE order: 3.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_ODE_non_constant_coefficients_of_type_Euler"
Maple gives the following as the ode type
[[_3rd_order, _exact, _linear, _homogeneous]]
\[ \boxed {x^{3} y^{\prime \prime \prime }-x^{2} y^{\prime \prime }-2 x y^{\prime }+6 y=0} \] With initial conditions \begin {align*} [y \left (1\right ) = 1, y^{\prime }\left (1\right ) = 0, y^{\prime \prime }\left (1\right ) = 0] \end {align*}
This is Euler ODE of higher order. Let \(y = x^{\lambda }\). Hence \begin {align*} y^{\prime } &= \lambda \,x^{\lambda -1}\\ y^{\prime \prime } &= \lambda \left (\lambda -1\right ) x^{\lambda -2}\\ y^{\prime \prime \prime } &= \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda -3} \end {align*}
Substituting these back into \[ x^{3} y^{\prime \prime \prime }-x^{2} y^{\prime \prime }-2 x y^{\prime }+6 y = 0 \] gives \[ -2 x \lambda \,x^{\lambda -1}-x^{2} \lambda \left (\lambda -1\right ) x^{\lambda -2}+x^{3} \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda -3}+6 x^{\lambda } = 0 \] Which simplifies to \[ -2 \lambda \,x^{\lambda }-\lambda \left (\lambda -1\right ) x^{\lambda }+\lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda }+6 x^{\lambda } = 0 \] And since \(x^{\lambda }\neq 0\) then dividing through by \(x^{\lambda }\), the above becomes
\[ -2 \lambda -\lambda \left (\lambda -1\right )+\lambda \left (\lambda -1\right ) \left (\lambda -2\right )+6 = 0 \] Simplifying gives the characteristic equation as \[ \lambda ^{3}-4 \lambda ^{2}+\lambda +6 = 0 \] Solving the above gives the following roots \begin {align*} \lambda _1 &= 2\\ \lambda _2 &= 3\\ \lambda _3 &= -1 \end {align*}
This table summarises the result
| root | multiplicity | type of root |
| \(-1\) | \(1\) | real root |
| \(2\) | \(1\) | real root |
| \(3\) | \(1\) | real root |
The solution is generated by going over the above table. For each real root \(\lambda \) of multiplicity one generates a \(c_1x^{\lambda }\) basis solution. Each real root of multiplicty two, generates \(c_1x^{\lambda }\) and \(c_2x^{\lambda } \ln \left (x \right )\) basis solutions. Each real root of multiplicty three, generates \(c_1x^{\lambda }\) and \(c_2x^{\lambda } \ln \left (x \right )\) and \(c_3x^{\lambda } \ln \left (x \right )^{2}\) basis solutions, and so on. Each complex root \(\alpha \pm i \beta \) of multiplicity one generates \(x^{\alpha } \left (c_1\cos (\beta \ln \left (x \right ))+c_2\sin (\beta \ln \left (x \right ))\right )\) basis solutions. And each complex root \(\alpha \pm i \beta \) of multiplicity two generates \(\ln \left (x \right ) x^{\alpha }\left (c_1\cos (\beta \ln \left (x \right ))+c_2\sin (\beta \ln \left (x \right ))\right )\) basis solutions. And each complex root \(\alpha \pm i \beta \) of multiplicity three generates \(\ln \left (x \right )^{2} x^{\alpha }\left (c_1\cos (\beta \ln \left (x \right ))+c_2\sin (\beta \ln \left (x \right ))\right )\) basis solutions. And so on. Using the above show that the solution is
\[ y = \frac {c_{1}}{x}+c_{2} x^{2}+c_{3} x^{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= \frac {1}{x}\\ y_2 &= x^{2}\\ y_3 &= x^{3} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = \frac {c_{1}}{x}+c_{2} x^{2}+c_{3} x^{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 \(y = 1\) and \(x = 1\) in the above gives \begin {align*} 1 = c_{1} +c_{2} +c_{3}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -\frac {c_{1}}{x^{2}}+2 c_{2} x +3 c_{3} x^{2} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 1\) in the above gives \begin {align*} 0 = -c_{1} +2 c_{2} +3 c_{3}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = \frac {2 c_{1}}{x^{3}}+2 c_{2} +6 c_{3} x \end {align*}
substituting \(y^{\prime \prime } = 0\) and \(x = 1\) in the above gives \begin {align*} 0 = 2 c_{1} +2 c_{2} +6 c_{3}\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 {1}{2}}\\ c_{2}&=1\\ c_{3}&=-{\frac {1}{2}} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = -\frac {x^{4}-2 x^{3}-1}{2 x} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {x^{4}-2 x^{3}-1}{2 x} \\ \end{align*}
