Differential Eq. Last Step Solution Separating Variables

[knowLittle]

Solve each differential equation. