problem 1568

Internal problem ID [9890]
Internal ﬁle name [OUTPUT/8837_Monday_June_06_2022_05_35_58_AM_60527840/index.tex]

Book: Diﬀerential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 4, linear fourth order
Problem number: 1568.
ODE order: 4.
ODE degree: 1.

The type(s) of ODE detected by this program : "higher_order_ODE_non_constant_coeﬃcients_of_type_Euler"

\[ \boxed {x^{4} y^{\prime \prime \prime \prime }+8 x^{3} y^{\prime \prime \prime }+12 x^{2} y^{\prime \prime }+a y=0} \] 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}\\ y^{\prime \prime \prime \prime } &= \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) \left (\lambda -3\right ) x^{\lambda -4} \end {align*}

Substituting these back into \[ x^{4} y^{\prime \prime \prime \prime }+8 x^{3} y^{\prime \prime \prime }+12 x^{2} y^{\prime \prime }+a y = 0 \] gives \[ 12 x^{2} \lambda \left (\lambda -1\right ) x^{\lambda -2}+8 x^{3} \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda -3}+x^{4} \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) \left (\lambda -3\right ) x^{\lambda -4}+a \,x^{\lambda } = 0 \] Which simpliﬁes to \[ 12 \lambda \left (\lambda -1\right ) x^{\lambda }+8 \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda }+\lambda \left (\lambda -1\right ) \left (\lambda -2\right ) \left (\lambda -3\right ) x^{\lambda }+a \,x^{\lambda } = 0 \] And since \(x^{\lambda }\neq 0\) then dividing through by \(x^{\lambda }\), the above becomes

\[ 12 \lambda \left (\lambda -1\right )+8 \lambda \left (\lambda -1\right ) \left (\lambda -2\right )+\lambda \left (\lambda -1\right ) \left (\lambda -2\right ) \left (\lambda -3\right )+a = 0 \] Simplifying gives the characteristic equation as \[ \lambda ^{4}+2 \lambda ^{3}-\lambda ^{2}+a -2 \lambda = 0 \] Solving the above gives the following roots \begin {align*} \lambda _1 &= -\frac {1}{2}+\frac {\sqrt {5+4 \sqrt {-a +1}}}{2}\\ \lambda _2 &= -\frac {1}{2}-\frac {\sqrt {5+4 \sqrt {-a +1}}}{2}\\ \lambda _3 &= -\frac {1}{2}+\frac {\sqrt {5-4 \sqrt {-a +1}}}{2}\\ \lambda _4 &= -\frac {1}{2}-\frac {\sqrt {5-4 \sqrt {-a +1}}}{2} \end {align*}


root	multiplicity	type of root

\(-\frac {1}{2}-\frac {\sqrt {5-4 \sqrt {-a +1}}}{2}\)	\(1\)	real root

\(-\frac {1}{2}-\frac {\sqrt {5+4 \sqrt {-a +1}}}{2}\)	\(1\)	real root

\(-\frac {1}{2}+\frac {\sqrt {5+4 \sqrt {-a +1}}}{2}\)	\(1\)	real root

\(-\frac {1}{2}+\frac {\sqrt {5-4 \sqrt {-a +1}}}{2}\)	\(1\)	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

\[ y = c_{1} x^{-\frac {1}{2}-\frac {\sqrt {5-4 \sqrt {-a +1}}}{2}}+c_{2} x^{-\frac {1}{2}-\frac {\sqrt {5+4 \sqrt {-a +1}}}{2}}+c_{3} x^{-\frac {1}{2}+\frac {\sqrt {5+4 \sqrt {-a +1}}}{2}}+c_{4} x^{-\frac {1}{2}+\frac {\sqrt {5-4 \sqrt {-a +1}}}{2}} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= x^{-\frac {1}{2}-\frac {\sqrt {5-4 \sqrt {-a +1}}}{2}}\\ y_2 &= x^{-\frac {1}{2}-\frac {\sqrt {5+4 \sqrt {-a +1}}}{2}}\\ y_3 &= x^{-\frac {1}{2}+\frac {\sqrt {5+4 \sqrt {-a +1}}}{2}}\\ y_4 &= x^{-\frac {1}{2}+\frac {\sqrt {5-4 \sqrt {-a +1}}}{2}} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{-\frac {1}{2}-\frac {\sqrt {5-4 \sqrt {-a +1}}}{2}}+c_{2} x^{-\frac {1}{2}-\frac {\sqrt {5+4 \sqrt {-a +1}}}{2}}+c_{3} x^{-\frac {1}{2}+\frac {\sqrt {5+4 \sqrt {-a +1}}}{2}}+c_{4} x^{-\frac {1}{2}+\frac {\sqrt {5-4 \sqrt {-a +1}}}{2}} \\ \end{align*}

Veriﬁcation of solutions

5.33.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{4} \left (\frac {d}{d x}\frac {d}{d x}y^{\prime \prime }\right )+8 x^{3} \left (\frac {d}{d x}y^{\prime \prime }\right )+12 x^{2} \left (\frac {d}{d x}y^{\prime }\right )+a y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & \frac {d}{d x}\frac {d}{d x}y^{\prime \prime } \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {8}{x}, P_{3}\left (x \right )=\frac {12}{x^{2}}, P_{4}\left (x \right )=0, P_{5}\left (x \right )=\frac {a}{x^{4}}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=8 \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=12 \\ {} & \circ & x^{3}\cdot P_{4}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{3}\cdot P_{4}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=0 \\ {} & \circ & x^{4}\cdot P_{5}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{4}\cdot P_{5}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=a \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite DE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{3}\cdot \left (\frac {d}{d x}y^{\prime \prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{3}\cdot \left (\frac {d}{d x}y^{\prime \prime }\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) \left (k +r -2\right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{4}\cdot \left (\frac {d}{d x}\frac {d}{d x}y^{\prime \prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{4}\cdot \left (\frac {d}{d x}\frac {d}{d x}y^{\prime \prime }\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) \left (k +r -2\right ) \left (k +r -3\right ) x^{k +r} \\ & {} & \textrm {Rewrite DE with series expansions}\hspace {3pt} \\ {} & {} & \moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k^{4}+4 k^{3} r +6 k^{2} r^{2}+4 k \,r^{3}+r^{4}+2 k^{3}+6 k^{2} r +6 k \,r^{2}+2 r^{3}-k^{2}-2 k r -r^{2}+a -2 k -2 r \right ) x^{k +r}=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (k^{4}+2 k^{3}-k^{2}+a -2 k \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k}=0 \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k}=0\right ] \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`

\[ y \left (x \right ) = \frac {c_{1} x^{-\frac {\sqrt {5-4 \sqrt {-a +1}}}{2}}+c_{2} x^{\frac {\sqrt {5-4 \sqrt {-a +1}}}{2}}+c_{3} x^{-\frac {\sqrt {5+4 \sqrt {-a +1}}}{2}}+c_{4} x^{\frac {\sqrt {5+4 \sqrt {-a +1}}}{2}}}{\sqrt {x}} \]

\[ y(x)\to \frac {c_1 x^{-\frac {1}{2} \sqrt {5-4 \sqrt {1-a}}}+c_2 x^{\frac {1}{2} \sqrt {5-4 \sqrt {1-a}}}+c_3 x^{-\frac {1}{2} \sqrt {4 \sqrt {1-a}+5}}+c_4 x^{\frac {1}{2} \sqrt {4 \sqrt {1-a}+5}}}{\sqrt {x}} \]

5.33 problem 1568

5.33.1 Maple step by step solution