problem 53 (e)

Internal problem ID [14760]
Internal ﬁle name [OUTPUT/14440_Monday_April_08_2024_06_25_00_AM_14190798/index.tex]

Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton. Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.7, page 195
Problem number: 53 (e).
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "kovacic", "second_order_change_of_variable_on_x_method_1", "second_order_change_of_variable_on_x_method_2"

\[ \boxed {\left (x^{2}+1\right )^{2} y^{\prime \prime }+2 x \left (x^{2}+1\right ) y^{\prime }+4 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 1] \end {align*}

15.52.1 Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime \prime } + p(x)y^{\prime } + q(x) y &= F \end {align*}

Where here \begin {align*} p(x) &=\frac {2 x^{3}+2 x}{\left (x^{2}+1\right )^{2}}\\ q(x) &=\frac {4}{\left (x^{2}+1\right )^{2}}\\ F &=0 \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+\frac {\left (2 x^{3}+2 x \right ) y^{\prime }}{\left (x^{2}+1\right )^{2}}+\frac {4 y}{\left (x^{2}+1\right )^{2}} = 0 \end {align*}

The domain of \(p(x)=\frac {2 x^{3}+2 x}{\left (x^{2}+1\right )^{2}}\) is \[ \{-\infty

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

In normal form the ode \begin {align*} \left (x^{2}+1\right )^{2} y^{\prime \prime }+\left (2 x^{3}+2 x \right ) y^{\prime }+4 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 x}{x^{2}+1}\\ q \left (x \right )&=\frac {4}{\left (x^{2}+1\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 x}{x^{2}+1}d x \right )}d x\\ &= \int e^{-\ln \left (x^{2}+1\right )} \,dx\\ &= \int \frac {1}{x^{2}+1}d x\\ &= \arctan \left (x \right )\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 {4}{\left (x^{2}+1\right )^{2}}}{\frac {1}{\left (x^{2}+1\right )^{2}}}\\ &= 4\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 )+4 y \left (\tau \right )&=0 \end {align*}

The above ode is now solved for \(y \left (\tau \right )\).This is second order with constant coeﬃcients homogeneous ODE. In standard form the ODE is \[ A y''(\tau ) + B y'(\tau ) + C y(\tau ) = 0 \] Where in the above \(A=1, B=0, C=4\). Let the solution be \(y \left (\tau \right )=e^{\lambda \tau }\). Substituting this into the ODE gives \[ \lambda ^{2} {\mathrm e}^{\lambda \tau }+4 \,{\mathrm e}^{\lambda \tau } = 0 \tag {1} \] Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda \tau }\) gives \[ \lambda ^{2}+4 = 0 \tag {2} \] Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula \[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \] Substituting \(A=1, B=0, C=4\) into the above gives \begin {align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {0^2 - (4) \left (1\right )\left (4\right )}\\ &= \pm 2 i \end {align*}

Which simpliﬁes to \begin{align*} \lambda _1 &= 2 i \\ \lambda _2 &= -2 i \\ \end{align*} Since roots are complex conjugate of each others, then let the roots be \[ \lambda _{1,2} = \alpha \pm i \beta \] Where \(\alpha =0\) and \(\beta =2\). Therefore the ﬁnal solution, when using Euler relation, can be written as \[ y \left (\tau \right ) = e^{\alpha \tau } \left ( c_{1} \cos (\beta \tau ) + c_{2} \sin (\beta \tau ) \right ) \] Which becomes \[ y \left (\tau \right ) = e^{0}\left (c_{1} \cos \left (2 \tau \right )+c_{2} \sin \left (2 \tau \right )\right ) \] Or \[ y \left (\tau \right ) = c_{1} \cos \left (2 \tau \right )+c_{2} \sin \left (2 \tau \right ) \] The above solution is now transformed back to \(y\) using (6) which results in \begin {align*} y &= \frac {-c_{1} x^{2}+2 c_{2} x +c_{1}}{x^{2}+1} \end {align*}

Looking at the above solution \begin {align*} y = \frac {-c_{1} x^{2}+2 c_{2} x +c_{1}}{x^{2}+1} \tag {1} \end {align*}

Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = c_{1}\tag {1A} \end {align*}

Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {-2 c_{1} x +2 c_{2}}{x^{2}+1}-\frac {2 \left (-c_{1} x^{2}+2 c_{2} x +c_{1} \right ) x}{\left (x^{2}+1\right )^{2}} \end {align*}

substituting \(y^{\prime } = 1\) and \(x = 0\) in the above gives \begin {align*} 1 = 2 c_{2}\tag {2A} \end {align*}

Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=0\\ c_{2}&={\frac {1}{2}} \end {align*}

Substituting these values back in above solution results in \begin {align*} y = \frac {x}{x^{2}+1} \end {align*}

15.52.3 Solving as second order change of variable on x method 1 ode

In normal form the ode \begin {align*} \left (x^{2}+1\right )^{2} y^{\prime \prime }+\left (2 x^{3}+2 x \right ) y^{\prime }+4 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 x}{x^{2}+1}\\ q \left (x \right )&=\frac {4}{\left (x^{2}+1\right )^{2}} \end {align*}

