Internal problem ID [9813]
Internal file name [OUTPUT/8756_Monday_June_06_2022_05_24_12_AM_93538667/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1487.
ODE order: 3.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_missing_y"
Maple gives the following as the ode type
[[_3rd_order, _missing_y]]
\[ \boxed {\left (2 x -1\right ) y^{\prime \prime \prime }+\left (4+x \right ) y^{\prime \prime }+2 y^{\prime }=0} \] Since \(y\) is missing from the ode then we can use the substitution \(y^{\prime } = v \left (x \right )\) to reduce the order by one. The ODE becomes \begin {align*} \left (2 x -1\right ) v^{\prime \prime }\left (x \right )+\left (4+x \right ) v^{\prime }\left (x \right )+2 v \left (x \right ) = 0 \end {align*}
Writing the ode as \begin {align*} \left (2 x -1\right ) v^{\prime \prime }\left (x \right )+\left (4+x \right ) v^{\prime }\left (x \right )+2 v \left (x \right ) &= 0 \tag {1} \\ A v^{\prime \prime }\left (x \right ) + B v^{\prime }\left (x \right ) + C v \left (x \right ) &= 0 \tag {2} \end {align*}
Comparing (1) and (2) shows that \begin {align*} A &= 2 x -1 \\ B &= 4+x\tag {3} \\ C &= 2 \end {align*}
Applying the Liouville transformation on the dependent variable gives \begin {align*} z(x) &= v \left (x \right ) 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 {x^{2}-8 x +6}{4 \left (2 x -1\right )^{2}}\tag {6} \end {align*}
Comparing the above to (5) shows that \begin {align*} s &= x^{2}-8 x +6\\ t &= 4 \left (2 x -1\right )^{2} \end {align*}
Therefore eq. (4) becomes \begin {align*} z''(x) &= \left ( \frac {x^{2}-8 x +6}{4 \left (2 x -1\right )^{2}}\right ) z(x)\tag {7} \end {align*}
Equation (7) is now solved. After finding \(z(x)\) then \(v \left (x \right )\) is found using the inverse transformation \begin {align*} v \left (x \right ) &= 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) \\ &= 2 - 2 \\ &= 0 \end {align*}
The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=4 \left (2 x -1\right )^{2}\). There is a pole at \(x={\frac {1}{2}}\) of order \(2\). Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(0\) 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. Therefore \begin {align*} L &= [1, 2] \end {align*}
Attempting to find a solution using case \(n=1\).
Looking at poles of order 2. The partial fractions decomposition of \(r\) is \[ r = \frac {1}{16}+\frac {9}{64 \left (x -\frac {1}{2}\right )^{2}}-\frac {7}{16 \left (x -\frac {1}{2}\right )} \] For the pole at \(x={\frac {1}{2}}\) let \(b\) be the coefficient of \(\frac {1}{ \left (x -\frac {1}{2}\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b={\frac {9}{64}}\). Hence \begin {alignat*} {2} [\sqrt r]_c &= 0 \\ \alpha _c^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= {\frac {9}{8}}\\ \alpha _c^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= -{\frac {1}{8}} \end {alignat*}
Since the order of \(r\) at \(\infty \) is \(O_r(\infty ) = 0\) then \begin {alignat*} {3} v &= \frac {-O_r(\infty )}{2} &&= \frac {0}{2} &&= 0 \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}^{0} a_i x^i \tag {8} \end {align*}
