2.1.36 Problem 37

Internal problem ID [8748]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 37
Date solved : Sunday, March 30, 2025 at 01:29:58 PM
CAS classification : [_rational, _Riccati]

Solved using first_order_ode_riccati

Time used: 1.503 (sec)

Solve

$$\begin{array}{rl}x{y}^{\prime }-2y+b{y}^2& =c{x}^4\end{array}$$

In canonical form the ODE is

$$\begin{array}{rl}{y}^{\prime }& =F(x,y)\\ & =-\frac{-c{x}^4+b{y}^2-2y}{x}\end{array}$$

This is a Riccati ODE. Comparing the ODE to solve

$$y^{\prime }=c{x}^3-\frac{b{y}^2}{x}+\frac{2y}{x}$$

With Riccati ODE standard form

$$y^{\prime }=f_0(x)+f_1(x)y+f_2(x)y^2$$

Shows that $f_0(x)=c{x}^3$, $f_1(x)=\frac{2}{x}$ and $f_2(x)=-\frac{b}{x}$. Let

$$\begin{array}{rl}y& =\frac{-u^{\prime }}{f_{2}u}\\ \text{(1)}& & =\frac{-u^{\prime }}{-\frac{ub}{x}}\end{array}$$

Using the above substitution in the given ODE results (after some simplification)in a second order ODE to solve for $u(x)$ which is

$$\begin{array}{}\text{(2)}& f_2{u}^{″}(x)-(f_2^{\prime }+f_1{f}_2)u^{\prime }(x)+f_2^2{f}_{0}u(x)& =0\end{array}$$

But

$$\begin{array}{rl}{f}_2^{\prime }& =\frac{b}{x^2}\\ f_1{f}_2& =-\frac{2b}{x^2}\\ f_2^2{f}_0& =b^{2}xc\end{array}$$

Substituting the above terms back in equation (2) gives

$$\begin{array}{r}-\frac{b{u}^{\prime \prime }\left(x\right)}{x}+\frac{b{u}^{\prime }\left(x\right)}{x^2}+b^{2}xcu\left(x\right)=0\end{array}$$

Solved as second order ode using change of variable on x method 2

Time used: 0.749 (sec)

In normal form the ode

$$\begin{array}{}\text{(1)}& -\frac{b{u}^{\prime \prime }}{x}+\frac{b{u}^{\prime }}{x^2}+b^{2}xcu& =0\end{array}$$

Becomes

$$\begin{array}{}\text{(2)}& u^{\prime \prime }+p\left(x\right)u^{\prime }+q\left(x\right)u& =0\end{array}$$

Where

$$\begin{array}{rl}p\left(x\right)& =-\frac{1}{x}\\ q\left(x\right)& =-bc{x}^2\end{array}$$

Applying change of variables $\tau =g\left(x\right)$ to (2) gives

$$\begin{array}{}\text{(3)}& \frac{d^2}{d{\tau }^2}u\left(\tau \right)+p_1\left(\frac{d}{d\tau }u\left(\tau \right)\right)+q_{1}u\left(\tau \right)& =0\end{array}$$

Where $\tau$ is the new independent variable, and

$$\begin{array}{}\text{(4)}& p_1\left(\tau \right)& =\frac{{\tau }^{\prime \prime }\left(x\right)+p\left(x\right){\tau }^{\prime }\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\text{(5)}& q_1\left(\tau \right)& =\frac{q\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\end{array}$$

Let $p_1=0$. Eq (4) simplifies to

$$\begin{array}{rl}{\tau }^{\prime \prime }\left(x\right)+p\left(x\right){\tau }^{\prime }\left(x\right)& =0\end{array}$$

This ode is solved resulting in

$$\begin{array}{rl}\tau & =\int {\mathrm{e}}^{-\int p\left(x\right)dx}dx\\ & =\int {\mathrm{e}}^{-\int -\frac{1}{x}dx}dx\\ & =\int e^{\ln \left(x\right)}dx\\ & =\int xdx\\ \text{(6)}& & =\frac{x^2}{2}\end{array}$$

Using (6) to evaluate $q_1$ from (5) gives

$$\begin{array}{rl}{q}_1\left(\tau \right)& =\frac{q\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\\ & =\frac{-bc{x}^2}{x^2}\\ \text{(7)}& & =-bc\end{array}$$

Substituting the above in (3) and noting that now $p_1=0$ results in

$$\begin{array}{rl}\frac{d^2}{d{\tau }^2}u\left(\tau \right)+q_{1}u\left(\tau \right)& =0\\ \frac{d^2}{d{\tau }^2}u\left(\tau \right)-bcu\left(\tau \right)& =0\end{array}$$

The above ode is now solved for $u\left(\tau \right)$.This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

$$A{u}^{″}(\tau )+B{u}^{\prime }(\tau )+Cu(\tau )=0$$

Where in the above $A=1,B=0,C=-bc$. Let the solution be $u=e^{\lambda \tau }$. Substituting this into the ODE gives

$$\begin{array}{}\text{(1)}& {\lambda }^2{\mathrm{e}}^{\tau \lambda }-bc{\mathrm{e}}^{\tau \lambda }=0\end{array}$$

Since exponential function is never zero, then dividing Eq(2) throughout by $e^{\lambda \tau }$ gives

$$\begin{array}{}\text{(2)}& -bc+{\lambda }^2=0\end{array}$$

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula

$${\lambda }_{1,2}=\frac{-B}{2A}\pm \frac{1}{2A}\sqrt{B^2-4AC}$$

Substituting $A=1,B=0,C=-bc$ into the above gives

$$\begin{array}{rl}{\lambda }_{1,2}& =\frac{0}{(2)\left(1\right)}\pm \frac{1}{(2)\left(1\right)}\sqrt{0^2-(4)\left(1\right)(-bc)}\\ & =\pm \sqrt{bc}\end{array}$$

Hence

$$\begin{array}{rl}{\lambda }_1& =+\sqrt{bc}\\ {\lambda }_2& =-\sqrt{bc}\end{array}$$

Which simplifies to

$$\begin{array}{rl}{\lambda }_1& =-\frac{\sqrt{bc}((-1+i)\sqrt{\mathrm{signum}\left(bc\right)}-1-i)}{2}\\ {\lambda }_2& =\frac{\sqrt{bc}((-1+i)\sqrt{\mathrm{signum}\left(bc\right)}-1-i)}{2}\end{array}$$

The roots are complex but they are not conjugate of each others. Hence simplification using Euler relation is not possible here. Therefore the final solution is

$$\begin{array}{rl}u& =c_1{e}^{{\lambda }_1\tau }+c_2{e}^{{\lambda }_2\tau }\\ & =c_1{e}^{-\frac{\tau \sqrt{bc}((-1+i)\sqrt{\mathrm{signum}\left(bc\right)}-1-i)}{2}}+c_2{e}^{\frac{\tau \sqrt{bc}((-1+i)\sqrt{\mathrm{signum}\left(bc\right)}-1-i)}{2}}\end{array}$$

Will add steps showing solving for IC soon.

The above solution is now transformed back to $u\left(x\right)$ using (6) which results in

$$u\left(x\right)=c_1{\mathrm{e}}^{-\frac{x^2\sqrt{bc}((-1+i)\sqrt{\mathrm{signum}\left(bc\right)}-1-i)}{4}}+c_2{\mathrm{e}}^{\frac{x^2\sqrt{bc}((-1+i)\sqrt{\mathrm{signum}\left(bc\right)}-1-i)}{4}}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}u\left(x\right)& =c_1{\mathrm{e}}^{-\frac{x^2\sqrt{bc}((-1+i)\sqrt{\mathrm{signum}\left(bc\right)}-1-i)}{4}}+c_2{\mathrm{e}}^{\frac{x^2\sqrt{bc}((-1+i)\sqrt{\mathrm{signum}\left(bc\right)}-1-i)}{4}}\end{array}$$

