Cubic equations

Question

I have a solution to a differential equation $y^3/3+2y^2+4y = t + k$. I want it in the form $y = f(t)$. Wolfram alpha tells me the answer is $y = \sqrt[3]{k - 3t -8} - 2 $. Now how did it do that?

Answer 1

Multiplying your equation by $3$ it becomes: $$ y^3+6y^2+12y=3t+3k $$ completing the cube at LHS we have: $$ y^3+6y^2+12y+8=(y+2)^3=3t+3k+8 $$

so:

$$ y=\sqrt[3]{3t+3k+8}-2 $$

Note that Wolfram-alpha gives all the three solution in $\mathbb{C}$ and the real solution is write as

$$ y=-\sqrt[3]{-3t-3k-8}-2 $$

(you have some typos with the minus signs in your question)