2.6.5 Problem 8 (eq 68)

Internal problem ID [18498]
Book : Elementary Differential Equations. By Thornton C. Fry. D Van Nostrand. NY. First Edition (1929)
Section : Chapter IV. Methods of solution: First order equations. section 33. Problems at page 91
Problem number : 8 (eq 68)
Date solved : Monday, March 31, 2025 at 05:38:16 PM
CAS classification : [_separable]

Solved using first_order_ode_separable

Time used: 0.107 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime }& =x(a{y}^2+b)\end{array}$$

The ode

$$\begin{array}{}\text{(1)}& y^{\prime }=x(a{y}^2+b)\end{array}$$

is separable as it can be written as

$$\begin{array}{rl}{y}^{\prime }& =x(a{y}^2+b)\\ & =f(x)g(y)\end{array}$$

Where

$$\begin{array}{rl}f(x)& =x\\ g(y)& =a{y}^2+b\end{array}$$

Integrating gives

$$\begin{array}{rl}\int \frac{1}{g(y)}dy& =\int f(x)dx\\ \int \frac{1}{a{y}^2+b}dy& =\int xdx\end{array}$$

$$\frac{\arctan \left(\frac{ay}{\sqrt{ba}}\right)}{\sqrt{ba}}=\frac{x^2}{2}+c_1$$

We now need to find the singular solutions, these are found by finding for what values $g(y)$ is zero, since we had to divide by this above. Solving $g(y)=0$ or

$$a{y}^2+b=0$$

for $y$ gives

$$\begin{array}{rl}y& =\frac{\sqrt{-ba}}{a}\\ y& =-\frac{\sqrt{-ba}}{a}\end{array}$$

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

$$\begin{array}{rl}\frac{\arctan \left(\frac{ay}{\sqrt{ba}}\right)}{\sqrt{ba}}& =\frac{x^2}{2}+c_1\\ y& =\frac{\sqrt{-ba}}{a}\\ y& =-\frac{\sqrt{-ba}}{a}\end{array}$$

Solving for $y$ gives

$$\begin{array}{rl}y& =\frac{\sqrt{-ba}}{a}\\ y& =\frac{\tan (\frac{x^2\sqrt{ba}}{2}+c_1\sqrt{ba})\sqrt{ba}}{a}\\ y& =-\frac{\sqrt{-ba}}{a}\end{array}$$

Which simplifies to

$$\begin{array}{rl}y& =\frac{\sqrt{-ba}}{a}\\ y& =\frac{\tan \left(\sqrt{ba}(\frac{x^2}{2}+c_1)\right)\sqrt{ba}}{a}\\ y& =-\frac{\sqrt{-ba}}{a}\end{array}$$

Summary of solutions found

$$\begin{array}{rl}y& =\frac{\sqrt{-ba}}{a}\\ y& =\frac{\tan \left(\sqrt{ba}(\frac{x^2}{2}+c_1)\right)\sqrt{ba}}{a}\\ y& =-\frac{\sqrt{-ba}}{a}\end{array}$$

Solved using first_order_ode_exact

Time used: 0.171 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime }& =x(a{y}^2+b)\end{array}$$

To solve an ode of the form

$$\begin{array}{}\text{(A)}& M(x,y)+N(x,y)\frac{dy}{dx}=0\end{array}$$

We assume there exists a function $\varphi (x,y)=c$ where $c$ is constant, that satisfies the ode. Taking derivative of $\varphi$ w.r.t. $x$ gives

$$\frac{d}{dx}\varphi (x,y)=0$$

Hence

$$\begin{array}{}\text{(B)}& \frac{\partial \varphi }{\partial x}+\frac{\partial \varphi }{\partial y}\frac{dy}{dx}=0\end{array}$$

Comparing (A,B) shows that

$$\begin{array}{rl}\frac{\partial \varphi }{\partial x}& =M\\ \frac{\partial \varphi }{\partial y}& =N\end{array}$$