Solved as second order ode using change of variable on x method 1

Time used: 0.107 (sec)

Solve

$$\begin{array}{rl}-\frac{b{u}^{\prime \prime }}{x}+\frac{b{u}^{\prime }}{x^2}+b^{2}xcu& =0\end{array}$$

In normal form the ode

$$\begin{array}{}\text{(1)}& -\frac{b{u}^{\prime \prime }}{x}+\frac{b{u}^{\prime }}{x^2}+b^{2}xcu& =0\end{array}$$

Becomes

$$\begin{array}{}\text{(2)}& u^{\prime \prime }+p\left(x\right)u^{\prime }+q\left(x\right)u& =0\end{array}$$

Where

$$\begin{array}{rl}p\left(x\right)& =-\frac{1}{x}\\ q\left(x\right)& =-bc{x}^2\end{array}$$

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

$$\begin{array}{}\text{(3)}& \frac{d^2}{d{\tau }^2}u\left(\tau \right)+p_1\left(\frac{d}{d\tau }u\left(\tau \right)\right)+q_{1}u\left(\tau \right)& =0\end{array}$$

Where $\tau$ is the new independent variable, and

$$\begin{array}{}\text{(4)}& p_1\left(\tau \right)& =\frac{{\tau }^{\prime \prime }\left(x\right)+p\left(x\right){\tau }^{\prime }\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\text{(5)}& q_1\left(\tau \right)& =\frac{q\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\end{array}$$

Let $q_1=c^2$ where $c$ is some constant. Therefore from (5)

$$\begin{array}{rl}{\tau }^{\prime }& =\frac{1}{c}\sqrt{q}\\ \text{(6)}& & =\frac{\sqrt{-bc{x}^2}}{c}{\tau }^{″}& =-\frac{bcx}{c\sqrt{-bc{x}^2}}\end{array}$$

Substituting the above into (4) results in

$$\begin{array}{rl}{p}_1\left(\tau \right)& =\frac{{\tau }^{\prime \prime }\left(x\right)+p\left(x\right){\tau }^{\prime }\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\\ & =\frac{-\frac{bcx}{c\sqrt{-bc{x}^2}}-\frac{1}{x}\frac{\sqrt{-bc{x}^2}}{c}}{{\left(\frac{\sqrt{-bc{x}^2}}{c}\right)}^2}\\ & =0\end{array}$$

Therefore ode (3) now becomes

$$\begin{array}{rl}u{\left(\tau \right)}^{″}+p_{1}u{\left(\tau \right)}^{\prime }+q_{1}u\left(\tau \right)& =0\\ \text{(7)}& \frac{d^2}{d{\tau }^2}u\left(\tau \right)+c^{2}u\left(\tau \right)& =0\end{array}$$

The above ode is now solved for $u\left(\tau \right)$. Since the ode is now constant coefficients, it can be easily solved to give

$$\begin{array}{rl}u\left(\tau \right)& =c_1\cos \left(c\tau \right)+c_2\sin \left(c\tau \right)\end{array}$$

Now from (6)

$$\begin{array}{rl}\tau & =\int \frac{1}{c}\sqrt{q}dx\\ & =\frac{\int \sqrt{-bc{x}^2}dx}{c}\\ & =\frac{x\sqrt{-bc{x}^2}}{2c}\end{array}$$

Substituting the above into the solution obtained gives

$$u=c_1\cos \left(\frac{x\sqrt{-bc{x}^2}}{2}\right)+c_2\sin \left(\frac{x\sqrt{-bc{x}^2}}{2}\right)$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}u& =c_1\cos \left(\frac{x\sqrt{-bc{x}^2}}{2}\right)+c_2\sin \left(\frac{x\sqrt{-bc{x}^2}}{2}\right)\end{array}$$

