Internal problem ID [10903]
Internal file name [OUTPUT/10160_Sunday_December_31_2023_11_03_04_AM_75682043/index.tex]
Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev.
Second edition
Section: Chapter 2, Second-Order Differential Equations. section 2.1.2-3 Equation of form
\((a x + b)y''+f(x)y'+g(x)y=0\)
Problem number: 80.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "kovacic"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x y^{\prime \prime }+\left (a \,x^{2}+b x \right ) y^{\prime }-\left (a c \,x^{2}+\left (b c +c^{2}+a \right ) x +b +2 c \right ) y=0} \]
Writing the ode as \begin {align*} x y^{\prime \prime }+\left (a x +b \right ) y^{\prime } x +\left (-c^{2} x +\left (-a \,x^{2}-b x -2\right ) c -a x -b \right ) y &= 0 \tag {1} \\ A y^{\prime \prime } + B y^{\prime } + C y &= 0 \tag {2} \end {align*}
Comparing (1) and (2) shows that \begin {align*} A &= x \\ B &= x \left (a x +b \right )\tag {3} \\ C &= -c^{2} x +\left (-a \,x^{2}-b x -2\right ) c -a x -b \end {align*}
Applying the Liouville transformation on the dependent variable gives \begin {align*} z(x) &= y e^{\int \frac {B}{2 A} \,dx} \end {align*}
Then (2) becomes \begin {align*} z''(x) = r z(x)\tag {4} \end {align*}
Where \(r\) is given by \begin {align*} r &= \frac {s}{t}\tag {5} \\ &= \frac {2 A B' - 2 B A' + B^2 - 4 A C}{4 A^2} \end {align*}
Substituting the values of \(A,B,C\) from (3) in the above and simplifying gives \begin {align*} r &= \frac {a^{2} x^{3}+2 a b \,x^{2}+4 a c \,x^{2}+b^{2} x +4 b c x +4 c^{2} x +6 a x +4 b +8 c}{4 x}\tag {6} \end {align*}
Comparing the above to (5) shows that \begin {align*} s &= a^{2} x^{3}+2 a b \,x^{2}+4 a c \,x^{2}+b^{2} x +4 b c x +4 c^{2} x +6 a x +4 b +8 c\\ t &= 4 x \end {align*}
Therefore eq. (4) becomes \begin {align*} z''(x) &= \left ( \frac {a^{2} x^{3}+2 a b \,x^{2}+4 a c \,x^{2}+b^{2} x +4 b c x +4 c^{2} x +6 a x +4 b +8 c}{4 x}\right ) z(x)\tag {7} \end {align*}
Equation (7) is now solved. After finding \(z(x)\) then \(y\) is found using the inverse transformation \begin {align*} y &= z \left (x \right ) e^{-\int \frac {B}{2 A} \,dx} \end {align*}
The first step is to determine the case of Kovacic algorithm this ode belongs to. There are 3 cases depending on the order of poles of \(r\) and the order of \(r\) at \(\infty \). The following table summarizes these cases.
| Case |
Allowed pole order for \(r\) |
Allowed value for \(\mathcal {O}(\infty )\) |
| 1 |
\(\left \{ 0,1,2,4,6,8,\cdots \right \} \) |
\(\left \{ \cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \right \} \) |
|
2 |
Need to have at least one pole that is either order \(2\) or odd order greater than \(2\). Any other pole order is allowed as long as the above condition is satisfied. Hence the following set of pole orders are all allowed. \(\{1,2\}\),\(\{1,3\}\),\(\{2\}\),\(\{3\}\),\(\{3,4\}\),\(\{1,2,5\}\). |
no condition |
| 3 |
\(\left \{ 1,2\right \} \) |
\(\left \{ 2,3,4,5,6,7,\cdots \right \} \) |
The order of \(r\) at \(\infty \) is the degree of \(t\) minus the degree of \(s\). Therefore \begin {align*} O\left (\infty \right ) &= \text {deg}(t) - \text {deg}(s) \\ &= 1 - 3 \\ &= -2 \end {align*}
The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=4 x\). There is a pole at \(x=0\) of order \(1\). Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(-2\) then the necessary conditions for case one are met. Therefore \begin {align*} L &= [1] \end {align*}
Attempting to find a solution using case \(n=1\).
