Internal problem ID [14746]
Internal file name [OUTPUT/14426_Monday_April_08_2024_06_24_35_AM_60417259/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: 38.
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]]
\[ \boxed {x^{3} y^{\prime \prime \prime }-6 x^{2} y^{\prime \prime }+17 x y^{\prime }-17 y=0} \] With initial conditions \begin {align*} [y \left (1\right ) = -2, 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 \[ x^{3} y^{\prime \prime \prime }-6 x^{2} y^{\prime \prime }+17 x y^{\prime }-17 y = 0 \] gives \[ 17 x \lambda \,x^{\lambda -1}-6 x^{2} \lambda \left (\lambda -1\right ) x^{\lambda -2}+x^{3} \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda -3}-17 x^{\lambda } = 0 \] Which simplifies to \[ 17 \lambda \,x^{\lambda }-6 \lambda \left (\lambda -1\right ) x^{\lambda }+\lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda }-17 x^{\lambda } = 0 \] And since \(x^{\lambda }\neq 0\) then dividing through by \(x^{\lambda }\), the above becomes
\[ 17 \lambda -6 \lambda \left (\lambda -1\right )+\lambda \left (\lambda -1\right ) \left (\lambda -2\right )-17 = 0 \] Simplifying gives the characteristic equation as \[ \lambda ^{3}-9 \lambda ^{2}+25 \lambda -17 = 0 \] Solving the above gives the following roots \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= 4-i\\ \lambda _3 &= 4+i \end {align*}
This table summarises the result
| root | multiplicity | type of root |
| \(1\) | \(1\) | real root |
| \(4 \pm 1 i\) | \(1\) | complex conjugate 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
\[ y = c_{1} x +x^{4} \left (c_{2} \cos \left (\ln \left (x \right )\right )+c_{3} \sin \left (\ln \left (x \right )\right )\right ) \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= x\\ y_2 &= x^{4} \cos \left (\ln \left (x \right )\right )\\ y_3 &= x^{4} \sin \left (\ln \left (x \right )\right ) \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = c_{1} x +x^{4} \left (c_{2} \cos \left (\ln \left (x \right )\right )+c_{3} \sin \left (\ln \left (x \right )\right )\right ) \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 \(y = -2\) and \(x = 1\) in the above gives \begin {align*} -2 = c_{1} +c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = c_{1} +4 x^{3} \left (c_{2} \cos \left (\ln \left (x \right )\right )+c_{3} \sin \left (\ln \left (x \right )\right )\right )+x^{4} \left (-\frac {c_{2} \sin \left (\ln \left (x \right )\right )}{x}+\frac {c_{3} \cos \left (\ln \left (x \right )\right )}{x}\right ) \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 1\) in the above gives \begin {align*} 0 = c_{1} +4 c_{2} +c_{3}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = 12 x^{2} \left (c_{2} \cos \left (\ln \left (x \right )\right )+c_{3} \sin \left (\ln \left (x \right )\right )\right )+8 x^{3} \left (-\frac {c_{2} \sin \left (\ln \left (x \right )\right )}{x}+\frac {c_{3} \cos \left (\ln \left (x \right )\right )}{x}\right )+x^{4} \left (\frac {c_{2} \sin \left (\ln \left (x \right )\right )}{x^{2}}-\frac {c_{2} \cos \left (\ln \left (x \right )\right )}{x^{2}}-\frac {c_{3} \cos \left (\ln \left (x \right )\right )}{x^{2}}-\frac {c_{3} \sin \left (\ln \left (x \right )\right )}{x^{2}}\right ) \end {align*}
substituting \(y^{\prime \prime } = 0\) and \(x = 1\) in the above gives \begin {align*} 0 = 11 c_{2} +7 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 {17}{5}}\\ c_{2}&={\frac {7}{5}}\\ c_{3}&=-{\frac {11}{5}} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {7 x^{4} \cos \left (\ln \left (x \right )\right )}{5}-\frac {11 x^{4} \sin \left (\ln \left (x \right )\right )}{5}-\frac {17 x}{5} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {7 x^{4} \cos \left (\ln \left (x \right )\right )}{5}-\frac {11 x^{4} \sin \left (\ln \left (x \right )\right )}{5}-\frac {17 x}{5} \\ \end{align*}
Verification of solutions