Solved as second order Bessel ode

Time used: 0.090 (sec)

Solve

$$\begin{array}{rl}-\frac{b{u}^{\prime \prime }}{x}+\frac{b{u}^{\prime }}{x^2}+b^{2}xcu& =0\end{array}$$

Writing the ode as

$$\begin{array}{}\text{(1)}& x^2{u}^{\prime \prime }-u^{\prime }x-bc{x}^{4}u=0\end{array}$$

Bessel ode has the form

$$\begin{array}{}\text{(2)}& x^2{u}^{\prime \prime }+u^{\prime }x+(-n^2+x^2)u=0\end{array}$$

The generalized form of Bessel ode is given by Bowman (1958) as the following

$$\begin{array}{}\text{(3)}& x^2{u}^{\prime \prime }+(1-2\alpha )x{u}^{\prime }+({\beta }^2{\gamma }^2{x}^{2\gamma }-n^2{\gamma }^2+{\alpha }^2)u=0\end{array}$$

With the standard solution

$$\begin{array}{}\text{(4)}& u& =x^{\alpha }(c_1\mathrm{BesselJ}(n,\beta x^{\gamma })+c_2\mathrm{BesselY}(n,\beta x^{\gamma }))\end{array}$$

Comparing (3) to (1) and solving for $\alpha ,\beta ,n,\gamma$ gives

$$\begin{array}{rl}\alpha & =1\\ \beta & =\frac{\sqrt{-bc}}{2}\\ n& =\frac{1}{2}\\ \gamma & =2\end{array}$$

Substituting all the above into (4) gives the solution as

$$\begin{array}{r}u=\frac{2c_{1}x\sin \left(\frac{\sqrt{-bc}{x}^2}{2}\right)}{\sqrt{\pi }\sqrt{\sqrt{-bc}{x}^2}}-\frac{2c_{2}x\cos \left(\frac{\sqrt{-bc}{x}^2}{2}\right)}{\sqrt{\pi }\sqrt{\sqrt{-bc}{x}^2}}\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}u& =\frac{2c_{1}x\sin \left(\frac{\sqrt{-bc}{x}^2}{2}\right)}{\sqrt{\pi }\sqrt{\sqrt{-bc}{x}^2}}-\frac{2c_{2}x\cos \left(\frac{\sqrt{-bc}{x}^2}{2}\right)}{\sqrt{\pi }\sqrt{\sqrt{-bc}{x}^2}}\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.389 (sec)

Solve

$$\begin{array}{rl}-\frac{b{u}^{\prime \prime }}{x}+\frac{b{u}^{\prime }}{x^2}+b^{2}xcu& =0\end{array}$$

Writing the ode as

$$\begin{array}{}\text{(1)}& -\frac{b{u}^{\prime \prime }}{x}+\frac{b{u}^{\prime }}{x^2}+b^{2}xcu& =0\text{(2)}& A{u}^{\prime \prime }+B{u}^{\prime }+Cu& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =-\frac{b}{x}\\ \text{(3)}& B& =\frac{b}{x^2}C& =b^{2}xc\end{array}$$

Applying the Liouville transformation on the dependent variable gives

$$\begin{array}{rl}z(x)& =u{e}^{\int \frac{B}{2A}dx}\end{array}$$

Then (2) becomes

$$\begin{array}{}\text{(4)}& z^{″}(x)=rz(x)\end{array}$$

Where $r$ is given by

$$\begin{array}{rl}\text{(5)}& r& =\frac{s}{t}& =\frac{2A{B}^{\prime }-2B{A}^{\prime }+B^2-4AC}{4A^2}\end{array}$$

Substituting the values of $A,B,C$ from (3) in the above and simplifying gives

$$\begin{array}{}\text{(6)}& r& =\frac{4bc{x}^4+3}{4x^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =4bc{x}^4+3\\ t& =4x^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{4bc{x}^4+3}{4x^2}\right)z(x)\end{array}$$

Equation (7) is now solved. After finding $z(x)$ then $u$ is found using the inverse transformation

$$\begin{array}{rl}u& =z\left(x\right)e^{-\int \frac{B}{2A}dx}\end{array}$$

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 $\mathrm{\infty }$. The following table summarizes these cases.

