Internal problem ID [9731]
Internal file name [OUTPUT/8673_Monday_June_06_2022_04_55_27_AM_92109750/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1404.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "kovacic", "second_order_change_of_variable_on_y_method_1", "linear_second_order_ode_solved_by_an_integrating_factor"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime }+\frac {\left (2 x^{2}+1\right ) y^{\prime }}{x^{3}}+\frac {\left (-2 x^{2}+1\right ) y}{4 x^{6}}=0} \]
The ode satisfies this form \[ y^{\prime \prime }+p \left (x \right ) y^{\prime }+\frac {\left (p \left (x \right )^{2}+p^{\prime }\left (x \right )\right ) y}{2} = f \left (x \right ) \] Where \( p(x) = \frac {2 x^{2}+1}{x^{3}}\). Therefore, there is an integrating factor given by \begin {align*} M(x) &= e^{\frac {1}{2} \int p \, dx} \\ &= e^{ \int \frac {2 x^{2}+1}{x^{3}} \, dx} \\ &= x \,{\mathrm e}^{-\frac {1}{4 x^{2}}} \end {align*}
Multiplying both sides of the ODE by the integrating factor \(M(x)\) makes the left side of the ODE a complete differential \begin{align*} \left ( M(x) y \right )'' &= 0 \\ \left ( x \,{\mathrm e}^{-\frac {1}{4 x^{2}}} y \right )'' &= 0 \\ \end{align*} Integrating once gives \[ \left ( x \,{\mathrm e}^{-\frac {1}{4 x^{2}}} y \right )' = c_{1} \] Integrating again gives \[ \left ( x \,{\mathrm e}^{-\frac {1}{4 x^{2}}} y \right ) = c_{1} x +c_{2} \] Hence the solution is \begin{align*} y &= \frac {c_{1} x +c_{2}}{x \,{\mathrm e}^{-\frac {1}{4 x^{2}}}} \\ \end{align*} Or \[ y = c_{1} {\mathrm e}^{\frac {1}{4 x^{2}}}+\frac {c_{2} {\mathrm e}^{\frac {1}{4 x^{2}}}}{x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{\frac {1}{4 x^{2}}}+\frac {c_{2} {\mathrm e}^{\frac {1}{4 x^{2}}}}{x} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{\frac {1}{4 x^{2}}}+\frac {c_{2} {\mathrm e}^{\frac {1}{4 x^{2}}}}{x} \] Verified OK.
In normal form the given ode is written as \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 {8 x^{5}+4 x^{3}}{4 x^{6}}\\ q \left (x \right )&=\frac {-2 x^{2}+1}{4 x^{6}} \end {align*}
Calculating the Liouville ode invariant \(Q\) given by \begin {align*} Q &= q - \frac {p'}{2}- \frac {p^2}{4} \\ &= \frac {-2 x^{2}+1}{4 x^{6}} - \frac {\left (\frac {8 x^{5}+4 x^{3}}{4 x^{6}}\right )'}{2}- \frac {\left (\frac {8 x^{5}+4 x^{3}}{4 x^{6}}\right )^2}{4} \\ &= \frac {-2 x^{2}+1}{4 x^{6}} - \frac {\left (\frac {40 x^{4}+12 x^{2}}{4 x^{6}}-\frac {3 \left (8 x^{5}+4 x^{3}\right )}{2 x^{7}}\right )}{2}- \frac {\left (\frac {\left (8 x^{5}+4 x^{3}\right )^{2}}{16 x^{12}}\right )}{4} \\ &= \frac {-2 x^{2}+1}{4 x^{6}} - \left (\frac {40 x^{4}+12 x^{2}}{8 x^{6}}-\frac {3 \left (8 x^{5}+4 x^{3}\right )}{4 x^{7}}\right )-\frac {\left (8 x^{5}+4 x^{3}\right )^{2}}{64 x^{12}}\\ &= 0 \end {align*}
Since the Liouville ode invariant does not depend on the independent variable \(x\) then the transformation \begin {align*} y = v \left (x \right ) z \left (x \right )\tag {3} \end {align*}
is used to change the original ode to a constant coefficients ode in \(v\). In (3) the term \(z \left (x \right )\) is given by \begin {align*} z \left (x \right )&={\mathrm e}^{-\left (\int \frac {p \left (x \right )}{2}d x \right )}\\ &= e^{-\int \frac {\frac {8 x^{5}+4 x^{3}}{4 x^{6}}}{2} }\\ &= \frac {{\mathrm e}^{\frac {1}{4 x^{2}}}}{x}\tag {5} \end {align*}
Hence (3) becomes \begin {align*} y = \frac {v \left (x \right ) {\mathrm e}^{\frac {1}{4 x^{2}}}}{x}\tag {4} \end {align*}