Verification of solutions
\[ y = -\frac {x^{4}-2 x^{3}-1}{2 x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{3} \left (\frac {d}{d x}y^{\prime \prime }\right )-x^{2} \left (\frac {d}{d x}y^{\prime }\right )-2 x y^{\prime }+6 y=0, y \left (1\right )=1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=0, \left (\frac {d}{d x}y^{\prime }\right )\bigg | {\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & \frac {d}{d x}y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 3rd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime \prime }=-\frac {6 y}{x^{3}}+\frac {x \left (\frac {d}{d x}y^{\prime }\right )+2 y^{\prime }}{x^{2}} \\ \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} \\ {} & {} & \frac {d}{d x}y^{\prime \prime }-\frac {\frac {d}{d x}y^{\prime }}{x}-\frac {2 y^{\prime }}{x^{2}}+\frac {6 y}{x^{3}}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & x^{3} \left (\frac {d}{d x}y^{\prime \prime }\right )-x^{2} \left (\frac {d}{d x}y^{\prime }\right )-2 x y^{\prime }+6 y=0 \\ \bullet & {} & \textrm {Make a change of variables}\hspace {3pt} \\ {} & {} & t =\ln \left (x \right ) \\ \square & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {1st}\hspace {3pt} \hspace {3pt}\textrm {derivative of}\hspace {3pt} \hspace {3pt}\textrm {y}\hspace {3pt} \hspace {3pt}\textrm {with respect to}\hspace {3pt} \hspace {3pt}\textrm {x}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & y^{\prime }=\left (\frac {d}{d t}y \left (t \right )\right ) t^{\prime }\left (x \right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\frac {d}{d t}y \left (t \right )}{x} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {2nd}\hspace {3pt} \hspace {3pt}\textrm {derivative of}\hspace {3pt} \hspace {3pt}\textrm {y}\hspace {3pt} \hspace {3pt}\textrm {with respect to}\hspace {3pt} \hspace {3pt}\textrm {x}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\left (\frac {d}{d t}\frac {d}{d t}y \left (t \right )\right ) {t^{\prime }\left (x \right )}^{2}+\left (\frac {d}{d x}t^{\prime }\left (x \right )\right ) \left (\frac {d}{d t}y \left (t \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\frac {\frac {d}{d t}\frac {d}{d t}y \left (t \right )}{x^{2}}-\frac {\frac {d}{d t}y \left (t \right )}{x^{2}} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {3rd}\hspace {3pt} \hspace {3pt}\textrm {derivative of}\hspace {3pt} \hspace {3pt}\textrm {y}\hspace {3pt} \hspace {3pt}\textrm {with respect to}\hspace {3pt} \hspace {3pt}\textrm {x}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime \prime }=\left (\frac {d}{d t}\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) {t^{\prime }\left (x \right )}^{3}+3 t^{\prime }\left (x \right ) \left (\frac {d}{d x}t^{\prime }\left (x \right )\right ) \left (\frac {d}{d t}\frac {d}{d t}y \left (t \right )\right )+\left (\frac {d}{d x}t^{\prime \prime }\left (x \right )\right ) \left (\frac {d}{d t}y \left (t \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime \prime }=\frac {\frac {d}{d t}\frac {d^{2}}{d t^{2}}y \left (t \right )}{x^{3}}-\frac {3 \left (\frac {d}{d t}\frac {d}{d t}y \left (t \right )\right )}{x^{3}}+\frac {2 \left (\frac {d}{d t}y \left (t \right )\right )}{x^{3}} \\ & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & {} & x^{3} \left (\frac {\frac {d}{d t}\frac {d^{2}}{d t^{2}}y \left (t \right )}{x^{3}}-\frac {3 \left (\frac {d}{d t}\frac {d}{d t}y \left (t \right )\right )}{x^{3}}+\frac {2 \left (\frac {d}{d t}y \left (t \right )\right )}{x^{3}}\right )-x^{2} \left (\frac {\frac {d}{d t}\frac {d}{d t}y \left (t \right )}{x^{2}}-\frac {\frac {d}{d t}y \left (t \right )}{x^{2}}\right )-2 \frac {d}{d t}y \left (t \right )+6 y \left (t \right )=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {d}{d t}\frac {d^{2}}{d t^{2}}y \left (t \right )-4 \frac {d}{d t}\frac {d}{d t}y \left (t \right )+\frac {d}{d t}y \left (t \right )+6 y \left (t \right )=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 (t \right ) \\ {} & {} & y_{1}\left (t \right )=y \left (t \right ) \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (t \right ) \\ {} & {} & y_{2}\left (t \right )=\frac {d}{d t}y \left (t \right ) \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (t \right ) \\ {} & {} & y_{3}\left (t \right )=\frac {d}{d t}\frac {d}{d t}y \left (t \right ) \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} \frac {d}{d t}y_{3}\left (t \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y_{3}\left (t \right )=4 y_{3}\left (t \right )-y_{2}\left (t \right )-6 y_{1}\left (t \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (t \right )=\frac {d}{d t}y_{1}\left (t \right ), y_{3}\left (t \right )=\frac {d}{d t}y_{2}\left (t \right ), \frac {d}{d t}y_{3}\left (t \right )=4 y_{3}\left (t \right )-y_{2}\left (t \right )-6 y_{1}\left (t \right )\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (t \right )=\left [\begin {array}{c} y_{1}\left (t \right ) \\ y_{2}\left (t \right ) \\ y_{3}\left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & \frac {d}{d t}{\moverset {\rightarrow }{y}}\left (t \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -6 & -1 & 4 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \right ) \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -6 & -1 & 4 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & \frac {d}{d t}{\moverset {\rightarrow }{y}}\left (t \right )=A \cdot {\moverset {\rightarrow }{y}}\left (t \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 [2, \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [3, \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 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}^{-t}\cdot \left [\begin {array}{c} 1 \\ -1 \\ 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}}_{2}={\mathrm e}^{2 t}\cdot \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [3, \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{3 t}\cdot \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{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}+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}^{-t}\cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]+c_{2} {\mathrm e}^{2 t}\cdot \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]+c_{3} {\mathrm e}^{3 t}\cdot \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=c_{1} {\mathrm e}^{-t}+\frac {c_{2} {\mathrm e}^{2 t}}{4}+\frac {c_{3} {\mathrm e}^{3 t}}{9} \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} t =\ln \left (x \right ) \\ {} & {} & y=\frac {c_{1}}{x}+\frac {c_{2} x^{2}}{4}+\frac {c_{3} x^{3}}{9} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (1\right )=1 \\ {} & {} & 1=c_{1} +\frac {c_{2}}{4}+\frac {c_{3}}{9} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {c_{1}}{x^{2}}+\frac {c_{2} x}{2}+\frac {c_{3} x^{2}}{3} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=0 \\ {} & {} & 0=-c_{1} +\frac {c_{2}}{2}+\frac {c_{3}}{3} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\frac {2 c_{1}}{x^{3}}+\frac {c_{2}}{2}+\frac {2 c_{3} x}{3} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} \left (\frac {d}{d x}y^{\prime }\right )\bigg | {\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=0 \\ {} & {} & 0=2 c_{1} +\frac {c_{2}}{2}+\frac {2 c_{3}}{3} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =\frac {1}{2}, c_{2} =4, c_{3} =-\frac {9}{2}, x =x \right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {1}{2 x}+x^{2}-\frac {x^{3}}{2} \end {array} \]
Maple trace
`Methods for third order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type <- LODE of Euler type successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 18
dsolve([x^3*diff(y(x),x$3)-x^2*diff(y(x),x$2)-2*x*diff(y(x),x)+6*y(x)=0,y(1) = 1, D(y)(1) = 0, (D@@2)(y)(1) = 0],y(x), singsol=all)
\[ y \left (x \right ) = -\frac {x^{3}}{2}+x^{2}+\frac {1}{2 x} \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 23
DSolve[{x^3*y'''[x]-x^2*y''[x]-2*x*y'[x]+6*y[x]==0,{y[1]==1,y'[1]==0,y''[1]==0}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -\frac {x^3}{2}+x^2+\frac {1}{2 x} \]