Internal problem ID [12254]
Internal file name [OUTPUT/10907_Thursday_September_28_2023_01_08_28_AM_87767529/index.tex]
Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A.
Dobrushkin. CRC Press 2015
Section: Chapter 4, Second and Higher Order Linear Differential Equations. Problems page
221
Problem number: Problem 13.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
Unable to solve or complete the solution.
\[ \boxed {x^{2} y^{\prime \prime }-4 y^{\prime } x^{2}+\left (x^{2}+1\right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )-4 y^{\prime } x^{2}+\left (x^{2}+1\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=4 y^{\prime }-\frac {\left (x^{2}+1\right ) y}{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 }-4 y^{\prime }+\frac {\left (x^{2}+1\right ) y}{x^{2}}=0 \\ \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 )=-4, P_{3}\left (x \right )=\frac {x^{2}+1}{x^{2}}\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}}}=0 \\ {} & \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}}}=1 \\ {} & \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 {Multiply by denominators}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )-4 y^{\prime } x^{2}+\left (x^{2}+1\right ) y=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 ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r +1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -1 \\ {} & {} & x^{2}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =1}{\sum }}a_{k -1} \left (k -1+r \right ) x^{k +r} \\ {} & \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 -1+r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (r^{2}-r +1\right ) x^{r}+\left (\left (r^{2}+r +1\right ) a_{1}-4 a_{0} r \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (k^{2}+2 k r +r^{2}-k -r +1\right )-4 a_{k -1} \left (k -1+r \right )+a_{k -2}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r^{2}-r +1=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}, \frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & \left (r^{2}+r +1\right ) a_{1}-4 a_{0} r =0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=\frac {4 a_{0} r}{r^{2}+r +1} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k^{2}+\left (2 r -1\right ) k +r^{2}-r +1\right ) a_{k}-4 a_{k -1} k -4 a_{k -1} r +a_{k -2}+4 a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (\left (k +2\right )^{2}+\left (2 r -1\right ) \left (k +2\right )+r^{2}-r +1\right ) a_{k +2}-4 a_{k +1} \left (k +2\right )-4 a_{k +1} r +a_{k}+4 a_{k +1}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {4 k a_{k +1}+4 a_{k +1} r -a_{k}+4 a_{k +1}}{k^{2}+2 k r +r^{2}+3 k +3 r +3} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2} \\ {} & {} & a_{k +2}=\frac {4 k a_{k +1}+4 a_{k +1} \left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )-a_{k}+4 a_{k +1}}{k^{2}+2 k \left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )+\left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}+3 k +\frac {9}{2}-\frac {3 \,\mathrm {I} \sqrt {3}}{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}}, a_{k +2}=\frac {4 k a_{k +1}+4 a_{k +1} \left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )-a_{k}+4 a_{k +1}}{k^{2}+2 k \left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )+\left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}+3 k +\frac {9}{2}-\frac {3 \,\mathrm {I} \sqrt {3}}{2}}, a_{1}=\frac {4 a_{0} \left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )}{\left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}+\frac {3}{2}-\frac {\mathrm {I} \sqrt {3}}{2}}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2} \\ {} & {} & a_{k +2}=\frac {4 k a_{k +1}+4 a_{k +1} \left (\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )-a_{k}+4 a_{k +1}}{k^{2}+2 k \left (\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )+\left (\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}+3 k +\frac {9}{2}+\frac {3 \,\mathrm {I} \sqrt {3}}{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}}, a_{k +2}=\frac {4 k a_{k +1}+4 a_{k +1} \left (\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )-a_{k}+4 a_{k +1}}{k^{2}+2 k \left (\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )+\left (\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}+3 k +\frac {9}{2}+\frac {3 \,\mathrm {I} \sqrt {3}}{2}}, a_{1}=\frac {4 a_{0} \left (\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )}{\left (\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}+\frac {3}{2}+\frac {\mathrm {I} \sqrt {3}}{2}}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}}\right ), a_{k +2}=\frac {4 k a_{k +1}+4 a_{k +1} \left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )-a_{k}+4 a_{k +1}}{k^{2}+2 k \left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )+\left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}+3 k +\frac {9}{2}-\frac {3 \,\mathrm {I} \sqrt {3}}{2}}, a_{1}=\frac {4 a_{0} \left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )}{\left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}+\frac {3}{2}-\frac {\mathrm {I} \sqrt {3}}{2}}, b_{k +2}=\frac {4 k b_{k +1}+4 b_{k +1} \left (\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )-b_{k}+4 b_{k +1}}{k^{2}+2 k \left (\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )+\left (\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}+3 k +\frac {9}{2}+\frac {3 \,\mathrm {I} \sqrt {3}}{2}}, b_{1}=\frac {4 b_{0} \left (\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )}{\left (\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}+\frac {3}{2}+\frac {\mathrm {I} \sqrt {3}}{2}}\right ] \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel <- Bessel successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 41
dsolve(x^2*diff(y(x),x$2)-4*x^2*diff(y(x),x)+(x^2+1)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \sqrt {x}\, {\mathrm e}^{2 x} \left (c_{1} \operatorname {BesselI}\left (\frac {i \sqrt {3}}{2}, \sqrt {3}\, x \right )+c_{2} \operatorname {BesselK}\left (\frac {i \sqrt {3}}{2}, \sqrt {3}\, x \right )\right ) \]
✓ Solution by Mathematica
Time used: 0.053 (sec). Leaf size: 67
DSolve[x^2*y''[x]-4*x^2*y'[x]+(x^2+1)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{2 x} \sqrt {x} \left (c_1 \operatorname {BesselJ}\left (\frac {i \sqrt {3}}{2},-i \sqrt {3} x\right )+c_2 \operatorname {BesselY}\left (\frac {i \sqrt {3}}{2},-i \sqrt {3} x\right )\right ) \]