Internal problem ID [11052]
Internal file name [OUTPUT/10309_Wednesday_January_24_2024_10_07_01_PM_39341230/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 2.1.2-7 Equation of form
\((a_4 x^4+a_3 x^3+a_2 x^2 x+a_1 x+a_0) y''+f(x)y'+g(x)y=0\)
Problem number: 228.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "kovacic", "second_order_change_of_variable_on_x_method_2"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {\left (a \,x^{2}+b \right )^{2} y^{\prime \prime }+\left (2 a x +c \right ) \left (a \,x^{2}+b \right ) y^{\prime }+k y=0} \]
In normal form the ode \begin {align*} \left (a \,x^{2}+b \right )^{2} y^{\prime \prime }+\left (2 a x +c \right ) \left (a \,x^{2}+b \right ) y^{\prime }+k y&=0 \tag {1} \end {align*}
Becomes \begin {align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end {align*}
Where \begin {align*} p \left (x \right )&=\frac {2 a x +c}{a \,x^{2}+b}\\ q \left (x \right )&=\frac {k}{\left (a \,x^{2}+b \right )^{2}} \end {align*}
Applying change of variables \(\tau = g \left (x \right )\) to (2) gives \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end {align*}
Where \(\tau \) is the new independent variable, and \begin {align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end {align*}
Let \(p_{1} = 0\). Eq (4) simplifies to \begin {align*} \tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )&=0 \end {align*}
This ode is solved resulting in \begin {align*} \tau &= \int {\mathrm e}^{-\left (\int p \left (x \right )d x \right )}d x\\ &= \int {\mathrm e}^{-\left (\int \frac {2 a x +c}{a \,x^{2}+b}d x \right )}d x\\ &= \int e^{-\ln \left (a \,x^{2}+b \right )-\frac {c \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{\sqrt {b a}}} \,dx\\ &= \int \frac {{\mathrm e}^{-\frac {c \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{\sqrt {b a}}}}{a \,x^{2}+b}d x\\ &= -\frac {{\mathrm e}^{-\frac {c \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{\sqrt {b a}}}}{c}\tag {6} \end {align*}
Using (6) to evaluate \(q_{1}\) from (5) gives \begin {align*} q_{1} \left (\tau \right ) &= \frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &= \frac {\frac {k}{\left (a \,x^{2}+b \right )^{2}}}{\frac {{\mathrm e}^{-\frac {2 c \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{\sqrt {b a}}}}{\left (a \,x^{2}+b \right )^{2}}}\\ &= k \,{\mathrm e}^{\frac {2 c \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{\sqrt {b a}}}\tag {7} \end {align*}
Substituting the above in (3) and noting that now \(p_{1} = 0\) results in \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+q_{1} y \left (\tau \right )&=0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+k \,{\mathrm e}^{\frac {2 c \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{\sqrt {b a}}} y \left (\tau \right )&=0 \\ \end {align*}
But in terms of \(\tau \) \begin {align*} k \,{\mathrm e}^{\frac {2 c \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{\sqrt {b a}}}&=\frac {k}{c^{2} \tau ^{2}} \end {align*}
Hence the above ode becomes \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+\frac {k y \left (\tau \right )}{c^{2} \tau ^{2}}&=0 \end {align*}