Case	Allowed pole order for $r$	Allowed value for $\mathcal{O}(\mathrm{\infty })$
1	$\{0,1,2,4,6,8,\cdots \}$	$\{\cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \}$
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	$\{1,2\}$	$\{2,3,4,5,6,7,\cdots \}$

Table 2.7: Necessary conditions for each Kovacic case

The order of $r$ at $\mathrm{\infty }$ is the degree of $t$ minus the degree of $s$. Therefore

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =2-4\\ & =-2\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=4x^2$. There is a pole at $x=0$ of order $2$. Since there is no odd order pole larger than $2$ and the order at $\mathrm{\infty }$ is $-2$ 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

$$\begin{array}{rl}L& =[1,2]\end{array}$$

Attempting to find a solution using case $n=1$.

Looking at poles of order 2. The partial fractions decomposition of $r$ is

$$r=bc{x}^2+\frac{3}{4x^2}$$

For the pole at $x=0$ let $b$ be the coefficient of $\frac{1}{x^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=\frac{3}{4}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{3}{2}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =-\frac{1}{2}\end{array}$$

Since the order of $r$ at $\mathrm{\infty }$ is $O_r(\mathrm{\infty })=-2$ then

$$\begin{array}{rlrlrl}v& =\frac{-O_r(\mathrm{\infty })}{2}& & =\frac{2}{2}& & =1\end{array}$$

$[\sqrt{r}{]}_{\mathrm{\infty }}$ is the sum of terms involving $x^i$ for $0\le i\le v$ in the Laurent series for $\sqrt{r}$ at $\mathrm{\infty }$. Therefore

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sum _{i=0}^v{a}_i{x}^i\\ \text{(8)}& & =\sum _{i=0}^1{a}_i{x}^i\end{array}$$

Let $a$ be the coefficient of $x^v=x^1$ in the above sum. The Laurent series of $\sqrt{r}$ at $\mathrm{\infty }$ is

$$\begin{array}{}\text{(9)}& \sqrt{r}\approx \sqrt{b}\sqrt{c}x+\frac{3}{8\sqrt{b}\sqrt{c}{x}^3}-\frac{9}{128b^{3/2}{c}^{3/2}{x}^7}+\frac{27}{1024b^{5/2}{c}^{5/2}{x}^{11}}-\frac{405}{32768b^{7/2}{c}^{7/2}{x}^{15}}+\frac{1701}{262144b^{9/2}{c}^{9/2}{x}^{19}}-\frac{15309}{4194304b^{11/2}{c}^{11/2}{x}^{23}}+\frac{72171}{33554432b^{13/2}{c}^{13/2}{x}^{27}}+\dots \end{array}$$

Comparing Eq. (9) with Eq. (8) shows that

$$a=\sqrt{b}\sqrt{c}$$

From Eq. (9) the sum up to $v=1$ gives

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sum _{i=0}^1{a}_i{x}^i\\ \text{(10)}& & =\sqrt{b}\sqrt{c}x\end{array}$$