Let \(a\) be the coefficient of \(x^v=x^0\) in the above sum. The Laurent series of \(\sqrt r\) at \(\infty \) is \[ \sqrt r \approx \frac {1}{4}-\frac {7}{8 x}-\frac {27}{16 x^{2}}-\frac {187}{32 x^{3}}-\frac {1667}{64 x^{4}}-\frac {16707}{128 x^{5}}-\frac {179427}{256 x^{6}}-\frac {2018787}{512 x^{7}} + \dots \tag {9} \] Comparing Eq. (9) with Eq. (8) shows that \[ a = {\frac {1}{4}} \] From Eq. (9) the sum up to \(v=0\) gives \begin {align*} [\sqrt r]_\infty &= \sum _{i=0}^{0} a_i x^i \\ &= {\frac {1}{4}} \tag {10} \end {align*}
Now we need to find \(b\), where \(b\) be the coefficient of \(x^{v-1} = x^{-1}=\frac {1}{x}\) 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 \[ \left ( [\sqrt r]_\infty \right )^2 = {\frac {1}{16}} \] This shows that the coefficient of \(\frac {1}{x}\) in the above is \(0\). Now we need to find the coefficient of \(\frac {1}{x}\) in \(r\). How this is done depends on if \(v=0\) or not. Since \(v=0\) then starting from \(r=\frac {s}{t}\) and doing long division in the form \[ r = Q + \frac {R}{t} \] Where \(Q\) is the quotient and \(R\) is the remainder. Then the coefficient of \(\frac {1}{x}\) in \(r\) will be the coefficient in \(R\) of the term in \(x\) of degree of \(t\) minus one, divided by the leading coefficient in \(t\). Doing long division gives \begin {align*} r &= \frac {s}{t} \\ &= \frac {x^{2}-8 x +6}{16 x^{2}-16 x +4} \\ &= Q + \frac {R}{16 x^{2}-16 x +4}\\ &= \left ({\frac {1}{16}}\right ) + \left ( \frac {-7 x +\frac {23}{4}}{16 x^{2}-16 x +4}\right ) \\ &= \frac {1}{16}+\frac {-7 x +\frac {23}{4}}{16 x^{2}-16 x +4} \end {align*}
Since the degree of \(t\) is \(2\), then we see that the coefficient of the term \(x\) in the remainder \(R\) is \(-7\). Dividing this by leading coefficient in \(t\) which is \(16\) gives \(-{\frac {7}{16}}\). Now \(b\) can be found. \begin {align*} b &= \left (-{\frac {7}{16}}\right )-\left (0\right )\\ &= -{\frac {7}{16}} \end {align*}
Hence \begin {alignat*} {3} [\sqrt r]_\infty &= {\frac {1}{4}}\\ \alpha _{\infty }^{+} &= \frac {1}{2} \left ( \frac {b}{a} - v \right ) &&= \frac {1}{2} \left ( \frac {-{\frac {7}{16}}}{{\frac {1}{4}}} - 0 \right ) &&= -{\frac {7}{8}}\\ \alpha _{\infty }^{-} &= \frac {1}{2} \left ( -\frac {b}{a} - v \right ) &&= \frac {1}{2} \left ( -\frac {-{\frac {7}{16}}}{{\frac {1}{4}}} - 0 \right ) &&= {\frac {7}{8}} \end {alignat*}
The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) where \(r\) is \[ r=\frac {x^{2}-8 x +6}{4 \left (2 x -1\right )^{2}} \]
| pole \(c\) location | pole order | \([\sqrt r]_c\) | \(\alpha _c^{+}\) | \(\alpha _c^{-}\) |
| \(\frac {1}{2}\) | \(2\) | \(0\) | \(\frac {9}{8}\) | \(-{\frac {1}{8}}\) |
| Order of \(r\) at \(\infty \) | \([\sqrt r]_\infty \) | \(\alpha _\infty ^{+}\) | \(\alpha _\infty ^{-}\) |
| \(0\) | \(\frac {1}{4}\) | \(-{\frac {7}{8}}\) | \(\frac {7}{8}\) |
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 \(\alpha _\infty ^{-} = {\frac {7}{8}}\) then \begin {align*} d &= \alpha _\infty ^{-} - \left ( \alpha _{c_1}^{-} \right ) \\ &= {\frac {7}{8}} - \left ( -{\frac {1}{8}} \right ) \\ &= 1 \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*}
The above gives \begin {align*} \omega &= \left ( (-)[\sqrt r]_{c_1} + \frac { \alpha _{c_1}^{-} }{x- c_1}\right ) + (-) [\sqrt r]_\infty \\ &= -\frac {1}{8 \left (x -\frac {1}{2}\right )} + (-) \left ( {\frac {1}{4}} \right ) \\ &= -\frac {1}{8 \left (x -\frac {1}{2}\right )}-\frac {1}{4}\\ &= -\frac {x}{4 x -2} \end {align*}