The above ode is now solved for \(y \left (\tau \right )\). The ode can be written as \[ \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right ) c^{2} \tau ^{2}+k y \left (\tau \right ) = 0 \] Which shows it is a Euler ODE. This is Euler second order ODE. Let the solution be \(y \left (\tau \right ) = \tau ^r\), then \(y'=r \tau ^{r-1}\) and \(y''=r(r-1) \tau ^{r-2}\). Substituting these back into the given ODE gives \[ c^{2} \tau ^{2}(r(r-1))\tau ^{r-2}+0 r \tau ^{r-1}+k \,\tau ^{r} = 0 \] Simplifying gives \[ c^{2} r \left (r -1\right )\tau ^{r}+0\,\tau ^{r}+k \,\tau ^{r} = 0 \] Since \(\tau ^{r}\neq 0\) then dividing throughout by \(\tau ^{r}\) gives \[ c^{2} r \left (r -1\right )+0+k = 0 \] Or \[ c^{2} r^{2}-c^{2} r +k = 0 \tag {1} \] Equation (1) is the characteristic equation. Its roots determine the form of the general solution. Using the quadratic equation the roots are \begin {align*} r_1 &= -\frac {-c +\sqrt {c^{2}-4 k}}{2 c}\\ r_2 &= \frac {c +\sqrt {c^{2}-4 k}}{2 c} \end {align*}
Since the roots are real and distinct, then the general solution is \[ y \left (\tau \right )= c_{1} y_1 + c_{2} y_2 \] Where \(y_1 = \tau ^{r_1}\) and \(y_2 = \tau ^{r_2} \). Hence \[ y \left (\tau \right ) = c_{1} \tau ^{-\frac {-c +\sqrt {c^{2}-4 k}}{2 c}}+c_{2} \tau ^{\frac {c +\sqrt {c^{2}-4 k}}{2 c}} \] The above solution is now transformed back to \(y\) using (6) which results in \begin {align*} y &= c_{1} \left (-\frac {{\mathrm e}^{-\frac {c \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{\sqrt {b a}}}}{c}\right )^{-\frac {-c +\sqrt {c^{2}-4 k}}{2 c}}+c_{2} \left (-\frac {{\mathrm e}^{-\frac {c \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{\sqrt {b a}}}}{c}\right )^{\frac {c +\sqrt {c^{2}-4 k}}{2 c}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (-\frac {{\mathrm e}^{-\frac {c \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{\sqrt {b a}}}}{c}\right )^{-\frac {-c +\sqrt {c^{2}-4 k}}{2 c}}+c_{2} \left (-\frac {{\mathrm e}^{-\frac {c \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{\sqrt {b a}}}}{c}\right )^{\frac {c +\sqrt {c^{2}-4 k}}{2 c}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (-\frac {{\mathrm e}^{-\frac {c \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{\sqrt {b a}}}}{c}\right )^{-\frac {-c +\sqrt {c^{2}-4 k}}{2 c}}+c_{2} \left (-\frac {{\mathrm e}^{-\frac {c \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{\sqrt {b a}}}}{c}\right )^{\frac {c +\sqrt {c^{2}-4 k}}{2 c}} \] Verified OK.
Writing the ode as \begin {align*} \left (a \,x^{2}+b \right )^{2} y^{\prime \prime }+\left (2 a x +c \right ) \left (a \,x^{2}+b \right ) y^{\prime }+k 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 &= \left (a \,x^{2}+b \right )^{2} \\ B &= \left (2 a x +c \right ) \left (a \,x^{2}+b \right )\tag {3} \\ C &= k \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 {4 b a +c^{2}-4 k}{4 \left (a \,x^{2}+b \right )^{2}}\tag {6} \end {align*}
Comparing the above to (5) shows that \begin {align*} s &= 4 b a +c^{2}-4 k\\ t &= 4 \left (a \,x^{2}+b \right )^{2} \end {align*}
Therefore eq. (4) becomes \begin {align*} z''(x) &= \left ( \frac {4 b a +c^{2}-4 k}{4 \left (a \,x^{2}+b \right )^{2}}\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) \\ &= 4 - 0 \\ &= 4 \end {align*}
The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=4 \left (a \,x^{2}+b \right )^{2}\). There is a pole at \(x=\frac {\sqrt {-b a}}{a}\) of order \(2\). There is a pole at \(x=-\frac {\sqrt {-b a}}{a}\) of order \(2\). Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(4\) then the necessary conditions for case one are met. Since there is a pole of order \(2\) then necessary conditions for case two are met. Since pole order is not larger than \(2\) and the order at \(\infty \) is \(4\) then the necessary conditions for case three are met. Therefore \begin {align*} L &= [1, 2, 4, 6, 12] \end {align*}
Attempting to find a solution using case \(n=1\).
Unable to find solution using case one
Attempting to find a solution using case \(n=2\).
