Internal problem ID [14721]
Internal file name [OUTPUT/14401_Wednesday_April_03_2024_02_18_11_PM_13873342/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.7, page 195
Problem number: 13.
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, _with_linear_symmetries]]
\[ \boxed {x^{3} y^{\prime \prime \prime }+22 x^{2} y^{\prime \prime }+124 x y^{\prime }+140 y=0} \] 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 \[ x^{3} y^{\prime \prime \prime }+22 x^{2} y^{\prime \prime }+124 x y^{\prime }+140 y = 0 \] gives \[ 124 x \lambda \,x^{\lambda -1}+22 x^{2} \lambda \left (\lambda -1\right ) x^{\lambda -2}+x^{3} \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda -3}+140 x^{\lambda } = 0 \] Which simplifies to \[ 124 \lambda \,x^{\lambda }+22 \lambda \left (\lambda -1\right ) x^{\lambda }+\lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda }+140 x^{\lambda } = 0 \] And since \(x^{\lambda }\neq 0\) then dividing through by \(x^{\lambda }\), the above becomes
\[ 124 \lambda +22 \lambda \left (\lambda -1\right )+\lambda \left (\lambda -1\right ) \left (\lambda -2\right )+140 = 0 \] Simplifying gives the characteristic equation as \[ \lambda ^{3}+19 \lambda ^{2}+104 \lambda +140 = 0 \] Solving the above gives the following roots \begin {align*} \lambda _1 &= -10\\ \lambda _2 &= -7\\ \lambda _3 &= -2 \end {align*}
This table summarises the result
| root | multiplicity | type of root |
| \(-2\) | \(1\) | real root |
| \(-7\) | \(1\) | real root |
| \(-10\) | \(1\) | 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
\[ y = \frac {c_{1}}{x^{2}}+\frac {c_{2}}{x^{7}}+\frac {c_{3}}{x^{10}} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= \frac {1}{x^{2}}\\ y_2 &= \frac {1}{x^{7}}\\ y_3 &= \frac {1}{x^{10}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1}}{x^{2}}+\frac {c_{2}}{x^{7}}+\frac {c_{3}}{x^{10}} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1}}{x^{2}}+\frac {c_{2}}{x^{7}}+\frac {c_{3}}{x^{10}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{3} \left (\frac {d}{d x}y^{\prime \prime }\right )+22 x^{2} \left (\frac {d}{d x}y^{\prime }\right )+124 x y^{\prime }+140 y=0 \\ \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 {140 y}{x^{3}}-\frac {2 \left (11 x \left (\frac {d}{d x}y^{\prime }\right )+62 y^{\prime }\right )}{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 {22 \left (\frac {d}{d x}y^{\prime }\right )}{x}+\frac {124 y^{\prime }}{x^{2}}+\frac {140 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 )+22 x^{2} \left (\frac {d}{d x}y^{\prime }\right )+124 x y^{\prime }+140 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 )+22 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 )+124 \frac {d}{d t}y \left (t \right )+140 y \left (t \right )=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {d}{d t}\frac {d^{2}}{d t^{2}}y \left (t \right )+19 \frac {d}{d t}\frac {d}{d t}y \left (t \right )+104 \frac {d}{d t}y \left (t \right )+140 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 )=-19 y_{3}\left (t \right )-104 y_{2}\left (t \right )-140 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 )=-19 y_{3}\left (t \right )-104 y_{2}\left (t \right )-140 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 \\ -140 & -104 & -19 \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 \\ -140 & -104 & -19 \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 [-10, \left [\begin {array}{c} \frac {1}{100} \\ -\frac {1}{10} \\ 1 \end {array}\right ]\right ], \left [-7, \left [\begin {array}{c} \frac {1}{49} \\ -\frac {1}{7} \\ 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 [-10, \left [\begin {array}{c} \frac {1}{100} \\ -\frac {1}{10} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-10 t}\cdot \left [\begin {array}{c} \frac {1}{100} \\ -\frac {1}{10} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-7, \left [\begin {array}{c} \frac {1}{49} \\ -\frac {1}{7} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{-7 t}\cdot \left [\begin {array}{c} \frac {1}{49} \\ -\frac {1}{7} \\ 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 t}\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}^{-10 t}\cdot \left [\begin {array}{c} \frac {1}{100} \\ -\frac {1}{10} \\ 1 \end {array}\right ]+c_{2} {\mathrm e}^{-7 t}\cdot \left [\begin {array}{c} \frac {1}{49} \\ -\frac {1}{7} \\ 1 \end {array}\right ]+c_{3} {\mathrm e}^{-2 t}\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 \left (t \right )=\frac {c_{1} {\mathrm e}^{-10 t}}{100}+\frac {c_{2} {\mathrm e}^{-7 t}}{49}+\frac {c_{3} {\mathrm e}^{-2 t}}{4} \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} t =\ln \left (x \right ) \\ {} & {} & y=\frac {c_{1}}{100 x^{10}}+\frac {c_{2}}{49 x^{7}}+\frac {c_{3}}{4 x^{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: 0.015 (sec). Leaf size: 20
dsolve(x^3*diff(y(x),x$3)+22*x^2*diff(y(x),x$2)+124*x*diff(y(x),x)+140*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {c_{3} x^{8}+c_{1} x^{3}+c_{2}}{x^{10}} \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 24
DSolve[x^3*y'''[x]+22*x^2*y''[x]+124*x*y'[x]+140*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {c_3 x^8+c_2 x^3+c_1}{x^{10}} \]