Now that \(\omega \) is determined, the next step is find a corresponding minimal polynomial \(p(x)\) of degree \(d=1\) 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) &= x +a_{0}\tag {2A} \end {align*}
Substituting the above in eq. (1A) gives \begin {align*} \left (0\right ) + 2 \left (-\frac {1}{8 \left (x -\frac {1}{2}\right )}-\frac {1}{4}\right ) \left (1\right ) + \left ( \left (\frac {1}{8 \left (x -\frac {1}{2}\right )^{2}}\right ) + \left (-\frac {1}{8 \left (x -\frac {1}{2}\right )}-\frac {1}{4}\right )^2 - \left (\frac {x^{2}-8 x +6}{4 \left (2 x -1\right )^{2}}\right ) \right ) &= 0\\ \frac {a_{0}}{2 x -1} = 0 \end {align*}
Solving for the coefficients \(a_i\) in the above using method of undetermined coefficients gives \[ \{a_{0} = 0\} \] Substituting these coefficients in \(p(x)\) in eq. (2A) results in \begin {align*} p(x) &= x \end {align*}
Therefore the first solution to the ode \(z'' = r z\) is \begin {align*} z_1(x) &= p e^{ \int \omega \,dx}\\ & = \left (x\right ) {\mathrm e}^{\int \left (-\frac {1}{8 \left (x -\frac {1}{2}\right )}-\frac {1}{4}\right )d x}\\ & = \left (x\right ) {\mathrm e}^{-\frac {\ln \left (x -\frac {1}{2}\right )}{8}-\frac {x}{4}}\\ & = \frac {x 2^{\frac {1}{8}} {\mathrm e}^{-\frac {x}{4}}}{\left (2 x -1\right )^{\frac {1}{8}}} \end {align*}
The first solution to the original ode in \(v \left (x \right )\) is found from \begin{align*} v_1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dx} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {4+x}{2 x -1} \,dx} \\ &= z_1 e^{-\frac {x}{4}-\frac {9 \ln \left (2 x -1\right )}{8}} \\ &= z_1 \left (\frac {{\mathrm e}^{-\frac {x}{4}}}{\left (2 x -1\right )^{\frac {9}{8}}}\right ) \\ \end{align*} Which simplifies to \[ v_1 = \frac {x 2^{\frac {1}{8}} {\mathrm e}^{-\frac {x}{2}}}{\left (2 x -1\right )^{\frac {5}{4}}} \] The second solution \(v_2\) to the original ode is found using reduction of order \[ v_2 = v_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{v_1^2} \,dx \] Substituting gives \begin{align*} v_2 &= v_1 \int \frac { e^{\int -\frac {4+x}{2 x -1} \,dx}}{\left (v_1\right )^2} \,dx \\ &= v_1 \int \frac { e^{-\frac {x}{2}-\frac {9 \ln \left (2 x -1\right )}{4}}}{\left (v_1\right )^2} \,dx \\ &= v_1 \left (\int \frac {2^{\frac {3}{4}} {\mathrm e}^{\frac {x}{2}} \left (2 x -1\right )^{\frac {1}{4}}}{2 x^{2}}d x\right ) \\ \end{align*} Therefore the solution is
\begin{align*} v \left (x \right ) &= c_{1} v_1 + c_{2} v_2 \\ &= c_{1} \left (\frac {x 2^{\frac {1}{8}} {\mathrm e}^{-\frac {x}{2}}}{\left (2 x -1\right )^{\frac {5}{4}}}\right ) + c_{2} \left (\frac {x 2^{\frac {1}{8}} {\mathrm e}^{-\frac {x}{2}}}{\left (2 x -1\right )^{\frac {5}{4}}}\left (\int \frac {2^{\frac {3}{4}} {\mathrm e}^{\frac {x}{2}} \left (2 x -1\right )^{\frac {1}{4}}}{2 x^{2}}d x\right )\right ) \\ \end{align*} But since \(y^{\prime } = v \left (x \right )\) then we now need to solve the ode \(y^{\prime } = \frac {c_{1} x 2^{\frac {1}{8}} {\mathrm e}^{-\frac {x}{2}}}{\left (2 x -1\right )^{\frac {5}{4}}}+\frac {c_{2} x 2^{\frac {7}{8}} {\mathrm e}^{-\frac {x}{2}} \left (\int \frac {{\mathrm e}^{\frac {x}{2}} \left (2 x -1\right )^{\frac {1}{4}}}{x^{2}}d x \right )}{2 \left (2 x -1\right )^{\frac {5}{4}}}\). Writing the ode as \begin {align*} y^{\prime }&=\frac {x 2^{\frac {1}{8}} {\mathrm e}^{-\frac {x}{2}} \left (c_{2} 2^{\frac {3}{4}} \left (\int \frac {{\mathrm e}^{\frac {x}{2}} \left (2 x -1\right )^{\frac {1}{4}}}{x^{2}}d x \right )+2 c_{1} \right )}{2 \left (2 x -1\right )^{\frac {5}{4}}}\\ y^{\prime }&= \omega \left ( x,y\right ) \end {align*}