Looking at poles of order 2. The partial fractions decomposition of \(r\) is \[ r = -\frac {4 b a +c^{2}-4 k}{16 a b \left (x -\sqrt {-\frac {b}{a}}\right )^{2}}-\frac {4 b a +c^{2}-4 k}{16 a b \left (x +\sqrt {-\frac {b}{a}}\right )^{2}}+\frac {-4 b a -c^{2}+4 k}{16 a^{2} \left (-\frac {b}{a}\right )^{\frac {3}{2}} \left (x -\sqrt {-\frac {b}{a}}\right )}-\frac {-4 b a -c^{2}+4 k}{16 a^{2} \left (-\frac {b}{a}\right )^{\frac {3}{2}} \left (x +\sqrt {-\frac {b}{a}}\right )} \] For the pole at \(x=\frac {\sqrt {-b a}}{a}\) let \(b\) be the coefficient of \(\frac {1}{ \left (x -\frac {\sqrt {-b a}}{a}\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b=0\). Hence \begin {align*} E_c &= \{2, 2+2\sqrt {1+4 b}, 2-2\sqrt {1+4 b} \} \\ &= \{0, 2, 4\} \end {align*}
For the pole at \(x=-\frac {\sqrt {-b a}}{a}\) let \(b\) be the coefficient of \(\frac {1}{ \left (x +\frac {\sqrt {-b a}}{a}\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b=0\). Hence \begin {align*} E_c &= \{2, 2+2\sqrt {1+4 b}, 2-2\sqrt {1+4 b} \} \\ &= \{0, 2, 4\} \end {align*}
Now since the order of \(r\) at \(\infty \) is \(4 > 2\) then \begin {align*} E_\infty = \{0,2,4\} \end {align*}
The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) for case 2 of Kovacic algorithm.
| pole \(c\) location | pole order | \(E_c\) |
| \(\frac {\sqrt {-b a}}{a}\) | \(2\) | \(\{0, 2, 4\}\) |
| \(-\frac {\sqrt {-b a}}{a}\) | \(2\) | \(\{0, 2, 4\}\) |
| Order of \(r\) at \(\infty \) | \(E_\infty \) |
| \(4\) | \(\{0, 2, 4\}\) |
Using the family \(\{e_1,e_2,\dots ,e_\infty \}\) given by \[ e_1=2,\hspace {3pt} e_2=2,\hspace {3pt} e_\infty =4 \] Gives a non negative integer \(d\) (the degree of the polynomial \(p(x)\)), which is generated using \begin {align*} d &= \frac {1}{2} \left ( e_\infty - \sum _{c \in \Gamma } e_c \right )\\ &= \frac {1}{2} \left ( 4 - \left (2+\left (2\right )\right )\right )\\ &= 0 \end {align*}
We now form the following rational function \begin {align*} \theta &= \frac {1}{2} \sum _{c \in \Gamma } \frac {e_c}{x-c} \\ &= \frac {1}{2} \left (\frac {2}{\left (x-\left (\frac {\sqrt {-b a}}{a}\right )\right )}+\frac {2}{\left (x-\left (-\frac {\sqrt {-b a}}{a}\right )\right )}\right ) \\ &= \frac {1}{x -\frac {\sqrt {-b a}}{a}}+\frac {1}{x +\frac {\sqrt {-b a}}{a}} \end {align*}
Now we search for a monic polynomial \(p(x)\) of degree \(d=0\) such that \[ p'''+3 \theta p'' + \left (3 \theta ^2 + 3 \theta ' - 4 r\right )p' + \left (\theta '' + 3 \theta \theta ' + \theta ^3 - 4 r \theta - 2 r' \right ) p = 0 \tag {1A} \] Since \(d=0\), then letting \[ p = 1\tag {2A} \] Substituting \(p\) and \(\theta \) into Eq. (1A) gives \[ 0 = 0 \] And solving for \(p\) gives \[ p = 1 \] Now that \(p(x)\) is found let \begin {align*} \phi &= \theta + \frac {p'}{p}\\ &= \frac {1}{x -\frac {\sqrt {-b a}}{a}}+\frac {1}{x +\frac {\sqrt {-b a}}{a}} \end {align*}
Let \(\omega \) be the solution of \begin {align*} \omega ^2 - \phi \omega + \left ( \frac {1}{2} \phi ' + \frac {1}{2} \phi ^2 - r \right ) &= 0 \end {align*}
Substituting the values for \(\phi \) and \(r\) into the above equation gives \[ w^{2}-\left (\frac {1}{x -\frac {\sqrt {-b a}}{a}}+\frac {1}{x +\frac {\sqrt {-b a}}{a}}\right ) w +\frac {4 a^{2} x^{2}-c^{2}+4 k}{4 \left (a \,x^{2}+b \right )^{2}} = 0 \] Solving for \(\omega \) gives \begin {align*} \omega &= \frac {2 a x +\sqrt {c^{2}-4 k}}{2 a \,x^{2}+2 b} \end {align*}
