Internal problem ID [14774]
Internal file name [OUTPUT/14454_Monday_April_08_2024_06_25_17_AM_53635310/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: 65.
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]]
\[ \boxed {x^{3} y^{\prime \prime \prime }+9 x^{2} y^{\prime \prime }+44 x y^{\prime }+58 y=0} \] With initial conditions \begin {align*} [y \left (1\right ) = 2, y^{\prime }\left (1\right ) = 10, y^{\prime \prime }\left (1\right ) = -2] \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 \[ x^{3} y^{\prime \prime \prime }+9 x^{2} y^{\prime \prime }+44 x y^{\prime }+58 y = 0 \] gives \[ 44 x \lambda \,x^{\lambda -1}+9 x^{2} \lambda \left (\lambda -1\right ) x^{\lambda -2}+x^{3} \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda -3}+58 x^{\lambda } = 0 \] Which simplifies to \[ 44 \lambda \,x^{\lambda }+9 \lambda \left (\lambda -1\right ) x^{\lambda }+\lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda }+58 x^{\lambda } = 0 \] And since \(x^{\lambda }\neq 0\) then dividing through by \(x^{\lambda }\), the above becomes
\[ 44 \lambda +9 \lambda \left (\lambda -1\right )+\lambda \left (\lambda -1\right ) \left (\lambda -2\right )+58 = 0 \] Simplifying gives the characteristic equation as \[ \lambda ^{3}+6 \lambda ^{2}+37 \lambda +58 = 0 \] Solving the above gives the following roots \begin {align*} \lambda _1 &= -2\\ \lambda _2 &= -2-5 i\\ \lambda _3 &= -2+5 i \end {align*}
This table summarises the result
| root | multiplicity | type of root |
| \(-2\) | \(1\) | real root |
| \(-2 \pm 5 i\) | \(1\) | 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
\[ y = \frac {c_{1}}{x^{2}}+\frac {c_{2} \cos \left (5 \ln \left (x \right )\right )+c_{3} \sin \left (5 \ln \left (x \right )\right )}{x^{2}} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= \frac {1}{x^{2}}\\ y_2 &= \frac {\cos \left (5 \ln \left (x \right )\right )}{x^{2}}\\ y_3 &= \frac {\sin \left (5 \ln \left (x \right )\right )}{x^{2}} \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^{2}}+\frac {c_{2} \cos \left (5 \ln \left (x \right )\right )+c_{3} \sin \left (5 \ln \left (x \right )\right )}{x^{2}} \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 } = -\frac {2 c_{1}}{x^{3}}-\frac {2 \left (c_{2} \cos \left (5 \ln \left (x \right )\right )+c_{3} \sin \left (5 \ln \left (x \right )\right )\right )}{x^{3}}+\frac {-\frac {5 c_{2} \sin \left (5 \ln \left (x \right )\right )}{x}+\frac {5 c_{3} \cos \left (5 \ln \left (x \right )\right )}{x}}{x^{2}} \end {align*}
substituting \(y^{\prime } = 10\) and \(x = 1\) in the above gives \begin {align*} 10 = -2 c_{2} +5 c_{3} -2 c_{1}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = \frac {6 c_{1}}{x^{4}}+\frac {6 c_{2} \cos \left (5 \ln \left (x \right )\right )+6 c_{3} \sin \left (5 \ln \left (x \right )\right )}{x^{4}}-\frac {4 \left (-\frac {5 c_{2} \sin \left (5 \ln \left (x \right )\right )}{x}+\frac {5 c_{3} \cos \left (5 \ln \left (x \right )\right )}{x}\right )}{x^{3}}+\frac {\frac {5 c_{2} \sin \left (5 \ln \left (x \right )\right )}{x^{2}}-\frac {25 c_{2} \cos \left (5 \ln \left (x \right )\right )}{x^{2}}-\frac {5 c_{3} \cos \left (5 \ln \left (x \right )\right )}{x^{2}}-\frac {25 c_{3} \sin \left (5 \ln \left (x \right )\right )}{x^{2}}}{x^{2}} \end {align*}
substituting \(y^{\prime \prime } = -2\) and \(x = 1\) in the above gives \begin {align*} -2 = -19 c_{2} -25 c_{3} +6 c_{1}\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 {106}{25}}\\ c_{2}&=-{\frac {56}{25}}\\ c_{3}&={\frac {14}{5}} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {70 \sin \left (5 \ln \left (x \right )\right )-56 \cos \left (5 \ln \left (x \right )\right )+106}{25 x^{2}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {70 \sin \left (5 \ln \left (x \right )\right )-56 \cos \left (5 \ln \left (x \right )\right )+106}{25 x^{2}} \\ \end{align*}