Now we need to find $b$, where $b$ be the coefficient of $x^{v-1}=x^0=1$ in $r$ minus the coefficient of same term but in ${([\sqrt{r}{]}_{\mathrm{\infty }})}^2$ where $[\sqrt{r}{]}_{\mathrm{\infty }}$ was found above in Eq (10). Hence

$${([\sqrt{r}{]}_{\mathrm{\infty }})}^2=bc{x}^2$$

This shows that the coefficient of $1$ in the above is $0$. Now we need to find the coefficient of $1$ in $r$. How this is done depends on if $v=0$ or not. Since $v=1$ which is not zero, then starting $r=\frac{s}{t}$, we do long division and write this in the form

$$r=Q+\frac{R}{t}$$

Where $Q$ is the quotient and $R$ is the remainder. Then the coefficient of $1$ in $r$ will be the coefficient this term in the quotient. Doing long division gives

$$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{4bc{x}^4+3}{4x^2}\\ & =Q+\frac{R}{4x^2}\\ & =\left(bc{x}^2\right)+\left(\frac{3}{4x^2}\right)\\ & =bc{x}^2+\frac{3}{4x^2}\end{array}$$

We see that the coefficient of the term $x$ in the quotient is $0$. Now $b$ can be found.

$$\begin{array}{rl}b& =\left(0\right)-\left(0\right)\\ & =0\end{array}$$

Hence

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sqrt{b}\sqrt{c}x\\ {\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)& & =\frac{1}{2}(\frac{0}{\sqrt{b}\sqrt{c}}-1)& & =-\frac{1}{2}\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)& & =\frac{1}{2}(-\frac{0}{\sqrt{b}\sqrt{c}}-1)& & =-\frac{1}{2}\end{array}$$

The following table summarizes the findings so far for poles and for the order of $r$ at $\mathrm{\infty }$ where $r$ is

$$r=\frac{4bc{x}^4+3}{4x^2}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$0$	$2$	$0$	$\frac{3}{2}$	$-\frac{1}{2}$

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$-2$	$\sqrt{b}\sqrt{c}x$	$-\frac{1}{2}$	$-\frac{1}{2}$

Now that the all $[\sqrt{r}{]}_c$ and its associated ${\alpha }_c^{\pm }$ have been determined for all the poles in the set $\mathrm{\Gamma }$ and $[\sqrt{r}{]}_{\mathrm{\infty }}$ and its associated ${\alpha }_{\mathrm{\infty }}^{\pm }$ have also been found, the next step is to determine possible non negative integer $d$ from these using

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{s(\mathrm{\infty })}-\sum _{c\in \mathrm{\Gamma }}{\alpha }_c^{s(c)}\end{array}$$

Where $s(c)$ is either $+$ or $-$ and $s(\mathrm{\infty })$ is the sign of ${\alpha }_{\mathrm{\infty }}^{\pm }$. This is done by trial over all set of families $s=(s(c){)}_{c\in \mathrm{\Gamma }\cup \mathrm{\infty }}$ until such $d$ is found to work in finding candidate $\omega$. Trying ${\alpha }_{\mathrm{\infty }}^{-}=-\frac{1}{2}$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}-\left({\alpha }_{c_1}^{-}\right)\\ & =-\frac{1}{2}-(-\frac{1}{2})\\ & =0\end{array}$$

Since $d$ an integer and $d\ge 0$ then it can be used to find $\omega$ using

$$\begin{array}{rl}\omega & =\sum _{c\in \mathrm{\Gamma }}(s(c)[\sqrt{r}{]}_c+\frac{{\alpha }_c^{s(c)}}{x-c})+s(\mathrm{\infty })[\sqrt{r}{]}_{\mathrm{\infty }}\end{array}$$

The above gives

$$\begin{array}{rl}\omega & =((-)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{-}}{x-c_1})+(-)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =-\frac{1}{2x}+(-)\left(\sqrt{b}\sqrt{c}x\right)\\ & =-\frac{1}{2x}-\sqrt{b}\sqrt{c}x\\ & =\frac{-2\sqrt{b}\sqrt{c}{x}^2-1}{2x}\end{array}$$

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

$$\begin{array}{}\text{(1A)}& p^{″}+2\omega p^{\prime }+({\omega }^{\prime }+{\omega }^2-r)p=0\end{array}$$