The condition of Lie symmetry is the linearized PDE given by \begin {align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end {align*}
The type of this ode is not in the lookup table. To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives \begin{align*} \tag{1E} \xi &= x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x b_{2}+y b_{3}+b_{1} \\ \end{align*} Where the unknown coefficients are \[ \{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\} \] Substituting equations (1E,2E) and \(\omega \) into (A) gives \begin{equation} \tag{5E} b_{2}+\frac {x 2^{\frac {1}{8}} {\mathrm e}^{-\frac {x}{2}} \left (c_{2} 2^{\frac {3}{4}} \left (\int \frac {{\mathrm e}^{\frac {x}{2}} \left (2 x -1\right )^{\frac {1}{4}}}{x^{2}}d x \right )+2 c_{1} \right ) \left (b_{3}-a_{2}\right )}{2 \left (2 x -1\right )^{\frac {5}{4}}}-\frac {x^{2} 2^{\frac {1}{4}} {\mathrm e}^{-x} \left (c_{2} 2^{\frac {3}{4}} \left (\int \frac {{\mathrm e}^{\frac {x}{2}} \left (2 x -1\right )^{\frac {1}{4}}}{x^{2}}d x \right )+2 c_{1} \right )^{2} a_{3}}{4 \left (2 x -1\right )^{\frac {5}{2}}}-\left (\frac {2^{\frac {1}{8}} {\mathrm e}^{-\frac {x}{2}} \left (c_{2} 2^{\frac {3}{4}} \left (\int \frac {{\mathrm e}^{\frac {x}{2}} \left (2 x -1\right )^{\frac {1}{4}}}{x^{2}}d x \right )+2 c_{1} \right )}{2 \left (2 x -1\right )^{\frac {5}{4}}}-\frac {x 2^{\frac {1}{8}} {\mathrm e}^{-\frac {x}{2}} \left (c_{2} 2^{\frac {3}{4}} \left (\int \frac {{\mathrm e}^{\frac {x}{2}} \left (2 x -1\right )^{\frac {1}{4}}}{x^{2}}d x \right )+2 c_{1} \right )}{4 \left (2 x -1\right )^{\frac {5}{4}}}+\frac {2^{\frac {7}{8}} {\mathrm e}^{-\frac {x}{2}} c_{2} {\mathrm e}^{\frac {x}{2}}}{2 x \left (2 x -1\right )}-\frac {5 x 2^{\frac {1}{8}} {\mathrm e}^{-\frac {x}{2}} \left (c_{2} 2^{\frac {3}{4}} \left (\int \frac {{\mathrm e}^{\frac {x}{2}} \left (2 x -1\right )^{\frac {1}{4}}}{x^{2}}d x \right )+2 c_{1} \right )}{4 \left (2 x -1\right )^{\frac {9}{4}}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right ) = 0 \end{equation} Putting the above in normal form gives \[ \text {Expression too large to display} \] Setting the numerator to zero gives \begin{equation} \tag{6E} \text {Expression too large to display} \end{equation} Simplifying the above gives \begin{equation} \tag{6E} \text {Expression too large to display} \end{equation} Since the PDE has radicals, simplifying gives \[ \text {Expression too large to display} \] Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \left \{x, y, \sqrt {2 x -1}, \left (2 x -1\right )^{\frac {1}{4}}, \left (2 x -1\right )^{\frac {3}{4}}, \int \frac {{\mathrm e}^{\frac {x}{2}} \left (2 x -1\right )^{\frac {1}{4}}}{x^{2}}d x, {\mathrm e}^{-x}, {\mathrm e}^{-\frac {x}{2}}, {\mathrm e}^{\frac {x}{2}}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, \sqrt {2 x -1} = v_{3}, \left (2 x -1\right )^{\frac {1}{4}} = v_{4}, \left (2 x -1\right )^{\frac {3}{4}} = v_{5}, \int \frac {{\mathrm e}^{\frac {x}{2}} \left (2 x -1\right )^{\frac {1}{4}}}{x^{2}}d x = v_{6}, {\mathrm e}^{-x} = v_{7}, {\mathrm e}^{-\frac {x}{2}} = v_{8}, {\mathrm e}^{\frac {x}{2}} = v_{9}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} \text {Expression too large to display} \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}, v_{6}, v_{7}, v_{8}, v_{9}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} \text {Expression too large to display} \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} 128 b_{2}&=0\\ -32 \,2^{\frac {1}{4}} c_{1}^{2} a_{3}&=0\\ -24 \,2^{\frac {1}{4}} c_{1}^{2} a_{3}&=0\\ 4 \,2^{\frac {1}{4}} c_{1}^{2} a_{3}&=0\\ 48 \,2^{\frac {1}{4}} c_{1}^{2} a_{3}&=0\\ -4 \,2^{\frac {1}{8}} c_{1} a_{1}&=0\\ 32 \,2^{\frac {1}{8}} c_{1} a_{2}&=0\\ -52 \,2^{\frac {1}{8}} c_{1} a_{3}&=0\\ -48 \,2^{\frac {1}{8}} c_{1} a_{3}&=0\\ -4 \,2^{\frac {1}{8}} c_{1} a_{3}&=0\\ 24 \,2^{\frac {1}{8}} c_{1} a_{3}&=0\\ 32 \,2^{\frac {1}{8}} c_{1} a_{3}&=0\\ 56 \,2^{\frac {1}{8}} c_{1} a_{3}&=0\\ -16 \,2^{\frac {3}{4}} c_{2}^{2} a_{3}&=0\\ -12 \,2^{\frac {3}{4}} c_{2}^{2} a_{3}&=0\\ 2 \,2^{\frac {3}{4}} c_{2}^{2} a_{3}&=0\\ 24 \,2^{\frac {3}{4}} c_{2}^{2} a_{3}&=0\\ -2 \,2^{\frac {7}{8}} c_{2} a_{1}&=0\\ 16 \,2^{\frac {7}{8}} c_{2} a_{2}&=0\\ -48 \,2^{\frac {7}{8}} c_{2} a_{3}&=0\\ -32 \,2^{\frac {7}{8}} c_{2} a_{3}&=0\\ -26 \,2^{\frac {7}{8}} c_{2} a_{3}&=0\\ -24 \,2^{\frac {7}{8}} c_{2} a_{3}&=0\\ -2 \,2^{\frac {7}{8}} c_{2} a_{3}&=0\\ 12 \,2^{\frac {7}{8}} c_{2} a_{3}&=0\\ 16 \,2^{\frac {7}{8}} c_{2} a_{3}&=0\\ 28 \,2^{\frac {7}{8}} c_{2} a_{3}&=0\\ 64 \,2^{\frac {7}{8}} c_{2} a_{3}&=0\\ -64 c_{1} c_{2} a_{3}&=0\\ -48 c_{1} c_{2} a_{3}&=0\\ 8 c_{1} c_{2} a_{3}&=0\\ 96 c_{1} c_{2} a_{3}&=0\\ -32 \,2^{\frac {7}{8}} c_{2} a_{2}-320 b_{2}&=0\\ -52 \,2^{\frac {1}{8}} c_{1} a_{1}+56 \,2^{\frac {1}{8}} c_{1} a_{2}-32 \,2^{\frac {1}{8}} c_{1} b_{3}&=0\\ -48 \,2^{\frac {1}{8}} c_{1} a_{1}+184 \,2^{\frac {1}{8}} c_{1} a_{2}-128 \,2^{\frac {1}{8}} c_{1} b_{3}&=0\\ 24 \,2^{\frac {1}{8}} c_{1} a_{1}-8 \,2^{\frac {1}{8}} c_{1} a_{2}+4 \,2^{\frac {1}{8}} c_{1} b_{3}&=0\\ 32 \,2^{\frac {1}{8}} c_{1} a_{1}-112 \,2^{\frac {1}{8}} c_{1} a_{2}+64 \,2^{\frac {1}{8}} c_{1} b_{3}&=0\\ 56 \,2^{\frac {1}{8}} c_{1} a_{1}-148 \,2^{\frac {1}{8}} c_{1} a_{2}+96 \,2^{\frac {1}{8}} c_{1} b_{3}&=0\\ -48 \,2^{\frac {7}{8}} c_{2} a_{1}+16 \,2^{\frac {7}{8}} c_{2} a_{2}+40 b_{2}&=0\\ -32 \,2^{\frac {7}{8}} c_{2} a_{1}+64 \,2^{\frac {7}{8}} c_{2} a_{2}+320 b_{2}&=0\\ -26 \,2^{\frac {7}{8}} c_{2} a_{1}+28 \,2^{\frac {7}{8}} c_{2} a_{2}-16 \,2^{\frac {7}{8}} c_{2} b_{3}&=0\\ -24 \,2^{\frac {7}{8}} c_{2} a_{1}+92 \,2^{\frac {7}{8}} c_{2} a_{2}-64 \,2^{\frac {7}{8}} c_{2} b_{3}&=0\\ 12 \,2^{\frac {7}{8}} c_{2} a_{1}-4 \,2^{\frac {7}{8}} c_{2} a_{2}+2 \,2^{\frac {7}{8}} c_{2} b_{3}&=0\\ 16 \,2^{\frac {7}{8}} c_{2} a_{1}-56 \,2^{\frac {7}{8}} c_{2} a_{2}+32 \,2^{\frac {7}{8}} c_{2} b_{3}&=0\\ 16 \,2^{\frac {7}{8}} c_{2} a_{1}-2 \,2^{\frac {7}{8}} c_{2} a_{2}-4 b_{2}&=0\\ 28 \,2^{\frac {7}{8}} c_{2} a_{1}-74 \,2^{\frac {7}{8}} c_{2} a_{2}+48 \,2^{\frac {7}{8}} c_{2} b_{3}&=0\\ 64 \,2^{\frac {7}{8}} c_{2} a_{1}-48 \,2^{\frac {7}{8}} c_{2} a_{2}-160 b_{2}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=0\\ a_{3}&=0\\ b_{1}&=b_{1}\\ b_{2}&=0\\ b_{3}&=0 \end {align*}
Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives \begin{align*} \xi &= 0 \\ \eta &= 1 \\ \end{align*} The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is \begin {align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end {align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Since \(\xi =0\) then in this special case \begin {align*} R = x \end {align*}
\(S\) is found from \begin {align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{1}} dy \end {align*}
Which results in \begin {align*} S&= y \end {align*}
Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by evaluating \begin {align*} \frac {dS}{dR} &= \frac { S_{x} + \omega (x,y) S_{y} }{ R_{x} + \omega (x,y) R_{y} }\tag {2} \end {align*}
Where in the above \(R_{x},R_{y},S_{x},S_{y}\) are all partial derivatives and \(\omega (x,y)\) is the right hand side of the original ode given by \begin {align*} \omega (x,y) &= \frac {x 2^{\frac {1}{8}} {\mathrm e}^{-\frac {x}{2}} \left (c_{2} 2^{\frac {3}{4}} \left (\int \frac {{\mathrm e}^{\frac {x}{2}} \left (2 x -1\right )^{\frac {1}{4}}}{x^{2}}d x \right )+2 c_{1} \right )}{2 \left (2 x -1\right )^{\frac {5}{4}}} \end {align*}
Evaluating all the partial derivatives gives \begin {align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= 0\\ S_{y} &= 1 \end {align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates. \begin {align*} \frac {dS}{dR} &= \frac {x 2^{\frac {1}{8}} {\mathrm e}^{-\frac {x}{2}} \left (c_{2} 2^{\frac {3}{4}} \left (\int \frac {{\mathrm e}^{\frac {x}{2}} \left (2 x -1\right )^{\frac {1}{4}}}{x^{2}}d x \right )+2 c_{1} \right )}{2 \left (2 x -1\right )^{\frac {5}{4}}}\tag {2A} \end {align*}
We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,y\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives \begin {align*} \frac {dS}{dR} &= \frac {R 2^{\frac {1}{8}} {\mathrm e}^{-\frac {R}{2}} \left (c_{2} 2^{\frac {3}{4}} \left (\int \frac {{\mathrm e}^{\frac {R}{2}} \left (2 R -1\right )^{\frac {1}{4}}}{R^{2}}d R \right )+2 c_{1} \right )}{2 \left (2 R -1\right )^{\frac {5}{4}}} \end {align*}
The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an ode, no matter how complicated it is, to one that can be solved by integration when the ode is in the canonical coordiates \(R,S\). Integrating the above gives \begin {align*} S \left (R \right ) = \int \frac {R 2^{\frac {1}{8}} {\mathrm e}^{-\frac {R}{2}} \left (c_{2} 2^{\frac {3}{4}} \left (\int \frac {{\mathrm e}^{\frac {R}{2}} \left (2 R -1\right )^{\frac {1}{4}}}{R^{2}}d R \right )+2 c_{1} \right )}{2 \left (2 R -1\right )^{\frac {5}{4}}}d R +c_{3}\tag {4} \end {align*}
To complete the solution, we just need to transform (4) back to \(x,y\) coordinates. This results in \begin {align*} y = \int \frac {x 2^{\frac {1}{8}} {\mathrm e}^{-\frac {x}{2}} \left (c_{2} 2^{\frac {3}{4}} \left (\int \frac {{\mathrm e}^{\frac {x}{2}} \left (2 x -1\right )^{\frac {1}{4}}}{x^{2}}d x \right )+2 c_{1} \right )}{2 \left (2 x -1\right )^{\frac {5}{4}}}d x +c_{3} \end {align*}
Which simplifies to \begin {align*} y = \int \frac {x 2^{\frac {1}{8}} {\mathrm e}^{-\frac {x}{2}} \left (c_{2} 2^{\frac {3}{4}} \left (\int \frac {{\mathrm e}^{\frac {x}{2}} \left (2 x -1\right )^{\frac {1}{4}}}{x^{2}}d x \right )+2 c_{1} \right )}{2 \left (2 x -1\right )^{\frac {5}{4}}}d x +c_{3} \end {align*}