But since $\frac{{\partial }^2\varphi }{\partial x\partial y}=\frac{{\partial }^2\varphi }{\partial y\partial x}$ then for the above to be valid, we require that

$$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$

If the above condition is satisfied, then the original ode is called exact. We still need to determine $\varphi (x,y)$ but at least we know now that we can do that since the condition $\frac{{\partial }^2\varphi }{\partial x\partial y}=\frac{{\partial }^2\varphi }{\partial y\partial x}$ is satisfied. If this condition is not satisfied then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The first step is to write the ODE in standard form to check for exactness, which is

$$\begin{array}{}\text{(1A)}& M(x,y)\mathrm{d}x+N(x,y)\mathrm{d}y=0\end{array}$$

Therefore

$$\begin{array}{rl}\mathrm{d}y& =\left(x(a{y}^2+b)\right)\mathrm{d}x\\ \text{(2A)}& (-x(a{y}^2+b))\mathrm{d}x+\mathrm{d}y& =0\end{array}$$

Comparing (1A) and (2A) shows that

$$\begin{array}{rl}M(x,y)& =-x(a{y}^2+b)\\ N(x,y)& =1\end{array}$$

The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisfied

$$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$

Using result found above gives

$$\begin{array}{rl}\frac{\partial M}{\partial y}& =\frac{\partial }{\partial y}(-x(a{y}^2+b))\\ & =-2xay\end{array}$$

And

$$\begin{array}{rl}\frac{\partial N}{\partial x}& =\frac{\partial }{\partial x}\left(1\right)\\ & =0\end{array}$$

Since $\frac{\partial M}{\partial y}\ne \frac{\partial N}{\partial x}$, then the ODE is not exact. Since the ODE is not exact, we will try to find an integrating factor to make it exact. Let

$$\begin{array}{rl}A& =\frac{1}{N}(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x})\\ & =1((-2xay)-\left(0\right))\\ & =-2xay\end{array}$$

Since $A$ depends on $y$, it can not be used to obtain an integrating factor. We will now try a second method to find an integrating factor. Let

$$\begin{array}{rl}B& =\frac{1}{M}(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})\\ & =-\frac{1}{x(a{y}^2+b)}(\left(0\right)-(-2xay))\\ & =-\frac{2ay}{a{y}^2+b}\end{array}$$

Since $B$ does not depend on $x$, it can be used to obtain an integrating factor. Let the integrating factor be $\mu$. Then

$$\begin{array}{rl}\mu & =e^{\int B\mathrm{d}y}\\ & =e^{\int -\frac{2ay}{a{y}^2+b}\mathrm{d}y}\end{array}$$

The result of integrating gives

$$\begin{array}{rl}\mu & =e^{-\ln (a{y}^2+b)}\\ & =\frac{1}{a{y}^2+b}\end{array}$$

$M$ and $N$ are now multiplied by this integrating factor, giving new $M$ and new $N$ which are called $\bar{M}$ and $\bar{N}$ so not to confuse them with the original $M$ and $N$.

$$\begin{array}{rl}\bar{M}& =\mu M\\ & =\frac{1}{a{y}^2+b}(-x(a{y}^2+b))\\ & =-x\end{array}$$

And

$$\begin{array}{rl}\bar{N}& =\mu N\\ & =\frac{1}{a{y}^2+b}\left(1\right)\\ & =\frac{1}{a{y}^2+b}\end{array}$$

So now a modified ODE is obtained from the original ODE which will be exact and can be solved using the standard method. The modified ODE is

$$\begin{array}{rl}\bar{M}+\bar{N}\frac{\mathrm{d}y}{\mathrm{d}x}& =0\\ (-x)+\left(\frac{1}{a{y}^2+b}\right)\frac{\mathrm{d}y}{\mathrm{d}x}& =0\end{array}$$

The following equations are now set up to solve for the function $\varphi (x,y)$

$$\begin{array}{}\text{(1)}& \frac{\partial \varphi }{\partial x}& =\bar{M}\text{(2)}& \frac{\partial \varphi }{\partial y}& =\bar{N}\end{array}$$

Integrating (1) w.r.t. $x$ gives

$$\begin{array}{rl}\int \frac{\partial \varphi }{\partial x}\mathrm{d}x& =\int \bar{M}\mathrm{d}x\\ \int \frac{\partial \varphi }{\partial x}\mathrm{d}x& =\int -x\mathrm{d}x\\ \text{(3)}& \varphi & =-\frac{x^2}{2}+f(y)\end{array}$$

Where $f(y)$ is used for the constant of integration since $\varphi$ is a function of both $x$ and $y$. Taking derivative of equation (3) w.r.t $y$ gives

$$\begin{array}{}\text{(4)}& \frac{\partial \varphi }{\partial y}=0+f^{\prime }(y)\end{array}$$