Looking at poles of order 1. For the pole at \(x = 0\) of order 1 then \begin {align*} [\sqrt r]_c &= 0 \\ \alpha _c^+ &= 1 \\ \alpha _c^- &= 1 \end {align*}
Since the order of \(r\) at \(\infty \) is \(O_r(\infty ) = -2\) then \begin {alignat*} {3} v &= \frac {-O_r(\infty )}{2} &&= \frac {2}{2} &&= 1 \end {alignat*}
\([\sqrt r]_\infty \) is the sum of terms involving \(x^i\) for \(0\leq i \leq v\) in the Laurent series for \(\sqrt r\) at \(\infty \). Therefore \begin {align*} [\sqrt r]_\infty &= \sum _{i=0}^{v} a_i x^i \\ &= \sum _{i=0}^{1} a_i x^i \tag {8} \end {align*}
Let \(a\) be the coefficient of \(x^v=x^1\) in the above sum. The Laurent series of \(\sqrt r\) at \(\infty \) is \[ \sqrt r \approx \frac {b}{2}+c +\frac {3}{2 x}+\frac {a x}{2}-\frac {b}{2 a \,x^{2}}-\frac {c}{a \,x^{2}}+\frac {b^{2}}{2 a^{2} x^{3}}+\frac {2 c^{2}}{a^{2} x^{3}}-\frac {b^{3}}{2 a^{3} x^{4}}-\frac {4 c^{3}}{a^{3} x^{4}}+\frac {15 b}{4 a^{2} x^{4}}+\frac {15 c}{2 a^{2} x^{4}}+\frac {b^{4}}{2 a^{4} x^{5}}+\frac {8 c^{4}}{a^{4} x^{5}}-\frac {11 b^{2}}{2 a^{3} x^{5}}-\frac {22 c^{2}}{a^{3} x^{5}}-\frac {b^{5}}{2 a^{5} x^{6}}-\frac {16 c^{5}}{a^{5} x^{6}}+\frac {15 b^{3}}{2 a^{4} x^{6}}+\frac {60 c^{3}}{a^{4} x^{6}}-\frac {81 b}{4 a^{3} x^{6}}-\frac {81 c}{2 a^{3} x^{6}}-\frac {9}{4 a \,x^{3}}+\frac {27}{4 a^{2} x^{5}}+\frac {2 b c}{a^{2} x^{3}}-\frac {3 b^{2} c}{a^{3} x^{4}}-\frac {6 b \,c^{2}}{a^{3} x^{4}}+\frac {4 b^{3} c}{a^{4} x^{5}}+\frac {12 b^{2} c^{2}}{a^{4} x^{5}}+\frac {16 b \,c^{3}}{a^{4} x^{5}}-\frac {22 b c}{a^{3} x^{5}}-\frac {5 b^{4} c}{a^{5} x^{6}}-\frac {20 b^{3} c^{2}}{a^{5} x^{6}}-\frac {40 b^{2} c^{3}}{a^{5} x^{6}}-\frac {40 b \,c^{4}}{a^{5} x^{6}}+\frac {45 b^{2} c}{a^{4} x^{6}}+\frac {90 b \,c^{2}}{a^{4} x^{6}} + \dots \tag {9} \] Comparing Eq. (9) with Eq. (8) shows that \[ a = \frac {a}{2} \] From Eq. (9) the sum up to \(v=1\) gives \begin {align*} [\sqrt r]_\infty &= \sum _{i=0}^{1} a_i x^i \\ &= \frac {b}{2}+c +\frac {a x}{2} \tag {10} \end {align*}
Now we need to find \(b\), where \(b\) be the coefficient of \(x^{v-1} = x^{0}=1\) in \(r\) minus the coefficient of same term but in \(\left ( [\sqrt r]_\infty \right )^2 \) where \([\sqrt r]_\infty \) was found above in Eq (10). Hence \[ \left ( [\sqrt r]_\infty \right )^2 = \frac {1}{4} b^{2}+b c +\frac {1}{2} x a b +c^{2}+x a c +\frac {1}{4} a^{2} x^{2} \] This shows that the coefficient of \(1\) in the above is \(\frac {1}{4} b^{2}+b c +c^{2}\). Now we need to find the coefficient of \(1\) in \(r\). How this is done depends on if \(v=0\) or not. Since \(v=1\) which is not zero, then starting \(r=\frac {s}{t}\), we do long division and write this in the form \[ r = Q + \frac {R}{t} \] Where \(Q\) is the quotient and \(R\) is the remainder. Then the coefficient of \(1\) in \(r\) will be the coefficient this term in the quotient. Doing long division gives \begin {align*} r &= \frac {s}{t} \\ &= \frac {a^{2} x^{3}+2 a b \,x^{2}+4 a c \,x^{2}+b^{2} x +4 b c x +4 c^{2} x +6 a x +4 b +8 c}{4 x} \\ &= Q + \frac {R}{4 x} \\ &= \left (\frac {a^{2} x^{2}}{4}+\left (\frac {1}{2} a b +a c \right ) x +\frac {b^{2}}{4}+b c +c^{2}+\frac {3 a}{2}\right ) + \left ( \frac {4 b +8 c}{4 x}\right ) \\ &= \frac {a^{2} x^{2}}{4}+\left (\frac {1}{2} a b +a c \right ) x +\frac {b^{2}}{4}+b c +c^{2}+\frac {3 a}{2}+\frac {4 b +8 c}{4 x} \end {align*}
