Internal problem ID [9574]
Internal file name [OUTPUT/8515_Monday_June_06_2022_03_35_45_AM_4435922/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1246.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "exact linear second order ode", "second_order_integrable_as_is"
Maple gives the following as the ode type
[[_2nd_order, _exact, _linear, _homogeneous]]
\[ \boxed {\left (x^{2}-1\right ) y^{\prime \prime }-2 \left (v -1\right ) x y^{\prime }-2 y v=0} \]
Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (\left (x^{2}-1\right ) y^{\prime \prime }+\left (-2 x v +2 x \right ) y^{\prime }-2 y v \right )d x &= 0 \\ -2 x v y+\left (x^{2}-1\right ) y^{\prime } = c_{1} \end {align*}
Which is now solved for \(y\).
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}
Where here \begin {align*} p(x) &=-\frac {2 x v}{x^{2}-1}\\ q(x) &=\frac {c_{1}}{x^{2}-1} \end {align*}
Hence the ode is \begin {align*} y^{\prime }-\frac {2 x v y}{x^{2}-1} = \frac {c_{1}}{x^{2}-1} \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int -\frac {2 x v}{x^{2}-1}d x} \\ &= {\mathrm e}^{-v \ln \left (x +1\right )-v \ln \left (x -1\right )} \\ \end{align*} Which simplifies to \[ \mu = \left (x +1\right )^{-v} \left (x -1\right )^{-v} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {c_{1}}{x^{2}-1}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\left (x +1\right )^{-v} \left (x -1\right )^{-v} y\right ) &= \left (\left (x +1\right )^{-v} \left (x -1\right )^{-v}\right ) \left (\frac {c_{1}}{x^{2}-1}\right )\\ \mathrm {d} \left (\left (x +1\right )^{-v} \left (x -1\right )^{-v} y\right ) &= \left (\frac {c_{1} \left (x +1\right )^{-v} \left (x -1\right )^{-v}}{x^{2}-1}\right )\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} \left (x +1\right )^{-v} \left (x -1\right )^{-v} y &= \int {\frac {c_{1} \left (x +1\right )^{-v} \left (x -1\right )^{-v}}{x^{2}-1}\,\mathrm {d} x}\\ \left (x +1\right )^{-v} \left (x -1\right )^{-v} y &= \int \frac {c_{1} \left (x +1\right )^{-v} \left (x -1\right )^{-v}}{x^{2}-1}d x + c_{2} \end {align*}
Dividing both sides by the integrating factor \(\mu =\left (x +1\right )^{-v} \left (x -1\right )^{-v}\) results in \begin {align*} y &= \left (x -1\right )^{v} \left (x +1\right )^{v} \left (\int \frac {c_{1} \left (x +1\right )^{-v} \left (x -1\right )^{-v}}{x^{2}-1}d x \right )+c_{2} \left (x -1\right )^{v} \left (x +1\right )^{v} \end {align*}
which simplifies to \begin {align*} y &= \left (x -1\right )^{v} \left (x +1\right )^{v} \left (c_{1} \left (\int \frac {\left (x +1\right )^{-v} \left (x -1\right )^{-v}}{x^{2}-1}d x \right )+c_{2} \right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (x -1\right )^{v} \left (x +1\right )^{v} \left (c_{1} \left (\int \frac {\left (x +1\right )^{-v} \left (x -1\right )^{-v}}{x^{2}-1}d x \right )+c_{2} \right ) \\ \end{align*}
Verification of solutions
\[ y = \left (x -1\right )^{v} \left (x +1\right )^{v} \left (c_{1} \left (\int \frac {\left (x +1\right )^{-v} \left (x -1\right )^{-v}}{x^{2}-1}d x \right )+c_{2} \right ) \] Verified OK.
Writing the ode as \[ \left (x^{2}-1\right ) y^{\prime \prime }+\left (-2 x v +2 x \right ) y^{\prime }-2 y v = 0 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (\left (x^{2}-1\right ) y^{\prime \prime }+\left (-2 x v +2 x \right ) y^{\prime }-2 y v \right )d x &= 0 \\ -2 x v y+y^{\prime } x^{2}-y^{\prime } = c_{1} \end {align*}
Which is now solved for \(y\).