Let

$$\begin{array}{}\text{(2A)}& p(x)& =1\end{array}$$

Substituting the above in eq. (1A) gives

$$\begin{array}{rl}\left(0\right)+2(-\frac{1}{2x}-\sqrt{b}\sqrt{c}x)\left(0\right)+((\frac{1}{2x^2}-\sqrt{b}\sqrt{c})+{(-\frac{1}{2x}-\sqrt{b}\sqrt{c}x)}^2-\left(\frac{4bc{x}^4+3}{4x^2}\right))& =0\\ 0=0\end{array}$$

The equation is satisfied since both sides are zero. Therefore the first solution to the ode $z^{″}=rz$ is

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & ={\mathrm{e}}^{\int (-\frac{1}{2x}-\sqrt{b}\sqrt{c}x)dx}\\ & =\frac{{\mathrm{e}}^{-\frac{\sqrt{b}\sqrt{c}{x}^2}{2}}}{\sqrt{x}}\end{array}$$

The first solution to the original ode in $u$ is found from

$$\begin{array}{rl}{u}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{\frac{b}{x^2}}{-\frac{b}{x}}dx}\\ & =z_1{e}^{\frac{\ln \left(x\right)}{2}}\\ & =z_1\left(\sqrt{x}\right)\end{array}$$

Which simplifies to

$$u_1={\mathrm{e}}^{-\frac{x^2\sqrt{bc}}{2}}$$

The second solution $u_2$ to the original ode is found using reduction of order

$$u_2=u_1\int \frac{e^{\int -\frac{B}{A}dx}}{u_1^2}dx$$

Substituting gives

$$\begin{array}{rl}{u}_2& =u_1\int \frac{e^{\int -\frac{\frac{b}{x^2}}{-\frac{b}{x}}dx}}{{\left(u_1\right)}^2}dx\\ & =u_1\int \frac{e^{\ln \left(x\right)}}{{\left(u_1\right)}^2}dx\\ & =u_1\left(\frac{\sqrt{bc}{\mathrm{e}}^{x^2\sqrt{bc}}}{2bc}\right)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}u& =c_1{u}_1+c_2{u}_2\\ & =c_1\left({\mathrm{e}}^{-\frac{x^2\sqrt{bc}}{2}}\right)+c_2\left({\mathrm{e}}^{-\frac{x^2\sqrt{bc}}{2}}\left(\frac{\sqrt{bc}{\mathrm{e}}^{x^2\sqrt{bc}}}{2bc}\right)\right)\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}u& =c_1{\mathrm{e}}^{-\frac{x^2\sqrt{bc}}{2}}+\frac{c_2{\mathrm{e}}^{\frac{x^2\sqrt{bc}}{2}}\sqrt{bc}}{2bc}\end{array}$$

Taking derivative gives

$$u^{\prime }\left(x\right)=-\frac{c_{1}x\sqrt{bc}((-1+i)\sqrt{\mathrm{signum}\left(bc\right)}-1-i){\mathrm{e}}^{-\frac{x^2\sqrt{bc}((-1+i)\sqrt{\mathrm{signum}\left(bc\right)}-1-i)}{4}}}{2}+\frac{c_{2}x\sqrt{bc}((-1+i)\sqrt{\mathrm{signum}\left(bc\right)}-1-i){\mathrm{e}}^{\frac{x^2\sqrt{bc}((-1+i)\sqrt{\mathrm{signum}\left(bc\right)}-1-i)}{4}}}{2}$$