Therefore the first solution to the ode \(z'' = r z\) is \begin {align*} z_1(x) &= e^{ \int \omega \,dx} \\ &= {\mathrm e}^{\int \frac {2 a x +\sqrt {c^{2}-4 k}}{2 a \,x^{2}+2 b}d x}\\ &= \sqrt {a \,x^{2}+b}\, {\mathrm e}^{\frac {\sqrt {c^{2}-4 k}\, \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{2 \sqrt {b a}}} \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 {\left (2 a x +c \right ) \left (a \,x^{2}+b \right )}{\left (a \,x^{2}+b \right )^{2}} \,dx} \\ &= z_1 e^{-\frac {\ln \left (a \,x^{2}+b \right )}{2}-\frac {c \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{2 \sqrt {b a}}} \\ &= z_1 \left (\frac {{\mathrm e}^{-\frac {c \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{2 \sqrt {b a}}}}{\sqrt {a \,x^{2}+b}}\right ) \\ \end{align*} Which simplifies to \[ y_1 = {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {b a}}\right ) \left (-c +\sqrt {c^{2}-4 k}\right )}{2 \sqrt {b a}}} \] 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 {\left (2 a x +c \right ) \left (a \,x^{2}+b \right )}{\left (a \,x^{2}+b \right )^{2}} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{-\ln \left (a \,x^{2}+b \right )-\frac {c \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{\sqrt {b a}}}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (-\frac {{\mathrm e}^{-\frac {\sqrt {c^{2}-4 k}\, \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{\sqrt {b a}}}}{\sqrt {c^{2}-4 k}}\right ) \\ \end{align*} Therefore the solution is
\begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left ({\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {b a}}\right ) \left (-c +\sqrt {c^{2}-4 k}\right )}{2 \sqrt {b a}}}\right ) + c_{2} \left ({\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {b a}}\right ) \left (-c +\sqrt {c^{2}-4 k}\right )}{2 \sqrt {b a}}}\left (-\frac {{\mathrm e}^{-\frac {\sqrt {c^{2}-4 k}\, \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{\sqrt {b a}}}}{\sqrt {c^{2}-4 k}}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {b a}}\right ) \left (-c +\sqrt {c^{2}-4 k}\right )}{2 \sqrt {b a}}}-\frac {c_{2} {\mathrm e}^{-\frac {\arctan \left (\frac {x a}{\sqrt {b a}}\right ) \left (c +\sqrt {c^{2}-4 k}\right )}{2 \sqrt {b a}}}}{\sqrt {c^{2}-4 k}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {b a}}\right ) \left (-c +\sqrt {c^{2}-4 k}\right )}{2 \sqrt {b a}}}-\frac {c_{2} {\mathrm e}^{-\frac {\arctan \left (\frac {x a}{\sqrt {b a}}\right ) \left (c +\sqrt {c^{2}-4 k}\right )}{2 \sqrt {b a}}}}{\sqrt {c^{2}-4 k}} \] 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 Group is reducible or imprimitive <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 162
dsolve((a*x^2+b)^2*diff(y(x),x$2)+(2*a*x+c)*(a*x^2+b)*diff(y(x),x)+k*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \left (\frac {-i \sqrt {a b}+a x}{i \sqrt {a b}+a x}\right )^{\frac {i \sqrt {a b}\, c \sqrt {-a b}+a^{2} \sqrt {\frac {c^{2}-4 k}{a^{2}}}\, b}{4 a b \sqrt {-a b}}}+c_{2} \left (\frac {-i \sqrt {a b}+a x}{i \sqrt {a b}+a x}\right )^{\frac {i \sqrt {a b}\, c \sqrt {-a b}-a^{2} \sqrt {\frac {c^{2}-4 k}{a^{2}}}\, b}{4 a b \sqrt {-a b}}} \]
✓ Solution by Mathematica
Time used: 2.188 (sec). Leaf size: 91
DSolve[(a*x^2+b)^2*y''[x]+(2*a*x+c)*(a*x^2+b)*y'[x]+k*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{-\frac {\left (\sqrt {c^2-4 k}+c\right ) \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{2 \sqrt {a} \sqrt {b}}} \left (c_2 e^{\frac {\sqrt {c^2-4 k} \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b}}}+c_1\right ) \]