Which gives \begin {align*} y = \int \frac {x 2^{\frac {1}{8}} {\mathrm e}^{-\frac {x}{2}} \left (c_{2} 2^{\frac {3}{4}} \left (\int \frac {{\mathrm e}^{\frac {x}{2}} \left (2 x -1\right )^{\frac {1}{4}}}{x^{2}}d x \right )+2 c_{1} \right )}{2 \left (2 x -1\right )^{\frac {5}{4}}}d x +c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \int \frac {x 2^{\frac {1}{8}} {\mathrm e}^{-\frac {x}{2}} \left (c_{2} 2^{\frac {3}{4}} \left (\int \frac {{\mathrm e}^{\frac {x}{2}} \left (2 x -1\right )^{\frac {1}{4}}}{x^{2}}d x \right )+2 c_{1} \right )}{2 \left (2 x -1\right )^{\frac {5}{4}}}d x +c_{3} \\ \end{align*}
Verification of solutions
\[ y = \int \frac {x 2^{\frac {1}{8}} {\mathrm e}^{-\frac {x}{2}} \left (c_{2} 2^{\frac {3}{4}} \left (\int \frac {{\mathrm e}^{\frac {x}{2}} \left (2 x -1\right )^{\frac {1}{4}}}{x^{2}}d x \right )+2 c_{1} \right )}{2 \left (2 x -1\right )^{\frac {5}{4}}}d x +c_{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (2 x -1\right ) y^{\prime \prime \prime }+\left (4+x \right ) y^{\prime \prime }+2 y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\prime \prime \prime } \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=\frac {1}{2}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {4+x}{2 x -1}, P_{3}\left (x \right )=\frac {2}{2 x -1}, P_{4}\left (x \right )=0\right ] \\ {} & \circ & \left (x -\frac {1}{2}\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {1}{2} \\ {} & {} & \left (\left (x -\frac {1}{2}\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {1}{2}}}}=\frac {9}{4} \\ {} & \circ & \left (x -\frac {1}{2}\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {1}{2} \\ {} & {} & \left (\left (x -\frac {1}{2}\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {1}{2}}}}=0 \\ {} & \circ & \left (x -\frac {1}{2}\right )^{3}\cdot P_{4}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {1}{2} \\ {} & {} & \left (\left (x -\frac {1}{2}\right )^{3}\cdot P_{4}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {1}{2}}}}=0 \\ {} & \circ & x =\frac {1}{2}\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=\frac {1}{2}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=\frac {1}{2} \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u +\frac {1}{2}\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & 2 u \left (\frac {d^{3}}{d u^{3}}y \left (u \right )\right )+\left (\frac {9}{2}+u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+2 \frac {d}{d u}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} \frac {d}{d u}y \left (u \right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d}{d u}y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \frac {d}{d u}y \left (u \right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}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^{2}}{d u^{2}}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} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u \cdot \left (\frac {d^{3}}{d u^{3}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & u \cdot \left (\frac {d^{3}}{d u^{3}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) \left (k +r -2\right ) u^{k +r -2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & u \cdot \left (\frac {d^{3}}{d u^{3}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2}{\sum }}a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \left (k +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & \frac {a_{0} r \left (1+4 r \right ) \left (-1+r \right ) u^{-2+r}}{2}+\left (\moverset {\infty }{\munderset {k =-1}{\sum }}\left (\frac {a_{k +2} \left (k +2+r \right ) \left (4 k +9+4 r \right ) \left (k +1+r \right )}{2}+a_{k +1} \left (k +1+r \right ) \left (k +2+r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \frac {r \left (1+4 r \right ) \left (-1+r \right )}{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1, -\frac {1}{4}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 2 \left (k +2+r \right ) \left (k +1+r \right ) \left (\left (k +r +\frac {9}{4}\right ) a_{k +2}+\frac {a_{k +1}}{2}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {2 a_{k +1}}{4 k +9+4 r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {2 a_{k +1}}{4 k +9} \\ \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 +2}=-\frac {2 a_{k +1}}{4 k +9}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x -\frac {1}{2}\right )^{k}, a_{k +2}=-\frac {2 a_{k +1}}{4 k +9}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +2}=-\frac {2 a_{k +1}}{4 k +13} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +1}, a_{k +2}=-\frac {2 a_{k +1}}{4 k +13}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x -\frac {1}{2}\right )^{k +1}, a_{k +2}=-\frac {2 a_{k +1}}{4 k +13}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{4} \\ {} & {} & a_{k +2}=-\frac {2 a_{k +1}}{4 k +8} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {1}{4} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -\frac {1}{4}}, a_{k +2}=-\frac {2 a_{k +1}}{4 k +8}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x -\frac {1}{2}\right )^{k -\frac {1}{4}}, a_{k +2}=-\frac {2 a_{k +1}}{4 k +8}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x -\frac {1}{2}\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x -\frac {1}{2}\right )^{k +1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} \left (x -\frac {1}{2}\right )^{k -\frac {1}{4}}\right ), a_{k +2}=-\frac {2 a_{k +1}}{4 k +9}, b_{k +2}=-\frac {2 b_{k +1}}{4 k +13}, c_{k +2}=-\frac {2 c_{k +1}}{4 k +8}\right ] \end {array} \]
Maple trace
`Methods for third order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying high order exact linear fully integrable <- high order exact linear fully integrable successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 38
dsolve((2*x-1)*diff(diff(diff(y(x),x),x),x)+(x+4)*diff(diff(y(x),x),x)+2*diff(y(x),x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (c_{3} +\int \frac {\left (2 c_{1} x +c_{2} \right ) {\mathrm e}^{\frac {x}{2}}}{\left (-1+2 x \right )^{\frac {3}{4}}}d x \right ) {\mathrm e}^{-\frac {x}{2}}}{\left (-1+2 x \right )^{\frac {1}{4}}} \]
✓ Solution by Mathematica
Time used: 60.683 (sec). Leaf size: 66
DSolve[2*y'[x] + (4 + x)*y''[x] + (-1 + 2*x)*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \int _1^xe^{-\frac {K[1]}{2}} \left (\frac {2 \sqrt {2} c_1 K[1]}{(2 K[1]-1)^{5/4}}+c_2 L_{-\frac {1}{4}}^{\frac {5}{4}}\left (\frac {K[1]}{2}-\frac {1}{4}\right )\right )dK[1]+c_3 \]