But equation (2) says that $\frac{\partial \varphi }{\partial y}=\frac{1}{a{y}^2+b}$. Therefore equation (4) becomes

$$\begin{array}{}\text{(5)}& \frac{1}{a{y}^2+b}=0+f^{\prime }(y)\end{array}$$

Solving equation (5) for $f^{\prime }(y)$ gives

$$f^{\prime }(y)=\frac{1}{a{y}^2+b}$$

Integrating the above w.r.t $y$ gives

$$\begin{array}{rl}\int f^{\prime }(y)\mathrm{d}y& =\int \left(\frac{1}{a{y}^2+b}\right)\mathrm{d}y\\ f(y)& =\frac{\arctan \left(\frac{ay}{\sqrt{ba}}\right)}{\sqrt{ba}}+c_2\end{array}$$

Where $c_2$ is constant of integration. Substituting result found above for $f(y)$ into equation (3) gives $\varphi$

$$\varphi =-\frac{x^2}{2}+\frac{\arctan \left(\frac{ay}{\sqrt{ba}}\right)}{\sqrt{ba}}+c_2$$

But since $\varphi$ itself is a constant function, then let $\varphi =c_3$ where $c_2$ is new constant and combining $c_2$ and $c_3$ constants into the constant $c_2$ gives the solution as

$$c_2=-\frac{x^2}{2}+\frac{\arctan \left(\frac{ay}{\sqrt{ba}}\right)}{\sqrt{ba}}$$

Solving for $y$ gives

$$\begin{array}{rl}y& =\frac{\tan (\frac{x^2\sqrt{ba}}{2}+c_2\sqrt{ba})\sqrt{ba}}{a}\end{array}$$

Which simplifies to

$$\begin{array}{rl}y& =\frac{\tan \left(\sqrt{ba}(\frac{x^2}{2}+c_2)\right)\sqrt{ba}}{a}\end{array}$$

Summary of solutions found

$$\begin{array}{rl}y& =\frac{\tan \left(\sqrt{ba}(\frac{x^2}{2}+c_2)\right)\sqrt{ba}}{a}\end{array}$$

Solved using first_order_ode_LIE

Time used: 0.394 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime }& =x(a{y}^2+b)\end{array}$$

Writing the ode as

$$\begin{array}{rl}{y}^{\prime }& =x(a{y}^2+b)\\ y^{\prime }& =\omega (x,y)\end{array}$$

The condition of Lie symmetry is the linearized PDE given by

$$\begin{array}{}\text{(A)}& {\eta }_x+\omega ({\eta }_y-{\xi }_x)-{\omega }^2{\xi }_y-{\omega }_x\xi -{\omega }_y\eta =0\end{array}$$

To determine $\xi ,\eta$ then (A) is solved using ansatz. Making bivariate polynomials of degree 2 to use as anstaz gives

$$\begin{array}{}\text{(1E)}& \xi & =x^2{a}_4+yx{a}_5+y^2{a}_6+x{a}_2+y{a}_3+a_1\text{(2E)}& \eta & =x^2{b}_4+yx{b}_5+y^2{b}_6+x{b}_2+y{b}_3+b_1\end{array}$$

Where the unknown coefficients are

$$\{a_1,a_2,a_3,a_4,a_5,a_6,b_1,b_2,b_3,b_4,b_5,b_6\}$$

Substituting equations (1E,2E) and $\omega$ into (A) gives

$$\begin{array}{}\text{(5E)}& 2x{b}_4+y{b}_5+b_2+x(a{y}^2+b)(-2x{a}_4+x{b}_5-y{a}_5+2y{b}_6-a_2+b_3)-x^2{(a{y}^2+b)}^2(x{a}_5+2y{a}_6+a_3)-(a{y}^2+b)(x^2{a}_4+yx{a}_5+y^2{a}_6+x{a}_2+y{a}_3+a_1)-2xay(x^2{b}_4+yx{b}_5+y^2{b}_6+x{b}_2+y{b}_3+b_1)=0\end{array}$$

