Bernoulli Substitution for differential equation

Question

$$x'+2t^2x=2t^2x^3 $$ I made the Bernoulli Substitution $$u=\frac{1}{x^2}$$ therefore $$u'=-2x^{-3}x'$$

then after some conversions I had the following equation $$u=4t^2u-4t^2$$ however I had the solution and the I put x again in but my problem was that I had a term like this $$x=\frac{1}{\sqrt{(ce^\frac{4t^3}{3} +1)}}$$ but the right solution should be $$x=\frac{1}{\sqrt{(e^{\frac{4t^3}{3}+c} +1)}}$$ I dont know where my mistake was can anyone help me with this? Thanks,Ciwan

Answer 1

$$x=\frac{1}{\sqrt{(ce^\frac{4t^3}{3} +1)}}$$and $$x=\frac{1}{\sqrt{(e^{\frac{4t^3}{3}+c} +1)}}$$

are actually the same thing! Note that $c$ in both equations are different though.

This is because $ce^\frac{4t^3}{3}=e^ke^\frac{4t^3}{3}=e^{\frac{4t^3}{3}+k}$, so they're equal so long as $c$ is an "arbitrary constant"

Answer 2

Your answer is correct.

Assume $c = e^k$,

then $ce^{\frac{4t^3}{3}} = e^ke^{\frac{4t^3}{3}} = e^{\frac{4t^3}{3}+k}$

where $c$ and $k$ are arbitrary constants

Answer 3

This equation is separable: $$\frac{x'(t)}{x(t)(-1+x(t)^2)}=2t^3$$ $$x \left( t \right) = \left( 1+{{\rm e}^{2/3\,{t}^{3}}}{\it \_C1} \right) ^{-1} $$