Applying this change of variable to the original ode results in \begin {align*} 4 \,{\mathrm e}^{\frac {1}{4 x^{2}}} x^{5} v^{\prime \prime }\left (x \right ) = 0 \end {align*}
Which is now solved for \(v \left (x \right )\) Integrating twice gives the solution \[ v \left (x \right )= c_{1} x + c_{2} \] Now that \(v \left (x \right )\) is known, then \begin {align*} y&= v \left (x \right ) z \left (x \right )\\ &= \left (c_{1} x +c_{2}\right ) \left (z \left (x \right )\right )\tag {7} \end {align*}
But from (5) \begin {align*} z \left (x \right )&= \frac {{\mathrm e}^{\frac {1}{4 x^{2}}}}{x} \end {align*}
Hence (7) becomes \begin {align*} y = \frac {\left (c_{1} x +c_{2} \right ) {\mathrm e}^{\frac {1}{4 x^{2}}}}{x} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (c_{1} x +c_{2} \right ) {\mathrm e}^{\frac {1}{4 x^{2}}}}{x} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (c_{1} x +c_{2} \right ) {\mathrm e}^{\frac {1}{4 x^{2}}}}{x} \] Verified OK.
Writing the ode as \begin {align*} 4 y^{\prime \prime } x^{6}+\left (8 x^{5}+4 x^{3}\right ) y^{\prime }+\left (-2 x^{2}+1\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 &= 4 x^{6} \\ B &= 8 x^{5}+4 x^{3}\tag {3} \\ C &= -2 x^{2}+1 \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 {0}{1}\tag {6} \end {align*}
Comparing the above to (5) shows that \begin {align*} s &= 0\\ t &= 1 \end {align*}
Therefore eq. (4) becomes \begin {align*} z''(x) &= 0 \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) \\ &= 0 - -\infty \\ &= \infty \end {align*}
There are no poles in \(r\). Therefore the set of poles \(\Gamma \) is empty. Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(infinity\) then the necessary conditions for case one are met. Therefore \begin {align*} L &= [1] \end {align*}
Since \(r = 0\) is not a function of \(x\), then there is no need run Kovacic algorithm to obtain a solution for transformed ode \(z''=r z\) as one solution is \[ z_1(x) = 1 \] Using the above, the solution for the original ode can now be found. 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 {8 x^{5}+4 x^{3}}{4 x^{6}} \,dx} \\ &= z_1 e^{-\ln \left (x \right )+\frac {1}{4 x^{2}}} \\ &= z_1 \left (\frac {{\mathrm e}^{\frac {1}{4 x^{2}}}}{x}\right ) \\ \end{align*} Which simplifies to \[ y_1 = \frac {{\mathrm e}^{\frac {1}{4 x^{2}}}}{x} \] 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 {8 x^{5}+4 x^{3}}{4 x^{6}} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{-2 \ln \left (x \right )+\frac {1}{2 x^{2}}}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (x\right ) \\ \end{align*} Therefore the solution is
\begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left (\frac {{\mathrm e}^{\frac {1}{4 x^{2}}}}{x}\right ) + c_{2} \left (\frac {{\mathrm e}^{\frac {1}{4 x^{2}}}}{x}\left (x\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} {\mathrm e}^{\frac {1}{4 x^{2}}}}{x}+c_{2} {\mathrm e}^{\frac {1}{4 x^{2}}} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} {\mathrm e}^{\frac {1}{4 x^{2}}}}{x}+c_{2} {\mathrm e}^{\frac {1}{4 x^{2}}} \] 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) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 19
dsolve(diff(diff(y(x),x),x) = -(2*x^2+1)/x^3*diff(y(x),x)-1/4*(-2*x^2+1)/x^6*y(x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {{\mathrm e}^{\frac {1}{4 x^{2}}} \left (c_{1} x +c_{2} \right )}{x} \]
✓ Solution by Mathematica
Time used: 0.032 (sec). Leaf size: 25
DSolve[y''[x] == -1/4*((1 - 2*x^2)*y[x])/x^6 - ((1 + 2*x^2)*y'[x])/x^3,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {e^{\frac {1}{4 x^2}} (c_2 x+c_1)}{x} \]