Putting the above in normal form gives

$$-a^2{x}^3{y}^4{a}_5-2a^2{x}^2{y}^5{a}_6-a^2{x}^2{y}^4{a}_3-2ab{x}^3{y}^2{a}_5-4ab{x}^2{y}^3{a}_6-2ab{x}^2{y}^2{a}_3-2a{x}^{3}y{b}_4-3a{x}^2{y}^2{a}_4-a{x}^2{y}^2{b}_5-2ax{y}^3{a}_5-a{y}^4{a}_6-b^2{x}^3{a}_5-2b^2{x}^{2}y{a}_6-2a{x}^{2}y{b}_2-2ax{y}^2{a}_2-ax{y}^2{b}_3-a{y}^3{a}_3-b^2{x}^2{a}_3-2axy{b}_1-a{y}^2{a}_1-3b{x}^2{a}_4+b{x}^2{b}_5-2bxy{a}_5+2bxy{b}_6-b{y}^2{a}_6-2bx{a}_2+bx{b}_3-by{a}_3-b{a}_1+2x{b}_4+y{b}_5+b_2=0$$

Setting the numerator to zero gives

$$\begin{array}{}\text{(6E)}& -a^2{x}^3{y}^4{a}_5-2a^2{x}^2{y}^5{a}_6-a^2{x}^2{y}^4{a}_3-2ab{x}^3{y}^2{a}_5-4ab{x}^2{y}^3{a}_6-2ab{x}^2{y}^2{a}_3-2a{x}^{3}y{b}_4-3a{x}^2{y}^2{a}_4-a{x}^2{y}^2{b}_5-2ax{y}^3{a}_5-a{y}^4{a}_6-b^2{x}^3{a}_5-2b^2{x}^{2}y{a}_6-2a{x}^{2}y{b}_2-2ax{y}^2{a}_2-ax{y}^2{b}_3-a{y}^3{a}_3-b^2{x}^2{a}_3-2axy{b}_1-a{y}^2{a}_1-3b{x}^2{a}_4+b{x}^2{b}_5-2bxy{a}_5+2bxy{b}_6-b{y}^2{a}_6-2bx{a}_2+bx{b}_3-by{a}_3-b{a}_1+2x{b}_4+y{b}_5+b_2=0\end{array}$$

Looking at the above PDE shows the following are all the terms with $\{x,y\}$ in them.

$$\{x,y\}$$

The following substitution is now made to be able to collect on all terms with $\{x,y\}$ in them

$$\{x=v_1,y=v_2\}$$

The above PDE (6E) now becomes

$$\begin{array}{}\text{(7E)}& -a^2{a}_5{v}_1^3{v}_2^4-2a^2{a}_6{v}_1^2{v}_2^5-a^2{a}_3{v}_1^2{v}_2^4-2ab{a}_5{v}_1^3{v}_2^2-4ab{a}_6{v}_1^2{v}_2^3-2ab{a}_3{v}_1^2{v}_2^2-3a{a}_4{v}_1^2{v}_2^2-2a{a}_5{v}_1{v}_2^3-a{a}_6{v}_2^4-2a{b}_4{v}_1^3{v}_2-a{b}_5{v}_1^2{v}_2^2-b^2{a}_5{v}_1^3-2b^2{a}_6{v}_1^2{v}_2-2a{a}_2{v}_1{v}_2^2-a{a}_3{v}_2^3-2a{b}_2{v}_1^2{v}_2-a{b}_3{v}_1{v}_2^2-b^2{a}_3{v}_1^2-a{a}_1{v}_2^2-2a{b}_1{v}_1{v}_2-3b{a}_4{v}_1^2-2b{a}_5{v}_1{v}_2-b{a}_6{v}_2^2+b{b}_5{v}_1^2+2b{b}_6{v}_1{v}_2-2b{a}_2{v}_1-b{a}_3{v}_2+b{b}_3{v}_1-b{a}_1+2b_4{v}_1+b_5{v}_2+b_2=0\end{array}$$

Collecting the above on the terms $v_i$ introduced, and these are

$$\{v_1,v_2\}$$

Equation (7E) now becomes

$$\begin{array}{}\text{(8E)}& -a^2{a}_5{v}_1^3{v}_2^4-2ab{a}_5{v}_1^3{v}_2^2-2a{b}_4{v}_1^3{v}_2-b^2{a}_5{v}_1^3-2a^2{a}_6{v}_1^2{v}_2^5-a^2{a}_3{v}_1^2{v}_2^4-4ab{a}_6{v}_1^2{v}_2^3+(-2ab{a}_3-3a{a}_4-a{b}_5)v_1^2{v}_2^2+(-2b^2{a}_6-2a{b}_2)v_1^2{v}_2+(-b^2{a}_3-3b{a}_4+b{b}_5)v_1^2-2a{a}_5{v}_1{v}_2^3+(-2a{a}_2-a{b}_3)v_1{v}_2^2+(-2a{b}_1-2b{a}_5+2b{b}_6)v_1{v}_2+(-2b{a}_2+b{b}_3+2b_4)v_1-a{a}_6{v}_2^4-a{a}_3{v}_2^3+(-a{a}_1-b{a}_6)v_2^2+(-b{a}_3+b_5)v_2-b{a}_1+b_2=0\end{array}$$