We see that the coefficient of the term \(1\) in the quotient is \(\frac {1}{4} b^{2}+b c +c^{2}+\frac {3}{2} a\). Now \(b\) can be found. \begin {align*} b &= \left (\frac {1}{4} b^{2}+b c +c^{2}+\frac {3}{2} a\right )-\left (\frac {1}{4} b^{2}+b c +c^{2}\right )\\ &= \frac {3 a}{2} \end {align*}
Hence \begin {alignat*} {3} [\sqrt r]_\infty &= \frac {b}{2}+c +\frac {a x}{2}\\ \alpha _{\infty }^{+} &= \frac {1}{2} \left ( \frac {b}{a} - v \right ) &&= \frac {1}{2} \left ( \frac {\frac {3 a}{2}}{\frac {a}{2}} - 1 \right ) &&= 1\\ \alpha _{\infty }^{-} &= \frac {1}{2} \left ( -\frac {b}{a} - v \right ) &&= \frac {1}{2} \left ( -\frac {\frac {3 a}{2}}{\frac {a}{2}} - 1 \right ) &&= -2 \end {alignat*}
The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) where \(r\) is \[ r=\frac {a^{2} x^{3}+2 a b \,x^{2}+4 a c \,x^{2}+b^{2} x +4 b c x +4 c^{2} x +6 a x +4 b +8 c}{4 x} \]
| pole \(c\) location | pole order | \([\sqrt r]_c\) | \(\alpha _c^{+}\) | \(\alpha _c^{-}\) |
| \(0\) | \(1\) | \(0\) | \(0\) | \(1\) |
| Order of \(r\) at \(\infty \) | \([\sqrt r]_\infty \) | \(\alpha _\infty ^{+}\) | \(\alpha _\infty ^{-}\) |
| \(-2\) | \(\frac {b}{2}+c +\frac {a x}{2}\) | \(1\) | \(-2\) |
Now that the all \([\sqrt r]_c\) and its associated \(\alpha _c^{\pm }\) have been determined for all the poles in the set \(\Gamma \) and \([\sqrt r]_\infty \) and its associated \(\alpha _\infty ^{\pm }\) have also been found, the next step is to determine possible non negative integer \(d\) from these using \begin {align*} d &= \alpha _\infty ^{s(\infty )} - \sum _{c \in \Gamma } \alpha _c^{s(c)} \end {align*}
Where \(s(c)\) is either \(+\) or \(-\) and \(s(\infty )\) is the sign of \(\alpha _\infty ^{\pm }\). This is done by trial over all set of families \(s=(s(c))_{c \in \Gamma \cup {\infty }}\) until such \(d\) is found to work in finding candidate \(\omega \). Trying \(\alpha _\infty ^{+} = 1\) then \begin {align*} d &= \alpha _\infty ^{+} - \left ( \alpha _{c_1}^{-} \right ) \\ &= 1 - \left ( 1 \right ) \\ &= 0 \end {align*}
Since \(d\) an integer and \(d \geq 0\) then it can be used to find \(\omega \) using \begin {align*} \omega &= \sum _{c \in \Gamma } \left ( s(c) [\sqrt r]_c + \frac {\alpha _c^{s(c)}}{x-c} \right ) + s(\infty ) [\sqrt r]_\infty \end {align*}
Substituting the above values in the above results in \begin {align*} \omega &= \left ( (-)[\sqrt r]_{c_1} + \frac { \alpha _{c_1}^{-} }{x- c_1}\right ) + (+) [\sqrt r]_\infty \\ &= \frac {1}{x} + \left ( \frac {b}{2}+c +\frac {a x}{2} \right ) \\ &= \frac {1}{x}+\frac {b}{2}+c +\frac {a x}{2}\\ &= \frac {1}{x}+\frac {b}{2}+c +\frac {a x}{2} \end {align*}
Now that \(\omega \) is determined, the next step is find a corresponding minimal polynomial \(p(x)\) of degree \(d=0\) to solve the ode. The polynomial \(p(x)\) needs to satisfy the equation \begin {align*} p'' + 2 \omega p' + \left ( \omega ' +\omega ^2 -r\right ) p = 0 \tag {1A} \end {align*}
Let \begin {align*} p(x) &= 1\tag {2A} \end {align*}