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}
Where here \begin {align*} p(x) &=-\frac {2 x v}{x^{2}-1}\\ q(x) &=\frac {c_{1}}{x^{2}-1} \end {align*}
Hence the ode is \begin {align*} y^{\prime }-\frac {2 x v y}{x^{2}-1} = \frac {c_{1}}{x^{2}-1} \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int -\frac {2 x v}{x^{2}-1}d x} \\ &= {\mathrm e}^{-v \ln \left (x +1\right )-v \ln \left (x -1\right )} \\ \end{align*} Which simplifies to \[ \mu = \left (x +1\right )^{-v} \left (x -1\right )^{-v} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {c_{1}}{x^{2}-1}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\left (x +1\right )^{-v} \left (x -1\right )^{-v} y\right ) &= \left (\left (x +1\right )^{-v} \left (x -1\right )^{-v}\right ) \left (\frac {c_{1}}{x^{2}-1}\right )\\ \mathrm {d} \left (\left (x +1\right )^{-v} \left (x -1\right )^{-v} y\right ) &= \left (\frac {c_{1} \left (x +1\right )^{-v} \left (x -1\right )^{-v}}{x^{2}-1}\right )\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} \left (x +1\right )^{-v} \left (x -1\right )^{-v} y &= \int {\frac {c_{1} \left (x +1\right )^{-v} \left (x -1\right )^{-v}}{x^{2}-1}\,\mathrm {d} x}\\ \left (x +1\right )^{-v} \left (x -1\right )^{-v} y &= \int \frac {c_{1} \left (x +1\right )^{-v} \left (x -1\right )^{-v}}{x^{2}-1}d x + c_{2} \end {align*}
Dividing both sides by the integrating factor \(\mu =\left (x +1\right )^{-v} \left (x -1\right )^{-v}\) results in \begin {align*} y &= \left (x -1\right )^{v} \left (x +1\right )^{v} \left (\int \frac {c_{1} \left (x +1\right )^{-v} \left (x -1\right )^{-v}}{x^{2}-1}d x \right )+c_{2} \left (x -1\right )^{v} \left (x +1\right )^{v} \end {align*}
which simplifies to \begin {align*} y &= \left (x -1\right )^{v} \left (x +1\right )^{v} \left (c_{1} \left (\int \frac {\left (x +1\right )^{-v} \left (x -1\right )^{-v}}{x^{2}-1}d x \right )+c_{2} \right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (x -1\right )^{v} \left (x +1\right )^{v} \left (c_{1} \left (\int \frac {\left (x +1\right )^{-v} \left (x -1\right )^{-v}}{x^{2}-1}d x \right )+c_{2} \right ) \\ \end{align*}
Verification of solutions
\[ y = \left (x -1\right )^{v} \left (x +1\right )^{v} \left (c_{1} \left (\int \frac {\left (x +1\right )^{-v} \left (x -1\right )^{-v}}{x^{2}-1}d x \right )+c_{2} \right ) \] Verified OK.
An ode of the form \begin {align*} p \left (x \right ) y^{\prime \prime }+q \left (x \right ) y^{\prime }+r \left (x \right ) y&=s \left (x \right ) \end {align*}
is exact if \begin {align*} p''(x) - q'(x) + r(x) &= 0 \tag {1} \end {align*}
For the given ode we have \begin {align*} p(x) &= x^{2}-1\\ q(x) &= -2 x v +2 x\\ r(x) &= -2 v\\ s(x) &= 0 \end {align*}
Hence \begin {align*} p''(x) &= 2\\ q'(x) &= -2 v +2 \end {align*}
Therefore (1) becomes \begin {align*} 2- \left (-2 v +2\right ) + \left (-2 v\right )&=0 \end {align*}
Hence the ode is exact. Since we now know the ode is exact, it can be written as \begin {align*} \left (p \left (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y\right )' &= s(x) \end {align*}
Integrating gives \begin {align*} p \left (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y&=\int {s \left (x \right )\, dx} \end {align*}
Substituting the above values for \(p,q,r,s\) gives \begin {align*} -2 x v y+\left (x^{2}-1\right ) y^{\prime }&=c_{1} \end {align*}
We now have a first order ode to solve which is \begin {align*} -2 x v y+\left (x^{2}-1\right ) y^{\prime } = c_{1} \end {align*}
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}
Where here \begin {align*} p(x) &=-\frac {2 x v}{x^{2}-1}\\ q(x) &=\frac {c_{1}}{x^{2}-1} \end {align*}
Hence the ode is \begin {align*} y^{\prime }-\frac {2 x v y}{x^{2}-1} = \frac {c_{1}}{x^{2}-1} \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int -\frac {2 x v}{x^{2}-1}d x} \\ &= {\mathrm e}^{-v \ln \left (x +1\right )-v \ln \left (x -1\right )} \\ \end{align*} Which simplifies to \[ \mu = \left (x +1\right )^{-v} \left (x -1\right )^{-v} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {c_{1}}{x^{2}-1}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\left (x +1\right )^{-v} \left (x -1\right )^{-v} y\right ) &= \left (\left (x +1\right )^{-v} \left (x -1\right )^{-v}\right ) \left (\frac {c_{1}}{x^{2}-1}\right )\\ \mathrm {d} \left (\left (x +1\right )^{-v} \left (x -1\right )^{-v} y\right ) &= \left (\frac {c_{1} \left (x +1\right )^{-v} \left (x -1\right )^{-v}}{x^{2}-1}\right )\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} \left (x +1\right )^{-v} \left (x -1\right )^{-v} y &= \int {\frac {c_{1} \left (x +1\right )^{-v} \left (x -1\right )^{-v}}{x^{2}-1}\,\mathrm {d} x}\\ \left (x +1\right )^{-v} \left (x -1\right )^{-v} y &= \int \frac {c_{1} \left (x +1\right )^{-v} \left (x -1\right )^{-v}}{x^{2}-1}d x + c_{2} \end {align*}