Setting each coefficients in (8E) to zero gives the following equations to solve

$$\begin{array}{rl}-a{a}_3& =0\\ -2a{a}_5& =0\\ -a{a}_6& =0\\ -2a{b}_4& =0\\ -a^2{a}_3& =0\\ -a^2{a}_5& =0\\ -2a^2{a}_6& =0\\ -b^2{a}_5& =0\\ -2ab{a}_5& =0\\ -4ab{a}_6& =0\\ -2a{a}_2-a{b}_3& =0\\ -b{a}_1+b_2& =0\\ -b{a}_3+b_5& =0\\ -a{a}_1-b{a}_6& =0\\ -2b^2{a}_6-2a{b}_2& =0\\ -2b{a}_2+b{b}_3+2b_4& =0\\ -b^2{a}_3-3b{a}_4+b{b}_5& =0\\ -2ab{a}_3-3a{a}_4-a{b}_5& =0\\ -2a{b}_1-2b{a}_5+2b{b}_6& =0\end{array}$$

Solving the above equations for the unknowns gives

$$\begin{array}{rl}{a}_1& =0\\ a_2& =0\\ a_3& =0\\ a_4& =0\\ a_5& =0\\ a_6& =0\\ b_1& =b_1\\ b_2& =0\\ b_3& =0\\ b_4& =0\\ b_5& =0\\ b_6& =\frac{a{b}_1}{b}\end{array}$$

Substituting the above solution in the anstaz (1E,2E) (using $1$ as arbitrary value for any unknown in the RHS) gives

$$\begin{array}{rl}\xi & =0\\ \eta & =\frac{a{y}^2+b}{b}\end{array}$$

The next step is to determine the canonical coordinates $R,S$. The canonical coordinates map $(x,y)\to (R,S)$ where $(R,S)$ 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

$$\begin{array}{}\text{(1)}& \frac{dx}{\xi }& =\frac{dy}{\eta }=dS\end{array}$$

The above comes from the requirements that $(\xi \frac{\partial }{\partial x}+\eta \frac{\partial }{\partial y})S(x,y)=1$. 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

$$\begin{array}{r}R=x\end{array}$$

$S$ is found from

$$\begin{array}{rl}S& =\int \frac{1}{\eta }dy\\ & =\int \frac{1}{\frac{a{y}^2+b}{b}}dy\end{array}$$

Which results in

$$\begin{array}{rl}S& =\frac{b\arctan \left(\frac{ay}{\sqrt{ba}}\right)}{\sqrt{ba}}\end{array}$$

Now that $R,S$ are found, we need to setup the ode in these coordinates. This is done by evaluating

$$\begin{array}{}\text{(2)}& \frac{dS}{dR}& =\frac{S_x+\omega (x,y)S_y}{R_x+\omega (x,y)R_y}\end{array}$$

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

$$\begin{array}{rl}\omega (x,y)& =x(a{y}^2+b)\end{array}$$

Evaluating all the partial derivatives gives

$$\begin{array}{rl}{R}_x& =1\\ R_y& =0\\ S_x& =0\\ S_y& =\frac{b}{a{y}^2+b}\end{array}$$

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.

$$\begin{array}{}\text{(2A)}& \frac{dS}{dR}& =bx\end{array}$$

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

$$\begin{array}{rl}\frac{dS}{dR}& =bR\end{array}$$

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 $R,S$.

Since the ode has the form $\frac{d}{dR}S\left(R\right)=f(R)$, then we only need to integrate $f(R)$.

$$\begin{array}{rl}\int dS& =\int bRdR\\ S\left(R\right)& =\frac{b{R}^2}{2}+c_5\end{array}$$

To complete the solution, we just need to transform the above back to $x,y$ coordinates. This results in

$$\begin{array}{r}\frac{\sqrt{b}\arctan \left(\frac{\sqrt{a}y}{\sqrt{b}}\right)}{\sqrt{a}}=\frac{b{x}^2}{2}+c_5\end{array}$$

Solving for $y$ gives

$$\begin{array}{rl}y& =\frac{\sqrt{b}\tan \left(\frac{\sqrt{a}(b{x}^2+2c_5)}{2\sqrt{b}}\right)}{\sqrt{a}}\end{array}$$

Summary of solutions found

$$\begin{array}{rl}y& =\frac{\sqrt{b}\tan \left(\frac{\sqrt{a}(b{x}^2+2c_5)}{2\sqrt{b}}\right)}{\sqrt{a}}\end{array}$$

Solved using first_order_ode_riccati

Time used: 1.382 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime }& =x(a{y}^2+b)\end{array}$$

In canonical form the ODE is

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

This is a Riccati ODE. Comparing the ODE to solve

$$y^{\prime }=ax{y}^2+bx$$

With Riccati ODE standard form

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

Shows that $f_0(x)=bx$, $f_1(x)=0$ and $f_2(x)=ax$. Let

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

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

$$\begin{array}{rl}{f}_2^{\prime }& =a\\ f_1{f}_2& =0\\ f_2^2{f}_0& =a^2{x}^{3}b\end{array}$$

Substituting the above terms back in equation (2) gives

$$\begin{array}{r}ax{u}^{\prime \prime }\left(x\right)-a{u}^{\prime }\left(x\right)+a^2{x}^{3}bu\left(x\right)=0\end{array}$$

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

Time used: 0.661 (sec)

In normal form the ode

$$\begin{array}{}\text{(1)}& ax{u}^{\prime \prime }-a{u}^{\prime }+a^2{x}^{3}bu& =0\end{array}$$

Becomes

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

Where

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

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

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

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

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

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

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

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

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

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 $A=1,B=0,C=ba$ into the above gives

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

Hence

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

Which simplifies to

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

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

$$\begin{array}{rl}u& =c_6{e}^{{\lambda }_1\tau }+c_7{e}^{{\lambda }_2\tau }\\ & =c_6{e}^{\frac{\tau ((-1+i)\sqrt{\mathrm{signum}\left(ba\right)}+1+i)\sqrt{ba}}{2}}+c_7{e}^{-\frac{\tau ((-1+i)\sqrt{\mathrm{signum}\left(ba\right)}+1+i)\sqrt{ba}}{2}}\end{array}$$

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

$$u\left(x\right)=c_6{\mathrm{e}}^{\frac{x^2((-1+i)\sqrt{\mathrm{signum}\left(ba\right)}+1+i)\sqrt{ba}}{4}}+c_7{\mathrm{e}}^{-\frac{x^2((-1+i)\sqrt{\mathrm{signum}\left(ba\right)}+1+i)\sqrt{ba}}{4}}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}u\left(x\right)& =c_6{\mathrm{e}}^{\frac{x^2((-1+i)\sqrt{\mathrm{signum}\left(ba\right)}+1+i)\sqrt{ba}}{4}}+c_7{\mathrm{e}}^{-\frac{x^2((-1+i)\sqrt{\mathrm{signum}\left(ba\right)}+1+i)\sqrt{ba}}{4}}\end{array}$$

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

Time used: 0.102 (sec)

Solve

$$\begin{array}{rl}ax{u}^{\prime \prime }-a{u}^{\prime }+a^2{x}^{3}bu& =0\end{array}$$

In normal form the ode

$$\begin{array}{}\text{(1)}& ax{u}^{\prime \prime }-a{u}^{\prime }+a^2{x}^{3}bu& =0\end{array}$$

Becomes

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

Where

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

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)

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

Substituting the above into (4) results in

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

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)

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

Substituting the above into the solution obtained gives

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

Will add steps showing solving for IC soon.

Summary of solutions found

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

Solved as second order Bessel ode

Time used: 0.082 (sec)

Solve

$$\begin{array}{rl}ax{u}^{\prime \prime }-a{u}^{\prime }+a^2{x}^{3}bu& =0\end{array}$$

Writing the ode as

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

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

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

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

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

Will add steps showing solving for IC soon.