Doing change of constants, the solution becomes

$$y=\frac{(-\frac{c_{1}x\sqrt{bc}((-1+i)\sqrt{\mathrm{signum}\left(bc\right)}-1-i){\mathrm{e}}^{-\frac{x^2\sqrt{bc}((-1+i)\sqrt{\mathrm{signum}\left(bc\right)}-1-i)}{4}}}{2}+\frac{x\sqrt{bc}((-1+i)\sqrt{\mathrm{signum}\left(bc\right)}-1-i){\mathrm{e}}^{\frac{x^2\sqrt{bc}((-1+i)\sqrt{\mathrm{signum}\left(bc\right)}-1-i)}{4}}}{2})x}{b(c_1{\mathrm{e}}^{-\frac{x^2\sqrt{bc}((-1+i)\sqrt{\mathrm{signum}\left(bc\right)}-1-i)}{4}}+{\mathrm{e}}^{\frac{x^2\sqrt{bc}((-1+i)\sqrt{\mathrm{signum}\left(bc\right)}-1-i)}{4}})}$$

Summary of solutions found

$$\begin{array}{rl}y& =\frac{(-\frac{c_{1}x\sqrt{bc}((-1+i)\sqrt{\mathrm{signum}\left(bc\right)}-1-i){\mathrm{e}}^{-\frac{x^2\sqrt{bc}((-1+i)\sqrt{\mathrm{signum}\left(bc\right)}-1-i)}{4}}}{2}+\frac{x\sqrt{bc}((-1+i)\sqrt{\mathrm{signum}\left(bc\right)}-1-i){\mathrm{e}}^{\frac{x^2\sqrt{bc}((-1+i)\sqrt{\mathrm{signum}\left(bc\right)}-1-i)}{4}}}{2})x}{b(c_1{\mathrm{e}}^{-\frac{x^2\sqrt{bc}((-1+i)\sqrt{\mathrm{signum}\left(bc\right)}-1-i)}{4}}+{\mathrm{e}}^{\frac{x^2\sqrt{bc}((-1+i)\sqrt{\mathrm{signum}\left(bc\right)}-1-i)}{4}})}\end{array}$$

✓ Maple. Time used: 0.004 (sec). Leaf size: 29

ode:=diff(y(x),x)*x-2*y(x)+b*y(x)^2 = c*x^4; 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{i\tan (-\frac{i{x}^2\sqrt{b}\sqrt{c}}{2}+c_1)x^2\sqrt{c}}{\sqrt{b}}$$

Maple trace

Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
trying separable 
trying inverse linear 
trying homogeneous types: 
trying Chini 
<- Chini successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & x\left(\frac{d}{dx}y\left(x\right)\right)-2y\left(x\right)+by{\left(x\right)}^2=c{x}^4\\ \bullet & & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Solve for the highest derivative}\\ & & \frac{d}{dx}y\left(x\right)=\frac{2y\left(x\right)-by{\left(x\right)}^2+c{x}^4}{x}\end{array}$$

✓ Mathematica. Time used: 0.242 (sec). Leaf size: 153

ode=x*D[y[x],x]-2*y[x]+b*y[x]^2==c*x^4; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$\begin{array}{r}y(x)\to \frac{\sqrt{c}{x}^2(-\cos \left(\frac{1}{2}\sqrt{-b}\sqrt{c}{x}^2\right)+c_1\sin \left(\frac{1}{2}\sqrt{-b}\sqrt{c}{x}^2\right))}{\sqrt{-b}(\sin \left(\frac{1}{2}\sqrt{-b}\sqrt{c}{x}^2\right)+c_1\cos \left(\frac{1}{2}\sqrt{-b}\sqrt{c}{x}^2\right))}\\ y(x)\to \frac{\sqrt{c}{x}^2\tan \left(\frac{1}{2}\sqrt{-b}\sqrt{c}{x}^2\right)}{\sqrt{-b}}\end{array}$$

✗ Sympy

from sympy import * 
x = symbols("x") 
b = symbols("b") 
c = symbols("c") 
y = Function("y") 
ode = Eq(b*y(x)**2 - c*x**4 + x*Derivative(y(x), x) - 2*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

RecursionError : maximum recursion depth exceeded