\[ y = \frac {7 x^{4} \cos \left (\ln \left (x \right )\right )}{5}-\frac {11 x^{4} \sin \left (\ln \left (x \right )\right )}{5}-\frac {17 x}{5} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{3} \left (\frac {d}{d x}y^{\prime \prime }\right )-6 x^{2} \left (\frac {d}{d x}y^{\prime }\right )+17 x y^{\prime }-17 y=0, y \left (1\right )=-2, 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 {17 y}{x^{3}}+\frac {6 \left (\frac {d}{d x}y^{\prime }\right ) x -17 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 {6 \left (\frac {d}{d x}y^{\prime }\right )}{x}+\frac {17 y^{\prime }}{x^{2}}-\frac {17 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 )-6 x^{2} \left (\frac {d}{d x}y^{\prime }\right )+17 x y^{\prime }-17 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 )-6 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 )+17 \frac {d}{d t}y \left (t \right )-17 y \left (t \right )=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {d}{d t}\frac {d^{2}}{d t^{2}}y \left (t \right )-9 \frac {d}{d t}\frac {d}{d t}y \left (t \right )+25 \frac {d}{d t}y \left (t \right )-17 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 )=9 y_{3}\left (t \right )-25 y_{2}\left (t \right )+17 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 )=9 y_{3}\left (t \right )-25 y_{2}\left (t \right )+17 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 \\ 17 & -25 & 9 \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 \\ 17 & -25 & 9 \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 [4-\mathrm {I}, \left [\begin {array}{c} \frac {15}{289}+\frac {8 \,\mathrm {I}}{289} \\ \frac {4}{17}+\frac {\mathrm {I}}{17} \\ 1 \end {array}\right ]\right ], \left [4+\mathrm {I}, \left [\begin {array}{c} \frac {15}{289}-\frac {8 \,\mathrm {I}}{289} \\ \frac {4}{17}-\frac {\mathrm {I}}{17} \\ 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 complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [4-\mathrm {I}, \left [\begin {array}{c} \frac {15}{289}+\frac {8 \,\mathrm {I}}{289} \\ \frac {4}{17}+\frac {\mathrm {I}}{17} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (4-\mathrm {I}\right ) t}\cdot \left [\begin {array}{c} \frac {15}{289}+\frac {8 \,\mathrm {I}}{289} \\ \frac {4}{17}+\frac {\mathrm {I}}{17} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Use Euler identity to write solution in terms of}\hspace {3pt} \sin \hspace {3pt}\textrm {and}\hspace {3pt} \cos \\ {} & {} & {\mathrm e}^{4 t}\cdot \left (\cos \left (t \right )-\mathrm {I} \sin \left (t \right )\right )\cdot \left [\begin {array}{c} \frac {15}{289}+\frac {8 \,\mathrm {I}}{289} \\ \frac {4}{17}+\frac {\mathrm {I}}{17} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{4 t}\cdot \left [\begin {array}{c} \left (\frac {15}{289}+\frac {8 \,\mathrm {I}}{289}\right ) \left (\cos \left (t \right )-\mathrm {I} \sin \left (t \right )\right ) \\ \left (\frac {4}{17}+\frac {\mathrm {I}}{17}\right ) \left (\cos \left (t \right )-\mathrm {I} \sin \left (t \right )\right ) \\ \cos \left (t \right )-\mathrm {I} \sin \left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{y}}_{2}\left (t \right )={\mathrm e}^{4 t}\cdot \left [\begin {array}{c} \frac {15 \cos \left (t \right )}{289}+\frac {8 \sin \left (t \right )}{289} \\ \frac {4 \cos \left (t \right )}{17}+\frac {\sin \left (t \right )}{17} \\ \cos \left (t \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (t \right )={\mathrm e}^{4 t}\cdot \left [\begin {array}{c} -\frac {15 \sin \left (t \right )}{289}+\frac {8 \cos \left (t \right )}{289} \\ -\frac {4 \sin \left (t \right )}{17}+\frac {\cos \left (t \right )}{17} \\ -\sin \left (t \right ) \end {array}\right ]\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}\left (t \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (t \right ) \\ \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}^{4 t}\cdot \left [\begin {array}{c} \frac {15 \cos \left (t \right )}{289}+\frac {8 \sin \left (t \right )}{289} \\ \frac {4 \cos \left (t \right )}{17}+\frac {\sin \left (t \right )}{17} \\ \cos \left (t \right ) \end {array}\right ]+c_{3} {\mathrm e}^{4 t}\cdot \left [\begin {array}{c} -\frac {15 \sin \left (t \right )}{289}+\frac {8 \cos \left (t \right )}{289} \\ -\frac {4 \sin \left (t \right )}{17}+\frac {\cos \left (t \right )}{17} \\ -\sin \left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=\frac {\left (\left (15 c_{2} +8 c_{3} \right ) \cos \left (t \right )+8 \sin \left (t \right ) \left (c_{2} -\frac {15 c_{3}}{8}\right )\right ) {\mathrm e}^{4 t}}{289}+c_{1} {\mathrm e}^{t} \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} t =\ln \left (x \right ) \\ {} & {} & y=\frac {\left (\left (15 c_{2} +8 c_{3} \right ) \cos \left (\ln \left (x \right )\right )+8 \sin \left (\ln \left (x \right )\right ) \left (c_{2} -\frac {15 c_{3}}{8}\right )\right ) x^{4}}{289}+c_{1} x \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=\frac {8 x^{4} \left (\left (c_{3} +\frac {15 c_{2}}{8}\right ) \cos \left (\ln \left (x \right )\right )-\frac {15 \left (-\frac {8 c_{2}}{15}+c_{3} \right ) \sin \left (\ln \left (x \right )\right )}{8}\right )}{289}+c_{1} x \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (1\right )=-2 \\ {} & {} & -2=\frac {8 c_{3}}{289}+\frac {15 c_{2}}{289}+c_{1} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {32 x^{3} \left (\left (c_{3} +\frac {15 c_{2}}{8}\right ) \cos \left (\ln \left (x \right )\right )-\frac {15 \left (-\frac {8 c_{2}}{15}+c_{3} \right ) \sin \left (\ln \left (x \right )\right )}{8}\right )}{289}+\frac {8 x^{4} \left (-\frac {\left (c_{3} +\frac {15 c_{2}}{8}\right ) \sin \left (\ln \left (x \right )\right )}{x}-\frac {15 \left (-\frac {8 c_{2}}{15}+c_{3} \right ) \cos \left (\ln \left (x \right )\right )}{8 x}\right )}{289}+c_{1} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=0 \\ {} & {} & 0=\frac {c_{3}}{17}+\frac {4 c_{2}}{17}+c_{1} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\frac {96 x^{2} \left (\left (c_{3} +\frac {15 c_{2}}{8}\right ) \cos \left (\ln \left (x \right )\right )-\frac {15 \left (-\frac {8 c_{2}}{15}+c_{3} \right ) \sin \left (\ln \left (x \right )\right )}{8}\right )}{289}+\frac {64 x^{3} \left (-\frac {\left (c_{3} +\frac {15 c_{2}}{8}\right ) \sin \left (\ln \left (x \right )\right )}{x}-\frac {15 \left (-\frac {8 c_{2}}{15}+c_{3} \right ) \cos \left (\ln \left (x \right )\right )}{8 x}\right )}{289}+\frac {8 x^{4} \left (\frac {\left (c_{3} +\frac {15 c_{2}}{8}\right ) \sin \left (\ln \left (x \right )\right )}{x^{2}}-\frac {\left (c_{3} +\frac {15 c_{2}}{8}\right ) \cos \left (\ln \left (x \right )\right )}{x^{2}}+\frac {15 \left (-\frac {8 c_{2}}{15}+c_{3} \right ) \cos \left (\ln \left (x \right )\right )}{8 x^{2}}+\frac {15 \left (-\frac {8 c_{2}}{15}+c_{3} \right ) \sin \left (\ln \left (x \right )\right )}{8 x^{2}}\right )}{289} \\ \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=-\frac {c_{3}}{17}+\frac {13 c_{2}}{17} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =-\frac {17}{5}, c_{2} =\frac {17}{5}, c_{3} =\frac {221}{5}, x =x \right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {17 x}{5}+\frac {\left (-11 \sin \left (\ln \left (x \right )\right )+7 \cos \left (\ln \left (x \right )\right )\right ) x^{4}}{5} \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: 0.046 (sec). Leaf size: 24
dsolve([x^3*diff(y(x),x$3)-6*x^2*diff(y(x),x$2)+17*x*diff(y(x),x)-17*y(x)=0,y(1) = -2, D(y)(1) = 0, (D@@2)(y)(1) = 0],y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (-11 \sin \left (\ln \left (x \right )\right )+7 \cos \left (\ln \left (x \right )\right )\right ) x^{4}}{5}-\frac {17 x}{5} \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 28
DSolve[{x^3*y'''[x]-6*x^2*y''[x]+17*x*y'[x]-17*y[x]==0,{y[1]==-2,y'[1]==0,y''[1]==0}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{5} x \left (-11 x^3 \sin (\log (x))+7 x^3 \cos (\log (x))-17\right ) \]