Summary of solutions found

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

Solved as second order ode using Kovacic algorithm

Time used: 0.401 (sec)

Solve

$$\begin{array}{rl}ax{u}^{\prime \prime }-a{u}^{\prime }+a^2{x}^{3}bu& =0\end{array}$$

Writing the ode as

$$\begin{array}{}\text{(1)}& ax{u}^{\prime \prime }-a{u}^{\prime }+a^2{x}^{3}bu& =0\text{(2)}& A{u}^{\prime \prime }+B{u}^{\prime }+Cu& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =ax\\ \text{(3)}& B& =-aC& =a^2{x}^{3}b\end{array}$$

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

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

Comparing the above to (5) shows that

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

Therefore eq. (4) becomes

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

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 2.11: 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

$$r=-a{x}^{2}b+\frac{3}{4x^2}$$

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

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

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

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

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

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

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

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

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

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

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

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

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

$$r=\frac{-4a{x}^{4}b+3}{4x^2}$$

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 $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$-2$	$i\sqrt{b}\sqrt{a}x$	$-\frac{1}{2}$	$-\frac{1}{2}$

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

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

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

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

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

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

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

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

Which simplifies to

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

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

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

Therefore the solution is

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

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}u& =c_1{\mathrm{e}}^{-\frac{i{x}^2\sqrt{ba}}{2}}-\frac{i{c}_2{\mathrm{e}}^{\frac{i{x}^2\sqrt{ba}}{2}}}{2\sqrt{ba}}\end{array}$$

Taking derivative gives

$$u^{\prime }\left(x\right)=\frac{c_{6}x((-1+i)\sqrt{\mathrm{signum}\left(ba\right)}+1+i)\sqrt{ba}{\mathrm{e}}^{\frac{x^2((-1+i)\sqrt{\mathrm{signum}\left(ba\right)}+1+i)\sqrt{ba}}{4}}}{2}-\frac{c_{7}x((-1+i)\sqrt{\mathrm{signum}\left(ba\right)}+1+i)\sqrt{ba}{\mathrm{e}}^{-\frac{x^2((-1+i)\sqrt{\mathrm{signum}\left(ba\right)}+1+i)\sqrt{ba}}{4}}}{2}$$

Doing change of constants, the solution becomes

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

Summary of solutions found

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

✓ Maple. Time used: 0.005 (sec). Leaf size: 28

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

$$y=\frac{\tan \left(\frac{\sqrt{ba}(x^2+2c_1)}{2}\right)\sqrt{ba}}{a}$$

Maple trace

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

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d}{dx}y\left(x\right)=x(ay{\left(x\right)}^2+b)\\ \bullet & & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Solve for the highest derivative}\\ & & \frac{d}{dx}y\left(x\right)=x(ay{\left(x\right)}^2+b)\\ \bullet & & \text{Separate variables}\\ & & \frac{\frac{d}{dx}y\left(x\right)}{ay{\left(x\right)}^2+b}=x\\ \bullet & & \text{Integrate both sides with respect to}x\\ & & \int \frac{\frac{d}{dx}y\left(x\right)}{ay{\left(x\right)}^2+b}dx=\int xdx+\mathit{C}\mathit{1}\\ \bullet & & \text{Evaluate integral}\\ & & \frac{\arctan \left(\frac{ay\left(x\right)}{\sqrt{ba}}\right)}{\sqrt{ba}}=\frac{x^2}{2}+\mathit{C}\mathit{1}\\ \bullet & & \text{Solve for}y\left(x\right)\\ & & y\left(x\right)=\frac{\tan (\frac{x^2\sqrt{ba}}{2}+\mathit{C}\mathit{1}\sqrt{ba})\sqrt{ba}}{a}\\ \bullet & & \text{Simplify}\\ & & y\left(x\right)=\frac{\tan \left(\sqrt{ba}(\frac{x^2}{2}+\mathit{C}\mathit{1})\right)\sqrt{ba}}{a}\\ \bullet & & \text{Redefine the integration constant(s)}\\ & & y\left(x\right)=\frac{\tan (\mathit{C}\mathit{1}+\frac{x^2\sqrt{ba}}{2})\sqrt{ba}}{a}\end{array}$$

✓ Mathematica. Time used: 8.114 (sec). Leaf size: 75

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

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

✓ Sympy. Time used: 1.585 (sec). Leaf size: 63

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

$$-\frac{x^2}{2}-\frac{\sqrt{-\frac{1}{ab}}\log (-b\sqrt{-\frac{1}{ab}}+y\left(x\right))}{2}+\frac{\sqrt{-\frac{1}{ab}}\log (b\sqrt{-\frac{1}{ab}}+y\left(x\right))}{2}=C_1$$