problem 1401

Internal problem ID [9728]
Internal ﬁle name [OUTPUT/8670_Monday_June_06_2022_04_45_35_AM_16488900/index.tex]

Book: Diﬀerential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1401.
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_change_of_variable_on_x_method_2"

\[ \boxed {y^{\prime \prime }+\frac {\left (3 x^{2}+a \right ) y^{\prime }}{x^{3}}+\frac {b y}{x^{6}}=0} \]

3.395.1 Solving as second order change of variable on x method 2 ode

In normal form the ode \begin {align*} y^{\prime \prime } x^{6}+\left (3 x^{2}+a \right ) y^{\prime } x^{3}+b 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 {3 x^{2}+a}{x^{3}}\\ q \left (x \right )&=\frac {b}{x^{6}} \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 {3 x^{2}+a}{x^{3}}d x \right )}d x\\ &= \int e^{-3 \ln \left (x \right )+\frac {a}{2 x^{2}}} \,dx\\ &= \int \frac {{\mathrm e}^{\frac {a}{2 x^{2}}}}{x^{3}}d x\\ &= -\frac {{\mathrm e}^{\frac {a}{2 x^{2}}}}{a}\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 {b}{x^{6}}}{\frac {{\mathrm e}^{\frac {a}{x^{2}}}}{x^{6}}}\\ &= b \,{\mathrm e}^{-\frac {a}{x^{2}}}\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 )+b \,{\mathrm e}^{-\frac {a}{x^{2}}} y \left (\tau \right )&=0 \\ \end {align*}

But in terms of \(\tau \) \begin {align*} b \,{\mathrm e}^{-\frac {a}{x^{2}}}&=\frac {b}{a^{2} \tau ^{2}} \end {align*}

Hence the above ode becomes \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+\frac {b y \left (\tau \right )}{a^{2} \tau ^{2}}&=0 \end {align*}

The above ode is now solved for \(y \left (\tau \right )\). The ode can be written as \[ \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right ) a^{2} \tau ^{2}+b y \left (\tau \right ) = 0 \] 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 \[ a^{2} \tau ^{2}(r(r-1))\tau ^{r-2}+0 r \tau ^{r-1}+b \,\tau ^{r} = 0 \] Simplifying gives \[ a^{2} r \left (r -1\right )\tau ^{r}+0\,\tau ^{r}+b \,\tau ^{r} = 0 \] Since \(\tau ^{r}\neq 0\) then dividing throughout by \(\tau ^{r}\) gives \[ a^{2} r \left (r -1\right )+0+b = 0 \] Or \[ a^{2} r^{2}-a^{2} r +b = 0 \tag {1} \] 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 {-a +\sqrt {a^{2}-4 b}}{2 a}\\ r_2 &= \frac {a +\sqrt {a^{2}-4 b}}{2 a} \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 \[ y \left (\tau \right ) = c_{1} \tau ^{-\frac {-a +\sqrt {a^{2}-4 b}}{2 a}}+c_{2} \tau ^{\frac {a +\sqrt {a^{2}-4 b}}{2 a}} \] The above solution is now transformed back to \(y\) using (6) which results in \begin {align*} y &= c_{1} \left (-\frac {{\mathrm e}^{\frac {a}{2 x^{2}}}}{a}\right )^{-\frac {-a +\sqrt {a^{2}-4 b}}{2 a}}+c_{2} \left (-\frac {{\mathrm e}^{\frac {a}{2 x^{2}}}}{a}\right )^{\frac {a +\sqrt {a^{2}-4 b}}{2 a}} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (-\frac {{\mathrm e}^{\frac {a}{2 x^{2}}}}{a}\right )^{-\frac {-a +\sqrt {a^{2}-4 b}}{2 a}}+c_{2} \left (-\frac {{\mathrm e}^{\frac {a}{2 x^{2}}}}{a}\right )^{\frac {a +\sqrt {a^{2}-4 b}}{2 a}} \\ \end{align*}

Veriﬁcation of solutions

\[ y = c_{1} \left (-\frac {{\mathrm e}^{\frac {a}{2 x^{2}}}}{a}\right )^{-\frac {-a +\sqrt {a^{2}-4 b}}{2 a}}+c_{2} \left (-\frac {{\mathrm e}^{\frac {a}{2 x^{2}}}}{a}\right )^{\frac {a +\sqrt {a^{2}-4 b}}{2 a}} \] 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`

\[ y \left (x \right ) = c_{1} {\mathrm e}^{-\frac {-a +\sqrt {a^{2}-4 b}}{4 x^{2}}}+c_{2} {\mathrm e}^{\frac {a +\sqrt {a^{2}-4 b}}{4 x^{2}}} \]

\[ y(x)\to e^{\frac {a-\sqrt {a^2-4 b}}{4 x^2}} \left (c_1 e^{\frac {\sqrt {a^2-4 b}}{2 x^2}}+c_2\right ) \]

3.395 problem 1401

3.395.1 Solving as second order change of variable on x method 2 ode