Dividing both sides by the integrating factor \(\mu =\left (x +1\right )^{-v} \left (x -1\right )^{-v}\) results in \begin {align*} y &= \left (x -1\right )^{v} \left (x +1\right )^{v} \left (\int \frac {c_{1} \left (x +1\right )^{-v} \left (x -1\right )^{-v}}{x^{2}-1}d x \right )+c_{2} \left (x -1\right )^{v} \left (x +1\right )^{v} \end {align*}
which simplifies to \begin {align*} y &= \left (x -1\right )^{v} \left (x +1\right )^{v} \left (c_{1} \left (\int \frac {\left (x +1\right )^{-v} \left (x -1\right )^{-v}}{x^{2}-1}d x \right )+c_{2} \right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (x -1\right )^{v} \left (x +1\right )^{v} \left (c_{1} \left (\int \frac {\left (x +1\right )^{-v} \left (x -1\right )^{-v}}{x^{2}-1}d x \right )+c_{2} \right ) \\ \end{align*}
Verification of solutions
\[ y = \left (x -1\right )^{v} \left (x +1\right )^{v} \left (c_{1} \left (\int \frac {\left (x +1\right )^{-v} \left (x -1\right )^{-v}}{x^{2}-1}d x \right )+c_{2} \right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}-1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (-2 x v +2 x \right ) y^{\prime }-2 y v =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 v y}{x^{2}-1}+\frac {2 x \left (v -1\right ) y^{\prime }}{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 \left (v -1\right ) y^{\prime }}{x^{2}-1}-\frac {2 v 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 \left (v -1\right )}{x^{2}-1}, P_{3}\left (x \right )=-\frac {2 v}{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-v \\ {} & \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 \left (v -1\right ) x y^{\prime }-2 y v =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 v +2 u +2 v -2\right ) \left (\frac {d}{d u}y \left (u \right )\right )-2 v 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 (r -v \right ) u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-2 a_{k +1} \left (k +r +1\right ) \left (k +1+r -v \right )+a_{k} \left (k +r +1\right ) \left (k +r -2 v \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 (r -v \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, v\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 v -2\right ) a_{k +1}+a_{k} \left (k +r -2 v \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 v \right )}{2 \left (k +1+r -v \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k -2 v \right )}{2 \left (k +1-v \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +1}=\frac {a_{k} \left (k -2 v \right )}{2 \left (k +1-v \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}, a_{k +1}=\frac {a_{k} \left (k -2 v \right )}{2 \left (k +1-v \right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =v \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k -v \right )}{2 \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =v \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +v}, a_{k +1}=\frac {a_{k} \left (k -v \right )}{2 \left (k +1\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 +v}, a_{k +1}=\frac {a_{k} \left (k -v \right )}{2 \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +1\right )^{k +v}\right ), a_{k +1}=\frac {a_{k} \left (k -2 v \right )}{2 \left (k +1-v \right )}, b_{k +1}=\frac {b_{k} \left (-v +k \right )}{2 \left (k +1\right )}\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)] One independent solution has integrals. Trying a hypergeometric solution free of integrals... -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 2F1 ODE -> Trying to convert hypergeometric functions to elementary form... <- elementary form is not straightforward to achieve - returning hypergeometric solution free of uncomputed integrals <- linear_1 successful`
✓ Solution by Maple
Time used: 0.078 (sec). Leaf size: 28
dsolve((x^2-1)*diff(diff(y(x),x),x)-2*(v-1)*x*diff(y(x),x)-2*v*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \left (\operatorname {hypergeom}\left (\left [\frac {1}{2}, v +1\right ], \left [\frac {3}{2}\right ], x^{2}\right ) c_{2} x +c_{1} \right ) \left (x^{2}-1\right )^{v} \]
✓ Solution by Mathematica
Time used: 0.041 (sec). Leaf size: 32
DSolve[-2*v*y[x] - 2*(-1 + v)*x*y'[x] + (-1 + x^2)*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \left (x^2-1\right )^{v/2} (c_1 P_v^v(x)+c_2 Q_v^v(x)) \]