Verification of solutions
\[ y = \frac {70 \sin \left (5 \ln \left (x \right )\right )-56 \cos \left (5 \ln \left (x \right )\right )+106}{25 x^{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{3} y^{\prime \prime \prime }+9 x^{2} y^{\prime \prime }+44 x y^{\prime }+58 y=0, y \left (1\right )=2, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=10, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=-2\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 {58 y}{x^{3}}-\frac {9 y^{\prime \prime } x +44 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} \\ {} & {} & y^{\prime \prime \prime }+\frac {9 y^{\prime \prime }}{x}+\frac {44 y^{\prime }}{x^{2}}+\frac {58 y}{x^{3}}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & x^{3} y^{\prime \prime \prime }+9 x^{2} y^{\prime \prime }+44 x y^{\prime }+58 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} \\ {} & {} & x^{3} \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 )+9 x^{2} \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 )+44 \frac {d}{d t}y \left (t \right )+58 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 )+37 \frac {d}{d t}y \left (t \right )+58 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 )-37 y_{2}\left (t \right )-58 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 )-37 y_{2}\left (t \right )-58 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 \\ -58 & -37 & -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 \\ -58 & -37 & -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 [-2, \left [\begin {array}{c} \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [-2-5 \,\mathrm {I}, \left [\begin {array}{c} -\frac {21}{841}-\frac {20 \,\mathrm {I}}{841} \\ -\frac {2}{29}+\frac {5 \,\mathrm {I}}{29} \\ 1 \end {array}\right ]\right ], \left [-2+5 \,\mathrm {I}, \left [\begin {array}{c} -\frac {21}{841}+\frac {20 \,\mathrm {I}}{841} \\ -\frac {2}{29}-\frac {5 \,\mathrm {I}}{29} \\ 1 \end {array}\right ]\right ]\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}}_{1}={\mathrm e}^{-2 t}\cdot \left [\begin {array}{c} \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-2-5 \,\mathrm {I}, \left [\begin {array}{c} -\frac {21}{841}-\frac {20 \,\mathrm {I}}{841} \\ -\frac {2}{29}+\frac {5 \,\mathrm {I}}{29} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (-2-5 \,\mathrm {I}\right ) t}\cdot \left [\begin {array}{c} -\frac {21}{841}-\frac {20 \,\mathrm {I}}{841} \\ -\frac {2}{29}+\frac {5 \,\mathrm {I}}{29} \\ 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}^{-2 t}\cdot \left (\cos \left (5 t \right )-\mathrm {I} \sin \left (5 t \right )\right )\cdot \left [\begin {array}{c} -\frac {21}{841}-\frac {20 \,\mathrm {I}}{841} \\ -\frac {2}{29}+\frac {5 \,\mathrm {I}}{29} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-2 t}\cdot \left [\begin {array}{c} \left (-\frac {21}{841}-\frac {20 \,\mathrm {I}}{841}\right ) \left (\cos \left (5 t \right )-\mathrm {I} \sin \left (5 t \right )\right ) \\ \left (-\frac {2}{29}+\frac {5 \,\mathrm {I}}{29}\right ) \left (\cos \left (5 t \right )-\mathrm {I} \sin \left (5 t \right )\right ) \\ \cos \left (5 t \right )-\mathrm {I} \sin \left (5 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}^{-2 t}\cdot \left [\begin {array}{c} -\frac {21 \cos \left (5 t \right )}{841}-\frac {20 \sin \left (5 t \right )}{841} \\ -\frac {2 \cos \left (5 t \right )}{29}+\frac {5 \sin \left (5 t \right )}{29} \\ \cos \left (5 t \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (t \right )={\mathrm e}^{-2 t}\cdot \left [\begin {array}{c} \frac {21 \sin \left (5 t \right )}{841}-\frac {20 \cos \left (5 t \right )}{841} \\ \frac {2 \sin \left (5 t \right )}{29}+\frac {5 \cos \left (5 t \right )}{29} \\ -\sin \left (5 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}^{-2 t}\cdot \left [\begin {array}{c} \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]+c_{2} {\mathrm e}^{-2 t}\cdot \left [\begin {array}{c} -\frac {21 \cos \left (5 t \right )}{841}-\frac {20 \sin \left (5 t \right )}{841} \\ -\frac {2 \cos \left (5 t \right )}{29}+\frac {5 \sin \left (5 t \right )}{29} \\ \cos \left (5 t \right ) \end {array}\right ]+c_{3} {\mathrm e}^{-2 t}\cdot \left [\begin {array}{c} \frac {21 \sin \left (5 t \right )}{841}-\frac {20 \cos \left (5 t \right )}{841} \\ \frac {2 \sin \left (5 t \right )}{29}+\frac {5 \cos \left (5 t \right )}{29} \\ -\sin \left (5 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 {{\mathrm e}^{-2 t} \left (84 c_{2} \cos \left (5 t \right )+80 c_{3} \cos \left (5 t \right )+80 c_{2} \sin \left (5 t \right )-84 c_{3} \sin \left (5 t \right )-841 c_{1} \right )}{3364} \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} t =\ln \left (x \right ) \\ {} & {} & y=-\frac {84 c_{2} \cos \left (5 \ln \left (x \right )\right )+80 c_{3} \cos \left (5 \ln \left (x \right )\right )+80 c_{2} \sin \left (5 \ln \left (x \right )\right )-84 c_{3} \sin \left (5 \ln \left (x \right )\right )-841 c_{1}}{3364 x^{2}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=-\frac {21 c_{2} \cos \left (5 \ln \left (x \right )\right )}{841 x^{2}}-\frac {20 c_{3} \cos \left (5 \ln \left (x \right )\right )}{841 x^{2}}-\frac {20 c_{2} \sin \left (5 \ln \left (x \right )\right )}{841 x^{2}}+\frac {21 c_{3} \sin \left (5 \ln \left (x \right )\right )}{841 x^{2}}+\frac {c_{1}}{4 x^{2}} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (1\right )=2 \\ {} & {} & 2=-\frac {21 c_{2}}{841}-\frac {20 c_{3}}{841}+\frac {c_{1}}{4} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {2 c_{2} \cos \left (5 \ln \left (x \right )\right )}{29 x^{3}}+\frac {5 c_{2} \sin \left (5 \ln \left (x \right )\right )}{29 x^{3}}+\frac {5 c_{3} \cos \left (5 \ln \left (x \right )\right )}{29 x^{3}}+\frac {2 c_{3} \sin \left (5 \ln \left (x \right )\right )}{29 x^{3}}-\frac {c_{1}}{2 x^{3}} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=10 \\ {} & {} & 10=-\frac {2 c_{2}}{29}+\frac {5 c_{3}}{29}-\frac {c_{1}}{2} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {31 c_{2} \cos \left (5 \ln \left (x \right )\right )}{29 x^{4}}-\frac {5 c_{2} \sin \left (5 \ln \left (x \right )\right )}{29 x^{4}}-\frac {5 c_{3} \cos \left (5 \ln \left (x \right )\right )}{29 x^{4}}-\frac {31 c_{3} \sin \left (5 \ln \left (x \right )\right )}{29 x^{4}}+\frac {3 c_{1}}{2 x^{4}} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=-2 \\ {} & {} & -2=\frac {31 c_{2}}{29}-\frac {5 c_{3}}{29}+\frac {3 c_{1}}{2} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =\frac {424}{25}, c_{2} =-\frac {224}{25}, c_{3} =\frac {518}{5}, x =x \right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {2 \left (53+35 \sin \left (5 \ln \left (x \right )\right )-28 \cos \left (5 \ln \left (x \right )\right )\right )}{25 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: 25
dsolve([x^3*diff(y(x),x$3)+9*x^2*diff(y(x),x$2)+44*x*diff(y(x),x)+58*y(x)=0,y(1) = 2, D(y)(1) = 10, (D@@2)(y)(1) = -2],y(x), singsol=all)
\[ y \left (x \right ) = \frac {\frac {106}{25}+\frac {14 \sin \left (5 \ln \left (x \right )\right )}{5}-\frac {56 \cos \left (5 \ln \left (x \right )\right )}{25}}{x^{2}} \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 28
DSolve[{x^3*y'''[x]+9*x^2*y''[x]+44*x*y'[x]+58*y[x]==0,{y[1]==2,y'[1]==10,y''[1]==-2}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {70 \sin (5 \log (x))-56 \cos (5 \log (x))+106}{25 x^2} \]