Analytically solving nonlinear second order ODE

Question

I need help with providing an answer to this nonlinear ODE

$a_1 + f_1(x) + f_2(x) y' - a_2\bigg((y')^2 - y''\bigg) = 0,$

where the $a_i$'s are constants and the $f_i$'s are arbitrary functions of $x$.

Any help is appreciated.

Answer 1

It is proved below that solving the ODE : $$a_1 + f_1(x) + f_2(x) y' - a_2\bigg((y')^2 - y''\bigg) = 0,$$ with any functions $f_1(x)$ and $f_2(x)$ requires to know how to solve the general second order linear ODE with any functions as coefficients.

This is only known for particular functions, but not for any functions.

So, do not expect to obtain a general solution. Only if you define the functions $f_1(x)$ and $f_2(x)$ you can expect analytically solve the ODE, in some particular cases. What is more, expecting it don't mean that it will be possible, depending the form of the given functions.

Answer 2

REduce the oRder with $p=\frac{dy}{dx}$

Answer 3

$$ a_2y''-a_2y'^2+f_2(x)y' +f_1(x) + a_1 = 0 $$ let $y' = p$ $$ a_2p' -a_2p^2+f_2(x)p + f_1(x)+a_1 =0 = a_2\left[p' -\left(p^2-\frac{f_2(x)}{a_2}p\right) + g_1(x)\right] = 0 $$ wlog $\frac{f_1(x) +a_1}{a_2} = g_1(x)$ $$ p' -\left(p-\frac{f_2(x)}{2a_2}\right)^2 + g_1(x) -\left(\frac{f_2(x)}{2a_2}\right)^2 =0 $$ set $v(x) = p-\frac{f_2(x)}{2a_2}$ we find $$ v' + \left(\frac{f_2(x)}{2a_2}\right)' -v^2 + g_1(x) -\left(\frac{f_2(x)}{2a_2}\right)^2 = 0\\ v' -v^2 = \left(\frac{f_2(x)}{2a_2}\right)^2 -\left(\frac{f_2(x)}{2a_2}\right)'-g_1(x) $$ Analytical solutions will depends on special cases of the rhs.