$$3(y-1)^2 = y''$$ where $y(0) = 3$, $y'(0) = 4$ and we assume $y \neq 1$.
I know how to solve IVPs in when the 2nd order DE is in the form:
$$p(x)y'' + q(x)y' + g(x)y = 0,$$
however, I think this problem shows a non-linear 2nd order DE based on the "vertex" form of $y$ which is $(y-1)^2$.
I tried substitution with $v = y'$ and expanding the squared expression, but it's getting me no where.
Does anyone know how to solve this?
Thanks
I believe the way to solve this is to multiply both sides by $y'$
$$3(y-1)^2y'=y'y''$$ $$(y-1)^3=\frac12(y')^2+C$$ $$(3-1)^3=\frac12(4)^2+C,C=0$$ $$y'=\sqrt2(y-1)^{3/2}$$ $$(y-1)^{-3/2}dy=\sqrt2dt$$ $$-\frac2{\sqrt{y-1}}=\sqrt2t+C$$ $$-\frac2{\sqrt{3-1}}=C, C=-\sqrt2$$ $$\sqrt{y-1}=-\frac2{\sqrt2t-\sqrt2}$$ $$y-1=\frac2{t^2-2t+1}$$ $$y=1+\frac2{t^2-2t+1}$$