Internal problem ID [5864]
Internal file name [OUTPUT/5112_Sunday_June_05_2022_03_25_02_PM_40711304/index.tex]
Book: THEORY OF DIFFERENTIAL EQUATIONS IN ENGINEERING AND MECHANICS.
K.T. CHAU, CRC Press. Boca Raton, FL. 2018
Section: Chapter 3. Ordinary Differential Equations. Section 3.5 HIGHER ORDER ODE. Page
181
Problem number: Example 3.37.
ODE order: 5.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_missing_y"
Maple gives the following as the ode type
[[_high_order, _missing_y]]
\[ \boxed {y^{\left (5\right )}-\frac {y^{\prime \prime \prime \prime }}{t}=0} \] Since \(y\) is missing from the ode then we can use the substitution \(y^{\prime } = v \left (t \right )\) to reduce the order by one. The ODE becomes \begin {align*} v^{\prime \prime \prime \prime }\left (t \right ) t -v^{\prime \prime \prime }\left (t \right ) = 0 \end {align*}
Since \(v \left (t \right )\) is missing from the ode then we can use the substitution \(v^{\prime }\left (t \right ) = w \left (t \right )\) to reduce the order by one. The ODE becomes \begin {align*} w^{\prime \prime \prime }\left (t \right ) t -w^{\prime \prime }\left (t \right ) = 0 \end {align*}
Since \(w \left (t \right )\) is missing from the ode then we can use the substitution \(w^{\prime }\left (t \right ) = r \left (t \right )\) to reduce the order by one. The ODE becomes \begin {align*} r^{\prime \prime }\left (t \right ) t -r^{\prime }\left (t \right ) = 0 \end {align*}
Integrating both sides of the ODE w.r.t \(t\) gives \begin {align*} \int \left (r^{\prime \prime }\left (t \right ) t -r^{\prime }\left (t \right )\right )d t &= 0 \\ r^{\prime }\left (t \right ) t -2 r \left (t \right ) = c_{1} \end {align*}
Which is now solved for \(r \left (t \right )\). In canonical form the ODE is \begin {align*} r' &= F(t,r)\\ &= f( t) g(r)\\ &= \frac {2 r +c_{1}}{t} \end {align*}
Where \(f(t)=\frac {1}{t}\) and \(g(r)=2 r +c_{1}\). Integrating both sides gives \begin{align*} \frac {1}{2 r +c_{1}} \,dr &= \frac {1}{t} \,d t \\ \int { \frac {1}{2 r +c_{1}} \,dr} &= \int {\frac {1}{t} \,d t} \\ \frac {\ln \left (r +\frac {c_{1}}{2}\right )}{2}&=\ln \left (t \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} \frac {\sqrt {4 r +2 c_{1}}}{2} &= {\mathrm e}^{\ln \left (t \right )+c_{2}} \end {align*}
Which simplifies to \begin {align*} \frac {\sqrt {4 r +2 c_{1}}}{2} &= c_{3} t \end {align*}
But since \(w^{\prime }\left (t \right ) = r \left (t \right )\) then we now need to solve the ode \(w^{\prime }\left (t \right ) = c_{3}^{2} {\mathrm e}^{2 c_{2}} t^{2}-\frac {c_{1}}{2}\). Integrating both sides gives \begin {align*} w \left (t \right ) &= \int { c_{3}^{2} {\mathrm e}^{2 c_{2}} t^{2}-\frac {c_{1}}{2}\,\mathop {\mathrm {d}t}}\\ &= \frac {c_{3}^{2} {\mathrm e}^{2 c_{2}} t^{3}}{3}-\frac {t c_{1}}{2}+c_{4} \end {align*}
But since \(v^{\prime }\left (t \right ) = w \left (t \right )\) then we now need to solve the ode \(v^{\prime }\left (t \right ) = \frac {c_{3}^{2} {\mathrm e}^{2 c_{2}} t^{3}}{3}-\frac {t c_{1}}{2}+c_{4}\). Integrating both sides gives \begin {align*} v \left (t \right ) &= \int { \frac {c_{3}^{2} {\mathrm e}^{2 c_{2}} t^{3}}{3}-\frac {t c_{1}}{2}+c_{4}\,\mathop {\mathrm {d}t}}\\ &= \frac {c_{3}^{2} {\mathrm e}^{2 c_{2}} t^{4}}{12}-\frac {t^{2} c_{1}}{4}+t c_{4} +c_{5} \end {align*}