Substituting the above in eq. (1A) gives \begin {align*} \left (0\right ) + 2 \left (\frac {1}{x}+\frac {b}{2}+c +\frac {a x}{2}\right ) \left (0\right ) + \left ( \left (-\frac {1}{x^{2}}+\frac {a}{2}\right ) + \left (\frac {1}{x}+\frac {b}{2}+c +\frac {a x}{2}\right )^2 - \left (\frac {a^{2} x^{3}+2 a b \,x^{2}+4 a c \,x^{2}+b^{2} x +4 b c x +4 c^{2} x +6 a x +4 b +8 c}{4 x}\right ) \right ) &= 0\\ 0 = 0 \end {align*}
The equation is satisfied since both sides are zero. Therefore the first solution to the ode \(z'' = r z\) is \begin {align*} z_1(x) &= p e^{ \int \omega \,dx} \\ &= {\mathrm e}^{\int \left (\frac {1}{x}+\frac {b}{2}+c +\frac {a x}{2}\right )d x}\\ &= x \,{\mathrm e}^{\frac {x \left (a x +2 b +4 c \right )}{4}} \end {align*}
The first solution to the original ode in \(y\) is found from \begin{align*} y_1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dx} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {x \left (a x +b \right )}{x} \,dx} \\ &= z_1 e^{-\frac {1}{4} a \,x^{2}-\frac {1}{2} b x} \\ &= z_1 \left ({\mathrm e}^{-\frac {x \left (a x +2 b \right )}{4}}\right ) \\ \end{align*} Which simplifies to \[ y_1 = x \,{\mathrm e}^{c x} \] The second solution \(y_2\) to the original ode is found using reduction of order \[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{y_1^2} \,dx \] Substituting gives \begin{align*} y_2 &= y_1 \int \frac { e^{\int -\frac {x \left (a x +b \right )}{x} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{-\frac {1}{2} a \,x^{2}-b x}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (\int \frac {{\mathrm e}^{-\frac {x \left (a x +2 b +4 c \right )}{2}}}{x^{2}}d x\right ) \\ \end{align*} Therefore the solution is
\begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left (x \,{\mathrm e}^{c x}\right ) + c_{2} \left (x \,{\mathrm e}^{c x}\left (\int \frac {{\mathrm e}^{-\frac {x \left (a x +2 b +4 c \right )}{2}}}{x^{2}}d x\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x \,{\mathrm e}^{c x}+c_{2} x \,{\mathrm e}^{c x} \left (\int \frac {{\mathrm e}^{-\frac {x \left (a x +2 b +4 c \right )}{2}}}{x^{2}}d x \right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \,{\mathrm e}^{c x}+c_{2} x \,{\mathrm e}^{c x} \left (\int \frac {{\mathrm e}^{-\frac {x \left (a x +2 b +4 c \right )}{2}}}{x^{2}}d x \right ) \] Verified OK.
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) Group is reducible, not completely reducible Solution has integrals. Trying a special function solution free of integrals... -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius <- Heun successful: received ODE is equivalent to the HeunB ODE, case c = 0 Special function solution also has integrals. Returning default Liouvillian solution. <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 8.546 (sec). Leaf size: 34
dsolve(x*diff(y(x),x$2)+(a*x^2+b*x)*diff(y(x),x)-(a*c*x^2+(a+b*c+c^2)*x+b+2*c)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{c x} x \left (c_{1} +c_{2} \left (\int \frac {{\mathrm e}^{-\frac {x \left (a x +2 b +4 c \right )}{2}}}{x^{2}}d x \right )\right ) \]
✓ Solution by Mathematica
Time used: 3.129 (sec). Leaf size: 49
DSolve[x*y''[x]+(a*x^2+b*x)*y'[x]-(a*c*x^2+(a+b*c+c^2)*x+b+2*c)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to x e^{c x} \left (c_2 \int _1^x\frac {e^{-\frac {1}{2} K[1] (2 b+4 c+a K[1])}}{K[1]^2}dK[1]+c_1\right ) \]