Solving differential equation of third degree If differential equation of the curves $$c(y+c)^2 = x^3$$ where 'c' is an arbitary constant is $$12y(y')^2 + ax = bx(y')^3$$
What is the value of a+b?
I tried differentiating the curve given and obtain a linear relation with the equation.
I got $$cyy'y'' + c(y')^3 + c^2y'y'' = 3x$$
But in the equation 'x' appears alongwith $(y')^3$.
I'm stuck here.
 A: Let $y'=z$. Differentiating given equ wrt $x$, we will get
$y+c=\frac{2xz}{3}$  ---------$2$
Now, put this in the original equation, so you will get $c=\frac{9x}{4z^2}$
Now, again put the value of $c$ in the equation $2$ mentioned above. After simplifying, you will get
$$12z^2y+27x=8z^3x$$ Now $a+b=35$
Hope this will be helpful !
A: First calculate the implicit derivative
$$2c(y+c)dy=3x^2dx.$$
Now devide by $dx$ and solve for $dy/dx$, to get the ODE having your curve as a solution:
$$\dfrac{dy}{dx}=\frac{3}{2c}\frac{x^2}{y+c}.$$
The given ODE in your question is: 
$$bx(y')^3-12y(y')^2-ax=0.$$
Use cardanos formula for the cubic eqatuion, to solve for $y'=dy/dx$. Compare with the ODE from the implicit derivative approach to determine the coefficients $a$ and $b$.  
A: $$c(y+c)^2=x^3$$
Differentiate :
$$2c(y+c)y'=3x^2$$
$2c(y+c)^2y'=3x^2(y+c)=2x^3y'$
$$y+c=\frac{2}{3}xy'\quad\to\quad c=\frac{2}{3}xy'-y$$
$c(y+c)^2=x^3=(\frac{2}{3}xy'-y)(\frac{2}{3}xy')^2$
$x=\frac{8}{27}xy'^3-\frac{4}{9}yy'^2$
$-\frac{8}{27}xy'^3+\frac{4}{9}yy'^2+x=0$
$$12yy'^2+27x-8xy'^3=0$$
Compared to 
$$12yy'^2+ax-bxy'^3=0$$ 
$$a=27\text{  and  }b=8 \quad a+b=35$$
