How to solve for $y$ in $\frac{y^2}{2}+y = \frac{x^3}{3}+\frac{x^2}{2}+c_1$?

Question

Please someone help me solve the following equation in terms of $y$:

$\frac{y^2}{2}+y = \frac{x^3}{3}+\frac{x^2}{2}+c_1$

The calculator gives me:

$y = \frac{1}{3}(\sqrt{3}\sqrt{c_1+2x^3+3x^2+3}-3), -\frac{1}{3}(\sqrt{3}\sqrt{c_1+2x^3+3x^2+3}-3)$

I do not know the procedure to get to the answer. Somebody help please. Thank you.

Answer 1

Since $$\frac{y^2}{2}+y = \frac{x^3}{3}+\frac{x^2}{2}+c_1\qquad\iff\qquad \frac12(y+1)^2-\frac12= \frac{x^3}{3}+\frac{x^2}{2}+c_1$$ It follows $$(y+1)^2=2\left(\frac{x^3}{3}+\frac{x^2}{2}+c_1\right)+1$$ Then $$y=-1\pm\sqrt{2\left(\frac{x^3}{3}+\frac{x^2}{2}+c_1\right)+1}$$

Answer 2

Multiply both sides by $6$ to make things a little nicer.

$$3y^2+6y=2x^3+3x^3+6c_1$$

$$3y^2+6y-2x^3-3x^3-6c_1=0$$

At this point you should realize that the variable $y$, what we are trying to solve for, is quadratic in the above equation. Although it looks quite messy, the equation is really algebraically in the form:

$$ay^2+by+c=0$$

We know how to deal with quadratics, one option is utilizing the quadratic formula with,

$$a=3$$

$$b=6$$

$$c=-2x^3-3x^3-6c_1$$

Another is to complete the square.