Internal problem ID [11591]
Internal file name [OUTPUT/10574_Thursday_May_18_2023_05_56_39_PM_4049348/index.tex]
Book: Differential Equations by Shepley L. Ross. Third edition. John Willey. New Delhi.
2004.
Section: Chapter 1, section 1.3. Exercises page 22
Problem number: 5.
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]]
\[ \boxed {x^{3} y^{\prime \prime \prime }-3 x^{2} y^{\prime \prime }+6 y^{\prime } x -6 y=0} \] With initial conditions \begin {align*} [y \left (2\right ) = 0, y^{\prime }\left (2\right ) = 2, y^{\prime \prime }\left (2\right ) = 6] \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 \[ x^{3} y^{\prime \prime \prime }-3 x^{2} y^{\prime \prime }+6 y^{\prime } x -6 y = 0 \] gives \[ 6 x \lambda \,x^{\lambda -1}-3 x^{2} \lambda \left (\lambda -1\right ) x^{\lambda -2}+x^{3} \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda -3}-6 x^{\lambda } = 0 \] Which simplifies to \[ 6 \lambda \,x^{\lambda }-3 \lambda \left (\lambda -1\right ) x^{\lambda }+\lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda }-6 x^{\lambda } = 0 \] And since \(x^{\lambda }\neq 0\) then dividing through by \(x^{\lambda }\), the above becomes
\[ 6 \lambda -3 \lambda \left (\lambda -1\right )+\lambda \left (\lambda -1\right ) \left (\lambda -2\right )-6 = 0 \] Simplifying gives the characteristic equation as \[ \lambda ^{3}-6 \lambda ^{2}+11 \lambda -6 = 0 \] Solving the above gives the following roots \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= 2\\ \lambda _3 &= 3 \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
\[ y = c_{3} x^{3}+c_{2} x^{2}+c_{1} x \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_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 = c_{3} x^{3}+c_{2} x^{2}+c_{1} x \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 \(x = 2\) in the above gives \begin {align*} 0 = 8 c_{3} +4 c_{2} +2 c_{1}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = 3 c_{3} x^{2}+2 c_{2} x +c_{1} \end {align*}
substituting \(y^{\prime } = 2\) and \(x = 2\) in the above gives \begin {align*} 2 = 12 c_{3} +4 c_{2} +c_{1}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = 6 c_{3} x +2 c_{2} \end {align*}
substituting \(y^{\prime \prime } = 6\) and \(x = 2\) in the above gives \begin {align*} 6 = 12 c_{3} +2 c_{2}\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}&=2\\ c_{2}&=-3\\ c_{3}&=1 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = x^{3}-3 x^{2}+2 x \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= x^{3}-3 x^{2}+2 x \\ \end{align*}
Verification of solutions
\[ y = x^{3}-3 x^{2}+2 x \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime } x^{3}-3 y^{\prime \prime } x^{2}+6 y^{\prime } x -6 y=0, y \left (2\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}2\right \}}}}=2, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}2\right \}}}}=6\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 {6 y}{x^{3}}+\frac {3 \left (y^{\prime \prime } x -2 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} \\ {} & {} & y^{\prime \prime \prime }-\frac {3 y^{\prime \prime }}{x}+\frac {6 y^{\prime }}{x^{2}}-\frac {6 y}{x^{3}}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime } x^{3}-3 y^{\prime \prime } x^{2}+6 y^{\prime } x -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} \\ {} & {} & y^{\prime \prime }=\left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) {t^{\prime }\left (x \right )}^{2}+t^{\prime \prime }\left (x \right ) \left (\frac {d}{d t}y \left (t \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {\frac {d^{2}}{d t^{2}}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} \\ {} & {} & y^{\prime \prime \prime }=\left (\frac {d^{3}}{d t^{3}}y \left (t \right )\right ) {t^{\prime }\left (x \right )}^{3}+3 t^{\prime }\left (x \right ) t^{\prime \prime }\left (x \right ) \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )+t^{\prime \prime \prime }\left (x \right ) \left (\frac {d}{d t}y \left (t \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=\frac {\frac {d^{3}}{d t^{3}}y \left (t \right )}{x^{3}}-\frac {3 \left (\frac {d^{2}}{d t^{2}}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} \\ {} & {} & \left (\frac {\frac {d^{3}}{d t^{3}}y \left (t \right )}{x^{3}}-\frac {3 \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )}{x^{3}}+\frac {2 \left (\frac {d}{d t}y \left (t \right )\right )}{x^{3}}\right ) x^{3}-3 \left (\frac {\frac {d^{2}}{d t^{2}}y \left (t \right )}{x^{2}}-\frac {\frac {d}{d t}y \left (t \right )}{x^{2}}\right ) x^{2}+6 \frac {d}{d t}y \left (t \right )-6 y \left (t \right )=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {d^{3}}{d t^{3}}y \left (t \right )-6 \frac {d^{2}}{d t^{2}}y \left (t \right )+11 \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^{2}}{d t^{2}}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 )=6 y_{3}\left (t \right )-11 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 )=6 y_{3}\left (t \right )-11 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 & -11 & 6 \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 & -11 & 6 \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=c_{1} x +\frac {1}{4} c_{2} x^{2}+\frac {1}{9} c_{3} x^{3} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (2\right )=0 \\ {} & {} & 0=2 c_{1} +c_{2} +\frac {8 c_{3}}{9} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=c_{1} +\frac {1}{2} c_{2} x +\frac {1}{3} c_{3} x^{2} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}2\right \}}}}=2 \\ {} & {} & 2=c_{1} +c_{2} +\frac {4 c_{3}}{3} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {c_{2}}{2}+\frac {2 c_{3} x}{3} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}2\right \}}}}=6 \\ {} & {} & 6=\frac {c_{2}}{2}+\frac {4 c_{3}}{3} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =2, c_{2} =-12, c_{3} =9, x =x \right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=x^{3}-3 x^{2}+2 x \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: 16
dsolve([x^3*diff(y(x),x$3)-3*x^2*diff(y(x),x$2)+6*x*diff(y(x),x)-6*y(x)=0,y(2) = 0, D(y)(2) = 2, (D@@2)(y)(2) = 6],y(x), singsol=all)
\[ y \left (x \right ) = x^{3}-3 x^{2}+2 x \]
✓ Solution by Mathematica
Time used: 0.013 (sec). Leaf size: 15
DSolve[{x^3*y'''[x]-3*x^2*y''[x]+6*x*y'[x]-6*y[x]==0,{y[2]==0,y'[2]==2,y''[2]==6}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to x \left (x^2-3 x+2\right ) \]