Internal problem ID [11724]
Internal file name [OUTPUT/11734_Thursday_April_11_2024_08_49_14_PM_67310780/index.tex]
Book: Differential Equations by Shepley L. Ross. Third edition. John Willey. New Delhi.
2004.
Section: Chapter 4, Section 4.1. Basic theory of linear differential equations. Exercises page
124
Problem number: 3.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "reduction_of_order", "second_order_change_of_variable_on_y_method_2", "second_order_ode_non_constant_coeff_transformation_on_B"
Maple gives the following as the ode type
[_Gegenbauer]
\[ \boxed {\left (x^{2}-1\right ) y^{\prime \prime }-2 x y^{\prime }+2 y=0} \] Given that one solution of the ode is \begin {align*} y_1 &= x \end {align*}
Given one basis solution \(y_{1}\left (x \right )\), then the second basis solution is given by \[ y_{2}\left (x \right ) = y_{1} \left (\int \frac {{\mathrm e}^{-\left (\int p d x \right )}}{y_{1}^{2}}d x \right ) \] Where \(p(x)\) is the coefficient of \(y^{\prime }\) when the ode is written in the normal form \[ y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y = f \left (x \right ) \] Looking at the ode to solve shows that \[ p \left (x \right ) = -\frac {2 x}{x^{2}-1} \] Therefore \begin{align*} y_{2}\left (x \right ) &= x \left (\int \frac {{\mathrm e}^{-\left (\int -\frac {2 x}{x^{2}-1}d x \right )}}{x^{2}}d x \right ) \\ y_{2}\left (x \right ) &= x \int \frac {{\mathrm e}^{\ln \left (x -1\right )+\ln \left (x +1\right )}}{x^{2}} , dx \\ y_{2}\left (x \right ) &= x \left (\int \frac {x^{2}-1}{x^{2}}d x \right ) \\ y_{2}\left (x \right ) &= x \left (x +\frac {1}{x}\right ) \\ \end{align*} Hence the solution is \begin{align*} y &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} x +c_{2} x \left (x +\frac {1}{x}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x +c_{2} x \left (x +\frac {1}{x}\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x +c_{2} x \left (x +\frac {1}{x}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}-1\right ) \left (\frac {d}{d x}y^{\prime }\right )-2 x y^{\prime }+2 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 }=\frac {2 x y^{\prime }}{x^{2}-1}-\frac {2 y}{x^{2}-1} \\ \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 }-\frac {2 x y^{\prime }}{x^{2}-1}+\frac {2 y}{x^{2}-1}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {2 x}{x^{2}-1}, P_{3}\left (x \right )=\frac {2}{x^{2}-1}\right ] \\ {} & \circ & \left (x +1\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=-1 \\ {} & \circ & \left (x +1\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=0 \\ {} & \circ & x =-1\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=-1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (x^{2}-1\right ) \left (\frac {d}{d x}y^{\prime }\right )-2 x y^{\prime }+2 y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -1\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{2}-2 u \right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (-2 u +2\right ) \left (\frac {d}{d u}y \left (u \right )\right )+2 y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -2 a_{0} r \left (-2+r \right ) u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-2 a_{k +1} \left (k +1+r \right ) \left (k +r -1\right )+a_{k} \left (k +r -1\right ) \left (k +r -2\right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -2 r \left (-2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 2\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +r -1\right ) \left (\left (-2 k -2 r -2\right ) a_{k +1}+a_{k} \left (k +r -2\right )\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +r -2\right )}{2 \left (k +1+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =2 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k -2\right )}{2 \left (k +1\right )} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=-a_{0} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =1 \\ {} & {} & a_{2}=-\frac {a_{1}}{4} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{2}=\frac {a_{0}}{4} \\ \bullet & {} & \textrm {Terminating series solution of the ODE for}\hspace {3pt} r =0\hspace {3pt}\textrm {. Use reduction of order to find the second linearly independent solution}\hspace {3pt} \\ {} & {} & y \left (u \right )=a_{0}\cdot \left (1-u +\frac {1}{4} u^{2}\right ) \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\frac {a_{0} \left (x -1\right )^{2}}{4}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +1}=\frac {a_{k} k}{2 \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +2}, a_{k +1}=\frac {a_{k} k}{2 \left (k +3\right )}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k +2}, a_{k +1}=\frac {a_{k} k}{2 \left (k +3\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\frac {a_{0} \left (x -1\right )^{2}}{4}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +1\right )^{k +2}\right ), b_{k +1}=\frac {b_{k} k}{2 \left (k +3\right )}\right ] \end {array} \]
Maple trace Kovacic algorithm successful
`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 A Liouvillian solution exists Reducible group (found an exponential solution) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 14
dsolve([(x^2-1)*diff(y(x),x$2)-2*x*diff(y(x),x)+2*y(x)=0,x],singsol=all)
\[ y \left (x \right ) = c_{2} x^{2}+c_{1} x +c_{2} \]
✓ Solution by Mathematica
Time used: 0.086 (sec). Leaf size: 39
DSolve[(x^2-1)*y''[x]-2*x*y'[x]+2*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {\sqrt {x^2-1} \left (c_1 (x-1)^2+c_2 x\right )}{\sqrt {1-x^2}} \]