Applying change of variables \(\tau = g \left (x \right )\) to (2) results \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*}

Let \(q_1=c^2\) where \(c\) is some constant. Therefore from (5) \begin {align*} \tau ' &= \frac {1}{c}\sqrt {q}\\ &= \frac {2}{c \left (x^{2}+1\right )}\tag {6} \\ \tau '' &= -\frac {4 x}{c \left (x^{2}+1\right )^{2}} \end {align*}

Substituting the above into (4) results in \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}}\\ &=\frac {-\frac {4 x}{c \left (x^{2}+1\right )^{2}}+\frac {2 x}{x^{2}+1}\frac {2}{c \left (x^{2}+1\right )}}{\left (\frac {2}{c \left (x^{2}+1\right )}\right )^2} \\ &=0 \end {align*}

Therefore ode (3) now becomes \begin {align*} y \left (\tau \right )'' + p_1 y \left (\tau \right )' + q_1 y \left (\tau \right ) &= 0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+c^{2} y \left (\tau \right ) &= 0 \tag {7} \end {align*}

The above ode is now solved for \(y \left (\tau \right )\). Since the ode is now constant coeﬃcients, it can be easily solved to give \begin {align*} y \left (\tau \right ) &= c_{1} \cos \left (c \tau \right )+c_{2} \sin \left (c \tau \right ) \end {align*}

Now from (6) \begin {align*} \tau &= \int \frac {1}{c} \sqrt q \,dx \\ &= \frac {\int \frac {2}{x^{2}+1}d x}{c}\\ &= \frac {2 \arctan \left (x \right )}{c} \end {align*}

Substituting the above into the solution obtained gives \[ y = \frac {-c_{1} x^{2}+2 c_{2} x +c_{1}}{x^{2}+1} \] Initial conditions are used to solve for the constants of integration.

Looking at the above solution \begin {align*} y = \frac {-c_{1} x^{2}+2 c_{2} x +c_{1}}{x^{2}+1} \tag {1} \end {align*}

substituting \(y^{\prime } = 1\) and \(x = 0\) in the above gives \begin {align*} 1 = 2 c_{2}\tag {2A} \end {align*}

Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=0\\ c_{2}&={\frac {1}{2}} \end {align*}

Substituting these values back in above solution results in \begin {align*} y = \frac {x}{x^{2}+1} \end {align*}

15.52.4 Solving using Kovacic algorithm

Writing the ode as \begin {align*} \left (x^{2}+1\right )^{2} y^{\prime \prime }+\left (2 x^{3}+2 x \right ) y^{\prime }+4 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 (x^{2}+1\right )^{2} \\ B &= 2 x^{3}+2 x\tag {3} \\ C &= 4 \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*}

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 {-3}{\left (x^{2}+1\right )^{2}}\tag {6} \end {align*}

Comparing the above to (5) shows that \begin {align*} s &= -3\\ t &= \left (x^{2}+1\right )^{2} \end {align*}

Therefore eq. (4) becomes \begin {align*} z''(x) &= \left ( -\frac {3}{\left (x^{2}+1\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	\(\left \{ 1,2\right \} \)	\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)

Table 616: Necessary conditions for each Kovacic case

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 \(t=\left (x^{2}+1\right )^{2}\). There is a pole at \(x=i\) of order \(2\). There is a pole at \(x=-i\) 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*}

Looking at poles of order 2. The partial fractions decomposition of \(r\) is \[ r = \frac {3}{4 \left (x -i\right )^{2}}+\frac {3}{4 \left (x +i\right )^{2}}+\frac {3 i}{4 \left (x -i\right )}-\frac {3 i}{4 \left (x +i\right )} \] For the pole at \(x=i\) let \(b\) be the coeﬃcient of \(\frac {1}{ \left (x -i\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b={\frac {3}{4}}\). Hence \begin {alignat*} {2} [\sqrt r]_c &= 0 \\ \alpha _c^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= {\frac {3}{2}}\\ \alpha _c^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= -{\frac {1}{2}} \end {alignat*}

For the pole at \(x=-i\) let \(b\) be the coeﬃcient of \(\frac {1}{ \left (x +i\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b={\frac {3}{4}}\). Hence \begin {alignat*} {2} [\sqrt r]_c &= 0 \\ \alpha _c^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= {\frac {3}{2}}\\ \alpha _c^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= -{\frac {1}{2}} \end {alignat*}

Since the order of \(r\) at \(\infty \) is \(4 > 2\) then \begin {alignat*} {2} [\sqrt r]_\infty &= 0 \\ \alpha _{\infty }^{+} &= 0 \\ \alpha _{\infty }^{-} &= 1 \end {alignat*}

The following table summarizes the ﬁndings so far for poles and for the order of \(r\) at \(\infty \) where \(r\) is \[ r=-\frac {3}{\left (x^{2}+1\right )^{2}} \]

