2nd order non linear differential equation A physical problem that I start to solve led me to this Second order non linear differential equation:
$$f(x)\biggl(\frac{\text d^2y}{\text dx^2}-f(x)y^\alpha\biggr)=a \frac{\text dy}{\text dx}+b $$
with this conditions: $y(0)=0$ and $y'(0)=0$
and where $y = y(x)$ and $a,b,\alpha$ are constants. The $f(x)$ is a simpler polynomial of the type:
$$ f(x) = c+dx$$ with $c,d$ constants. I try different substitutions (for exemple $u=y^\alpha$, $u=y^\alpha y'$) to find analytical solution but I could not to solve the problem. Can anyone help me or suggest some ideas to solve the equation. Thanks
 A: This is not an answer to the full problem, but may provide some qualitative information.
Case $c=d=0$. The only solution is the zero function, which corresponds to $b=0$. Otherwise, no solution is found.
Case $a=d=0$. Multiplication of the original differential equation by $y' = \frac{\text d}{\text d x} y$ leads to
$c y' y'' = c^2 y' y^\alpha + b y'$.
Integration of the previous equation gives the differential equation
$$
(y')^2 = y \left(A y^{\alpha} + B\right) , \qquad A = 2\tfrac{c}{1 + \alpha}, \qquad B =  2\tfrac{b}{c}
$$
where $\alpha > -1$ (the initial condition $y'(0) = 0$ has been used). The left-hand side is non-negative, and so is the right-hand side too, which allows us to rewrite the problem as
$$
y' = \pm \sqrt{ y \left(A y^{\alpha} + B\right)}
\qquad\text{with}\qquad
y \left(A y^{\alpha} + B\right) \geq 0 .
$$
The value $y=0$ is an equilibrium of the differential equation, which tells that the unique solution to the IVP is the zero function, obtained for $\alpha > -1$.
The local existence and uniqueness of solutions in the vicinity of $x=0^+$ is highly dependent on the values of the parameters. Since the problem comes from physics, there may be estimates or bounds on the parameters that can be used to analyze local existence and uniqueness before trying to solve the full IVP (quasi)-analytically or numerically.
A: Hint:
Assume $d\neq0$ for the key case:
Let $r=x+\dfrac{c}{d}$ ,
Then $dr\dfrac{\text d^2y}{\text dr^2}-d^2r^2y^\alpha=a\dfrac{\text dy}{\text dr}+b$
$dr\dfrac{\text d^2y}{\text dr^2}-a\dfrac{\text dy}{\text dr}-d^2r^2y^\alpha-b=0$
$\dfrac{\text d^2y}{\text dr^2}-\dfrac{a}{dr}\dfrac{\text dy}{\text dr}-dry^\alpha-\dfrac{b}{dr}=0$
Let $y=r^ku$ ,
Then $\dfrac{\text dy}{\text dr}=r^k\dfrac{\text du}{\text dr}+kr^{k-1}u$
$\dfrac{\text d^2y}{\text dr^2}=r^k\dfrac{\text d^2u}{\text dr^2}+kr^{k-1}\dfrac{\text du}{\text dr}+kr^{k-1}\dfrac{\text du}{\text dr}+k(k-1)r^{k-2}u=r^k\dfrac{\text d^2u}{\text dr^2}+2kr^{k-1}\dfrac{\text du}{\text dr}+k(k-1)r^{k-2}u$
$\therefore r^k\dfrac{\text d^2u}{\text dr^2}+2kr^{k-1}\dfrac{\text du}{\text dr}+k(k-1)r^{k-2}u-\dfrac{a}{dr}\left(r^k\dfrac{\text du}{\text dr}+kr^{k-1}u\right)-dr^{\alpha k+1}u^\alpha-\dfrac{b}{dr}=0$
$r^k\dfrac{\text d^2u}{\text dr^2}+\dfrac{(2dk-a)r^{k-1}}{d}\dfrac{\text du}{\text dr}+\dfrac{k(dk-a-d)r^{k-2}u}{d}-dr^{\alpha k+1}u^\alpha-\dfrac{b}{dr}=0$
Choose $k=\dfrac{a}{2d}$ , the ODE becomes
$r^\frac{a}{2d}\dfrac{\text d^2u}{\text dr^2}-\dfrac{a(a+2d)}{4d^2}r^{\frac{a}{2d}-2}u-dr^{\frac{a\alpha}{2d}+1}u^\alpha-\dfrac{b}{dr}=0$
$\dfrac{\text d^2u}{\text dr^2}-\dfrac{a(a+2d)}{4d^2}r^{-2}u-dr^{\frac{a(\alpha-1)}{2d}+1}u^\alpha-\dfrac{b}{dr^{\frac{a}{2d}+1}}=0$