But since \(y^{\prime } = v \left (t \right )\) then we now need to solve the ode \(y^{\prime } = \frac {c_{3}^{2} {\mathrm e}^{2 c_{2}} t^{4}}{12}-\frac {t^{2} c_{1}}{4}+t c_{4} +c_{5}\). Integrating both sides gives \begin {align*} y &= \int { \frac {c_{3}^{2} {\mathrm e}^{2 c_{2}} t^{4}}{12}-\frac {t^{2} c_{1}}{4}+t c_{4} +c_{5}\,\mathop {\mathrm {d}t}}\\ &= \frac {c_{3}^{2} {\mathrm e}^{2 c_{2}} t^{5}}{60}-\frac {t^{3} c_{1}}{12}+\frac {t^{2} c_{4}}{2}+t c_{5} +c_{6} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{3}^{2} {\mathrm e}^{2 c_{2}} t^{5}}{60}-\frac {t^{3} c_{1}}{12}+\frac {t^{2} c_{4}}{2}+t c_{5} +c_{6} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{3}^{2} {\mathrm e}^{2 c_{2}} t^{5}}{60}-\frac {t^{3} c_{1}}{12}+\frac {t^{2} c_{4}}{2}+t c_{5} +c_{6} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d t}\frac {d^{2}}{d t^{2}}y^{\prime \prime }\right ) t -\frac {d}{d t}\frac {d}{d t}y^{\prime \prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 5 \\ {} & {} & \frac {d}{d t}\frac {d^{2}}{d t^{2}}y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 5th derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}\frac {d^{2}}{d t^{2}}y^{\prime \prime }=\frac {\frac {d}{d t}\frac {d}{d t}y^{\prime \prime }}{t} \\ \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 t}\frac {d^{2}}{d t^{2}}y^{\prime \prime }-\frac {\frac {d}{d t}\frac {d}{d t}y^{\prime \prime }}{t}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d t}\frac {d^{2}}{d t^{2}}y^{\prime \prime }\right ) t -\frac {d}{d t}\frac {d}{d t}y^{\prime \prime }=0 \\ \bullet & {} & \textrm {Make a change of variables}\hspace {3pt} \\ {} & {} & s =\ln \left (t \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 {t}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & y^{\prime }=\left (\frac {d}{d s}y \left (s \right )\right ) s^{\prime }\left (t \right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\frac {d}{d s}y \left (s \right )}{t} \\ {} & \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 {t}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y^{\prime }=\left (\frac {d}{d s}\frac {d}{d s}y \left (s \right )\right ) {s^{\prime }\left (t \right )}^{2}+\left (\frac {d}{d t}s^{\prime }\left (t \right )\right ) \left (\frac {d}{d s}y \left (s \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y^{\prime }=\frac {\frac {d}{d s}\frac {d}{d s}y \left (s \right )}{t^{2}}-\frac {\frac {d}{d s}y \left (s \right )}{t^{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 {t}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y^{\prime \prime }=\left (\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )\right ) {s^{\prime }\left (t \right )}^{3}+3 s^{\prime }\left (t \right ) \left (\frac {d}{d t}s^{\prime }\left (t \right )\right ) \left (\frac {d}{d s}\frac {d}{d s}y \left (s \right )\right )+\left (\frac {d}{d t}s^{\prime \prime }\left (t \right )\right ) \left (\frac {d}{d s}y \left (s \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y^{\prime \prime }=\frac {\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )}{t^{3}}-\frac {3 \left (\frac {d}{d s}\frac {d}{d s}y \left (s \right )\right )}{t^{3}}+\frac {2 \left (\frac {d}{d s}y \left (s \right )\right )}{t^{3}} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {4th}\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 {t}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & \frac {d}{d t}\frac {d}{d t}y^{\prime \prime }=\left (\frac {d}{d s}\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )\right ) {s^{\prime }\left (t \right )}^{4}+3 {s^{\prime }\left (t \right )}^{2} \left (\frac {d}{d t}s^{\prime }\left (t \right )\right ) \left (\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )\right )+3 \left (\frac {d}{d t}s^{\prime }\left (t \right )\right )^{2} \left (\frac {d}{d s}\frac {d}{d s}y \left (s \right )\right )+3 \left (\left (\frac {d}{d t}s^{\prime \prime }\left (t \right )\right ) \left (\frac {d}{d s}\frac {d}{d s}y \left (s \right )\right )+\left (\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )\right ) s^{\prime }\left (t \right ) \left (\frac {d}{d t}s^{\prime }\left (t \right )\right )\right ) s^{\prime }\left (t \right )+\left (\frac {d}{d t}\frac {d}{d t}s^{\prime \prime }\left (t \right )\right ) \left (\frac {d}{d s}y \left (s \right )\right )+\left (\frac {d}{d s}\frac {d}{d s}y \left (s \right )\right ) s^{\prime }\left (t \right ) \left (\frac {d}{d t}s^{\prime \prime }\left (t \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}\frac {d}{d t}y^{\prime \prime }=\frac {\frac {d}{d s}\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )}{t^{4}}-\frac {3 \left (\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )\right )}{t^{4}}+\frac {5 \left (\frac {d}{d s}\frac {d}{d s}y \left (s \right )\right )}{t^{4}}+\frac {3 \left (\frac {2 \left (\frac {d}{d s}\frac {d}{d s}y \left (s \right )\right )}{t^{3}}-\frac {\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )}{t^{3}}\right )}{t}-\frac {6 \left (\frac {d}{d s}y \left (s \right )\right )}{t^{4}} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {5th}\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 {t}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & \frac {d}{d t}\frac {d^{2}}{d t^{2}}y^{\prime \prime }=\left (\frac {d}{d s}\frac {d^{2}}{d s^{2}}\frac {d^{2}}{d s^{2}}y \left (s \right )\right ) {s^{\prime }\left (t \right )}^{5}+4 {s^{\prime }\left (t \right )}^{3} \left (\frac {d}{d t}s^{\prime }\left (t \right )\right ) \left (\frac {d}{d s}\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )\right )+9 s^{\prime }\left (t \right ) \left (\frac {d}{d t}s^{\prime }\left (t \right )\right )^{2} \left (\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )\right )+3 \left (\left (\frac {d}{d t}s^{\prime \prime }\left (t \right )\right ) \left (\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )\right )+\left (\frac {d}{d s}\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )\right ) s^{\prime }\left (t \right ) \left (\frac {d}{d t}s^{\prime }\left (t \right )\right )\right ) {s^{\prime }\left (t \right )}^{2}+6 \left (\frac {d}{d t}s^{\prime }\left (t \right )\right ) \left (\frac {d}{d t}s^{\prime \prime }\left (t \right )\right ) \left (\frac {d}{d s}\frac {d}{d s}y \left (s \right )\right )+3 \left (\left (\frac {d}{d t}\frac {d}{d t}s^{\prime \prime }\left (t \right )\right ) \left (\frac {d}{d s}\frac {d}{d s}y \left (s \right )\right )+\left (\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )\right ) s^{\prime }\left (t \right ) \left (\frac {d}{d t}s^{\prime \prime }\left (t \right )\right )+\left (\frac {d}{d s}\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )\right ) {s^{\prime }\left (t \right )}^{2} \left (\frac {d}{d t}s^{\prime }\left (t \right )\right )+\left (\left (\frac {d}{d t}s^{\prime }\left (t \right )\right )^{2}+\left (\frac {d}{d t}s^{\prime \prime }\left (t \right )\right ) s^{\prime }\left (t \right )\right ) \left (\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )\right )\right ) s^{\prime }\left (t \right )+3 \left (\frac {d}{d t}s^{\prime }\left (t \right )\right ) \left (\left (\frac {d}{d t}s^{\prime \prime }\left (t \right )\right ) \left (\frac {d}{d s}\frac {d}{d s}y \left (s \right )\right )+\left (\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )\right ) s^{\prime }\left (t \right ) \left (\frac {d}{d t}s^{\prime }\left (t \right )\right )\right )+\left (\frac {d}{d t}\frac {d^{2}}{d t^{2}}s^{\prime \prime }\left (t \right )\right ) \left (\frac {d}{d s}y \left (s \right )\right )+\left (\frac {d}{d s}\frac {d}{d s}y \left (s \right )\right ) s^{\prime }\left (t \right ) \left (\frac {d}{d t}\frac {d}{d t}s^{\prime \prime }\left (t \right )\right )+\left (\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )\right ) {s^{\prime }\left (t \right )}^{2} \left (\frac {d}{d t}s^{\prime \prime }\left (t \right )\right )+\left (\left (\frac {d}{d t}\frac {d}{d t}s^{\prime \prime }\left (t \right )\right ) s^{\prime }\left (t \right )+\left (\frac {d}{d t}s^{\prime }\left (t \right )\right ) \left (\frac {d}{d t}s^{\prime \prime }\left (t \right )\right )\right ) \left (\frac {d}{d s}\frac {d}{d s}y \left (s \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}\frac {d^{2}}{d t^{2}}y^{\prime \prime }=\frac {\frac {d}{d s}\frac {d^{2}}{d s^{2}}\frac {d^{2}}{d s^{2}}y \left (s \right )}{t^{5}}-\frac {4 \left (\frac {d}{d s}\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )\right )}{t^{5}}+\frac {11 \left (\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )\right )}{t^{5}}+\frac {3 \left (\frac {2 \left (\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )\right )}{t^{3}}-\frac {\frac {d}{d s}\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )}{t^{3}}\right )}{t^{2}}-\frac {26 \left (\frac {d}{d s}\frac {d}{d s}y \left (s \right )\right )}{t^{5}}+\frac {3 \left (-\frac {6 \left (\frac {d}{d s}\frac {d}{d s}y \left (s \right )\right )}{t^{4}}+\frac {5 \left (\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )\right )}{t^{4}}-\frac {\frac {d}{d s}\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )}{t^{4}}\right )}{t}-\frac {3 \left (\frac {2 \left (\frac {d}{d s}\frac {d}{d s}y \left (s \right )\right )}{t^{3}}-\frac {\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )}{t^{3}}\right )}{t^{2}}+\frac {24 \left (\frac {d}{d s}y \left (s \right )\right )}{t^{5}} \\ & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & {} & \left (\frac {\frac {d}{d s}\frac {d^{2}}{d s^{2}}\frac {d^{2}}{d s^{2}}y \left (s \right )}{t^{5}}-\frac {4 \left (\frac {d}{d s}\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )\right )}{t^{5}}+\frac {11 \left (\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )\right )}{t^{5}}+\frac {3 \left (\frac {2 \left (\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )\right )}{t^{3}}-\frac {\frac {d}{d s}\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )}{t^{3}}\right )}{t^{2}}-\frac {26 \left (\frac {d}{d s}\frac {d}{d s}y \left (s \right )\right )}{t^{5}}+\frac {3 \left (-\frac {6 \left (\frac {d}{d s}\frac {d}{d s}y \left (s \right )\right )}{t^{4}}+\frac {5 \left (\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )\right )}{t^{4}}-\frac {\frac {d}{d s}\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )}{t^{4}}\right )}{t}-\frac {3 \left (\frac {2 \left (\frac {d}{d s}\frac {d}{d s}y \left (s \right )\right )}{t^{3}}-\frac {\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )}{t^{3}}\right )}{t^{2}}+\frac {24 \left (\frac {d}{d s}y \left (s \right )\right )}{t^{5}}\right ) t -\frac {\frac {d}{d s}\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )}{t^{4}}+\frac {3 \left (\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )\right )}{t^{4}}-\frac {5 \left (\frac {d}{d s}\frac {d}{d s}y \left (s \right )\right )}{t^{4}}-\frac {3 \left (\frac {2 \left (\frac {d}{d s}\frac {d}{d s}y \left (s \right )\right )}{t^{3}}-\frac {\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )}{t^{3}}\right )}{t}+\frac {6 \left (\frac {d}{d s}y \left (s \right )\right )}{t^{4}}=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d s}\frac {d^{2}}{d s^{2}}\frac {d^{2}}{d s^{2}}y \left (s \right )-11 \frac {d}{d s}\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )+41 \frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )-61 \frac {d}{d s}\frac {d}{d s}y \left (s \right )+30 \frac {d}{d s}y \left (s \right )}{t^{4}}=0 \\ \bullet & {} & \textrm {Isolate 5th derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d s}\frac {d^{2}}{d s^{2}}\frac {d^{2}}{d s^{2}}y \left (s \right )=11 \frac {d}{d s}\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )-41 \frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )+61 \frac {d}{d s}\frac {d}{d s}y \left (s \right )-30 \frac {d}{d s}y \left (s \right ) \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (s \right )\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 s}\frac {d^{2}}{d s^{2}}\frac {d^{2}}{d s^{2}}y \left (s \right )-11 \frac {d}{d s}\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )+41 \frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right )-61 \frac {d}{d s}\frac {d}{d s}y \left (s \right )+30 \frac {d}{d s}y \left (s \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 (s \right ) \\ {} & {} & y_{1}\left (s \right )=y \left (s \right ) \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (s \right ) \\ {} & {} & y_{2}\left (s \right )=\frac {d}{d s}y \left (s \right ) \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (s \right ) \\ {} & {} & y_{3}\left (s \right )=\frac {d}{d s}\frac {d}{d s}y \left (s \right ) \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{4}\left (s \right ) \\ {} & {} & y_{4}\left (s \right )=\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right ) \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{5}\left (s \right ) \\ {} & {} & y_{5}\left (s \right )=\frac {d}{d s}\frac {d}{d s}\frac {d^{2}}{d s^{2}}y \left (s \right ) \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} \frac {d}{d s}y_{5}\left (s \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & \frac {d}{d s}y_{5}\left (s \right )=11 y_{5}\left (s \right )-41 y_{4}\left (s \right )+61 y_{3}\left (s \right )-30 y_{2}\left (s \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (s \right )=\frac {d}{d s}y_{1}\left (s \right ), y_{3}\left (s \right )=\frac {d}{d s}y_{2}\left (s \right ), y_{4}\left (s \right )=\frac {d}{d s}y_{3}\left (s \right ), y_{5}\left (s \right )=\frac {d}{d s}y_{4}\left (s \right ), \frac {d}{d s}y_{5}\left (s \right )=11 y_{5}\left (s \right )-41 y_{4}\left (s \right )+61 y_{3}\left (s \right )-30 y_{2}\left (s \right )\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (s \right )=\left [\begin {array}{c} y_{1}\left (s \right ) \\ y_{2}\left (s \right ) \\ y_{3}\left (s \right ) \\ y_{4}\left (s \right ) \\ y_{5}\left (s \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & \frac {d}{d s}{\moverset {\rightarrow }{y}}\left (s \right )=\left [\begin {array}{ccccc} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & -30 & 61 & -41 & 11 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (s \right ) \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccccc} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & -30 & 61 & -41 & 11 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & \frac {d}{d s}{\moverset {\rightarrow }{y}}\left (s \right )=A \cdot {\moverset {\rightarrow }{y}}\left (s \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 [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]\right ], \left [2, \left [\begin {array}{c} \frac {1}{16} \\ \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [3, \left [\begin {array}{c} \frac {1}{81} \\ \frac {1}{27} \\ \frac {1}{9} \\ \frac {1}{3} \\ 1 \end {array}\right ]\right ], \left [5, \left [\begin {array}{c} \frac {1}{625} \\ \frac {1}{125} \\ \frac {1}{25} \\ \frac {1}{5} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}=\left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{s}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [2, \left [\begin {array}{c} \frac {1}{16} \\ \frac {1}{8} \\ \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 s}\cdot \left [\begin {array}{c} \frac {1}{16} \\ \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [3, \left [\begin {array}{c} \frac {1}{81} \\ \frac {1}{27} \\ \frac {1}{9} \\ \frac {1}{3} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{4}={\mathrm e}^{3 s}\cdot \left [\begin {array}{c} \frac {1}{81} \\ \frac {1}{27} \\ \frac {1}{9} \\ \frac {1}{3} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [5, \left [\begin {array}{c} \frac {1}{625} \\ \frac {1}{125} \\ \frac {1}{25} \\ \frac {1}{5} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{5}={\mathrm e}^{5 s}\cdot \left [\begin {array}{c} \frac {1}{625} \\ \frac {1}{125} \\ \frac {1}{25} \\ \frac {1}{5} \\ 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}+c_{4} {\moverset {\rightarrow }{y}}_{4}+c_{5} {\moverset {\rightarrow }{y}}_{5} \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{2} {\mathrm e}^{s}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]+c_{3} {\mathrm e}^{2 s}\cdot \left [\begin {array}{c} \frac {1}{16} \\ \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]+c_{4} {\mathrm e}^{3 s}\cdot \left [\begin {array}{c} \frac {1}{81} \\ \frac {1}{27} \\ \frac {1}{9} \\ \frac {1}{3} \\ 1 \end {array}\right ]+c_{5} {\mathrm e}^{5 s}\cdot \left [\begin {array}{c} \frac {1}{625} \\ \frac {1}{125} \\ \frac {1}{25} \\ \frac {1}{5} \\ 1 \end {array}\right ]+\left [\begin {array}{c} c_{1} \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y \left (s \right )=c_{2} {\mathrm e}^{s}+\frac {c_{3} {\mathrm e}^{2 s}}{16}+\frac {c_{4} {\mathrm e}^{3 s}}{81}+\frac {c_{5} {\mathrm e}^{5 s}}{625}+c_{1} \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} s =\ln \left (t \right ) \\ {} & {} & y=c_{2} t +\frac {1}{16} c_{3} t^{2}+\frac {1}{81} c_{4} t^{3}+\frac {1}{625} c_{5} t^{5}+c_{1} \end {array} \]
Maple trace
`Methods for high 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: 24
dsolve(diff(y(t),t$5)-1/t*diff(y(t),t$4)=0,y(t), singsol=all)
\[ y \left (t \right ) = c_{3} t^{5}+c_{5} t^{3}+c_{2} t^{2}+c_{4} t +c_{1} \]
✓ Solution by Mathematica
Time used: 0.025 (sec). Leaf size: 33
DSolve[y'''''[t]-1/t*y''''[t]==0,y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to \frac {c_1 t^5}{120}+c_5 t^3+c_4 t^2+c_3 t+c_2 \]