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}^{-}+\alpha _{c_2}^{+} \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*}

The above gives \begin {align*} \omega &= \left ( (-)[\sqrt r]_{c_1} + \frac { \alpha _{c_1}^{-} }{x- c_1}\right )+\left ( (+) [\sqrt r]_{c_2} + \frac { \alpha _{c_2}^{+} }{x- c_2}\right ) + (-) [\sqrt r]_\infty \\ &= -\frac {1}{2 \left (x -i\right )}+\frac {3}{2 \left (x +i\right )} + (-) \left ( 0 \right ) \\ &= -\frac {1}{2 \left (x -i\right )}+\frac {3}{2 \left (x +i\right )}\\ &= \frac {x -2 i}{x^{2}+1} \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*}

Substituting the above in eq. (1A) gives \begin {align*} \left (0\right ) + 2 \left (-\frac {1}{2 \left (x -i\right )}+\frac {3}{2 \left (x +i\right )}\right ) \left (0\right ) + \left ( \left (\frac {1}{2 \left (x -i\right )^{2}}-\frac {3}{2 \left (x +i\right )^{2}}\right ) + \left (-\frac {1}{2 \left (x -i\right )}+\frac {3}{2 \left (x +i\right )}\right )^2 - \left (-\frac {3}{\left (x^{2}+1\right )^{2}}\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}{2 \left (x -i\right )}+\frac {3}{2 \left (x +i\right )}\right )d x}\\ &= \frac {\left (x^{2}+1\right )^{\frac {3}{2}}}{\left (i x +1\right )^{2}} \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 {2 x^{3}+2 x}{\left (x^{2}+1\right )^{2}} \,dx} \\ &= z_1 e^{-\frac {\ln \left (x^{2}+1\right )}{2}} \\ &= z_1 \left (\frac {1}{\sqrt {x^{2}+1}}\right ) \\ \end{align*} Which simpliﬁes to \[ y_1 = \frac {x^{2}+1}{\left (i x +1\right )^{2}} \] 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 {2 x^{3}+2 x}{\left (x^{2}+1\right )^{2}} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{-\ln \left (x^{2}+1\right )}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (-\frac {x}{\left (x +i\right )^{2}}\right ) \\ \end{align*} Therefore the solution is

\begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left (\frac {x^{2}+1}{\left (i x +1\right )^{2}}\right ) + c_{2} \left (\frac {x^{2}+1}{\left (i x +1\right )^{2}}\left (-\frac {x}{\left (x +i\right )^{2}}\right )\right ) \\ \end{align*} Initial conditions are used to solve for the constants of integration.

Looking at the above solution \begin {align*} y = \frac {c_{1} \left (x^{2}+1\right )}{\left (i x +1\right )^{2}}+\frac {c_{2} \left (x^{2}+1\right ) x}{\left (-x +i\right )^{2} \left (x +i\right )^{2}} \tag {1} \end {align*}

Taking derivative of the solution gives \begin {align*} y^{\prime } = -\frac {2 i c_{1} \left (x^{2}+1\right )}{\left (i x +1\right )^{3}}+\frac {2 c_{1} x}{\left (i x +1\right )^{2}}+\frac {2 c_{2} x^{2}}{\left (-x +i\right )^{2} \left (x +i\right )^{2}}+\frac {c_{2} \left (x^{2}+1\right )}{\left (-x +i\right )^{2} \left (x +i\right )^{2}}+\frac {2 c_{2} \left (x^{2}+1\right ) x}{\left (-x +i\right )^{3} \left (x +i\right )^{2}}-\frac {2 c_{2} \left (x^{2}+1\right ) x}{\left (-x +i\right )^{2} \left (x +i\right )^{3}} \end {align*}

substituting \(y^{\prime } = 1\) and \(x = 0\) in the above gives \begin {align*} 1 = -2 c_{1} i+c_{2}\tag {2A} \end {align*}

Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=0\\ c_{2}&=1 \end {align*}

Substituting these values back in above solution results in \begin {align*} y = \frac {x}{x^{2}+1} \end {align*}

Maple trace

`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)] 
<- linear_1 successful`


pole \(c\) location	pole order	\([\sqrt r]_c\)	\(\alpha _c^{+}\)	\(\alpha _c^{-}\)

\(i\)	\(2\)	\(0\)	\(\frac {3}{2}\)	\(-{\frac {1}{2}}\)

\(-i\)	\(2\)	\(0\)	\(\frac {3}{2}\)	\(-{\frac {1}{2}}\)


Order of \(r\) at \(\infty \)	\([\sqrt r]_\infty \)	\(\alpha _\infty ^{+}\)	\(\alpha _\infty ^{-}\)

\(4\)	\(0\)	\(0\)	\(1\)

15.52 problem 53 (e)

15.52.1 Existence and uniqueness analysis

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

15.52.3 Solving as second order change of variable on x method 1 ode

15.52.4 Solving using Kovacic algorithm