Internal problem ID [9585]
Internal file name [OUTPUT/8526_Monday_June_06_2022_03_38_17_AM_49570942/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1257.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_y]]
\[ \boxed {x \left (x -1\right ) y^{\prime \prime }+\left (\left (a +1\right ) x +b \right ) y^{\prime }=0} \]
This is second order ode with missing dependent variable \(y\). Let \begin {align*} p(x) &= y^{\prime } \end {align*}
Then \begin {align*} p'(x) &= y^{\prime \prime } \end {align*}
Hence the ode becomes \begin {align*} \left (x^{2}-x \right ) p^{\prime }\left (x \right )+\left (a x +b +x \right ) p \left (x \right ) = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. In canonical form the ODE is \begin {align*} p' &= F(x,p)\\ &= f( x) g(p)\\ &= -\frac {p \left (a x +b +x \right )}{x \left (x -1\right )} \end {align*}
Where \(f(x)=-\frac {a x +b +x}{x \left (x -1\right )}\) and \(g(p)=p\). Integrating both sides gives \begin {align*} \frac {1}{p} \,dp &= -\frac {a x +b +x}{x \left (x -1\right )} \,d x\\ \int { \frac {1}{p} \,dp} &= \int {-\frac {a x +b +x}{x \left (x -1\right )} \,d x}\\ \ln \left (p \right )&=b \ln \left (x \right )-\left (a +b +1\right ) \ln \left (x -1\right )+c_{1}\\ p&={\mathrm e}^{b \ln \left (x \right )-\left (a +b +1\right ) \ln \left (x -1\right )+c_{1}}\\ &=c_{1} {\mathrm e}^{b \ln \left (x \right )-\left (a +b +1\right ) \ln \left (x -1\right )} \end {align*}
Which simplifies to \[ p \left (x \right ) = \frac {c_{1} x^{b} \left (x -1\right )^{-a} \left (x -1\right )^{-b}}{x -1} \] Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = \frac {c_{1} x^{b} \left (x -1\right )^{-a} \left (x -1\right )^{-b}}{x -1} \end {align*}
Integrating both sides gives \begin {align*} y = \int \frac {c_{1} x^{b} \left (x -1\right )^{-a} \left (x -1\right )^{-b}}{x -1}d x +c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \int \frac {c_{1} x^{b} \left (x -1\right )^{-a} \left (x -1\right )^{-b}}{x -1}d x +c_{2} \\ \end{align*}
Verification of solutions
\[ y = \int \frac {c_{1} x^{b} \left (x -1\right )^{-a} \left (x -1\right )^{-b}}{x -1}d x +c_{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}-x \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (a x +b +x \right ) y^{\prime }=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 {\left (a x +b +x \right ) y^{\prime }}{x \left (x -1\right )} \\ \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 {\left (a x +b +x \right ) y^{\prime }}{x \left (x -1\right )}=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 {a x +b +x}{x \left (x -1\right )}, P_{3}\left (x \right )=0\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}}}=-b \\ {} & \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}}}=0 \\ {} & \circ & x =0\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}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x \left (x -1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (a x +b +x \right ) y^{\prime }=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^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\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 -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (1-r +b \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (-k -r +b \right )+a_{k} \left (k +r \right ) \left (k +r +a \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (1-r +b \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, b +1\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +1+r \right ) \left (-k -r +b \right )+a_{k} \left (k +r \right ) \left (k +r +a \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k +r \right ) \left (k +r +a \right )}{\left (k +1+r \right ) \left (-k -r +b \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=-\frac {a_{k} k \left (k +a \right )}{\left (k +1\right ) \left (-k +b \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +1}=-\frac {a_{k} k \left (k +a \right )}{\left (k +1\right ) \left (-k +b \right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =b +1 \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k +b +1\right ) \left (k +b +1+a \right )}{\left (k +2+b \right ) \left (-k -1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =b +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +b +1}, a_{k +1}=-\frac {a_{k} \left (k +b +1\right ) \left (k +b +1+a \right )}{\left (k +2+b \right ) \left (-k -1\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k +b +1}\right ), c_{k +1}=-\frac {c_{k} k \left (k +a \right )}{\left (k +1\right ) \left (-k +b \right )}, d_{k +1}=-\frac {d_{k} \left (k +b +1\right ) \left (k +b +1+a \right )}{\left (k +2+b \right ) \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)] checking if the LODE is missing y <- LODE missing y successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 27
dsolve(x*(x-1)*diff(diff(y(x),x),x)+((a+1)*x+b)*diff(y(x),x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} +x^{b +1} \operatorname {hypergeom}\left (\left [b +1, b +a +1\right ], \left [b +2\right ], x\right ) c_{2} \]
✓ Solution by Mathematica
Time used: 0.07 (sec). Leaf size: 33
DSolve[(b + (1 + a)*x)*y'[x] + (-1 + x)*x*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {c_1 x^{b+1} \operatorname {Hypergeometric2F1}(b+1,a+b+1,b+2,x)}{b+1}+c_2 \]