3.3.22.1 Solution method

(This all Need to be revised, as I am using different transformation here than described above, I need to clarify all of this).

Find what is called the abel invariant and check if constant.

$$\mathrm{\Delta }=-\frac{{(-f_0^{\prime }{f}_3+f_0{f}_3^{\prime }+3f_0{f}_3{f}_1)}^3}{27f_3^4{f}_0^5}$$

The substitution $y=\frac{1}{u}$ is now applied. Therefore $y^{\prime }=-\frac{1}{u^2}{u}^{\prime }$. Substituting this in (1) gives

$$\begin{array}{rl}-\frac{1}{u^2}{u}^{\prime }& =f_0(x)+f_1(x)\frac{1}{u}+f_2(x)\frac{1}{u^2}+f_3(x)\frac{1}{u^3}\\ -u{u}^{\prime }& =u^3{f}_0(x)+u^2{f}_1(x)+u{f}_2(x)+f_3(x)\\ \text{(2)}& u{u}^{\prime }& =-u^3{f}_0(x)-u^2{f}_1(x)-u{f}_2(x)-f_3(x)\end{array}$$

Using the substitution $u=\frac{1}{E}(y+\frac{f_2}{3f_3})$ where $E=\exp (\int f_1-\frac{f_2^2}{3f_3}dx)$ in the above gives

$$\frac{1}{E}(y+\frac{f_2}{3f_3})u^{\prime }=-u^3{f}_0(x)-u^2{f}_1(x)-u{f}_2(x)-f_3(x)$$

Hence

$$\begin{array}{rl}{u}^{\prime }& =\frac{1}{E^2}\frac{dE}{dx}(y+\frac{f_2}{3f_3})+\frac{1}{E}(y^{\prime }+\frac{1}{3}\frac{f_2^{\prime }{f}_3-f_2{f}_3^{\prime }}{f_3^2})\\ & =\frac{1}{E^2}\frac{dE}{dx}(\frac{1}{u}+\frac{f_2}{3f_3})+\frac{1}{E}(-\frac{1}{u^2}{u}^{\prime }+\frac{1}{3}\frac{f_2^{\prime }{f}_3-f_2{f}_3^{\prime }}{f_3^2})\\ u^{\prime }+\frac{u^{\prime }}{E{u}^2}& =\frac{1}{E^2}\frac{dE}{dx}(\frac{1}{u}+\frac{f_2}{3f_3})+\frac{1}{3E}\frac{f_2^{\prime }{f}_3-f_2{f}_3^{\prime }}{f_3^2}\\ u^{\prime }(1+\frac{1}{E{u}^2})& =\frac{1}{E^2}\frac{dE}{dx}(\frac{1}{u}+\frac{f_2}{3f_3})+\frac{1}{3E}\frac{f_2^{\prime }{f}_3-f_2{f}_3^{\prime }}{f_3^2}\\ u^{\prime }& =\frac{E{u}^2}{1+E{u}^2}(\frac{1}{E^2}\frac{dE}{dx}(\frac{1}{u}+\frac{f_2}{3f_3})+\frac{1}{3E}\frac{f_2^{\prime }{f}_3-f_2{f}_3^{\prime }}{f_3^2})\\ u^{\prime }& =\frac{u^2}{1+E{u}^2}(\frac{1}{E}\frac{dE}{dx}(\frac{1}{u}+\frac{f_2}{3f_3})+\frac{1}{3}\frac{f_2^{\prime }{f}_3-f_2{f}_3^{\prime }}{f_3^2})\end{array}$$

Substituting the above into (2) gives

$$u\frac{u^2}{1+E{u}^2}(\frac{1}{E}\frac{dE}{dx}(\frac{1}{u}+\frac{f_2}{3f_3})+\frac{1}{3}\frac{f_2^{\prime }{f}_3-f_2{f}_3^{\prime }}{f_3^2})=-u^3{f}_0-u^2{f}_1-u{f}_2-f_3$$

Therefore

$$\begin{array}{rl}E& =\exp (\int f_1\left(x\right)-\frac{f_2^2\left(x\right)}{3f_3\left(x\right)}dx)\\ \xi & =\int f_3\left(x\right)E^{2}dx\\ u& =\frac{1}{E}(y+\frac{f_2\left(x\right)}{3f_3\left(x\right)})\end{array}$$

The above are used to convert the first kind Abel ode to canonical form. (To finish).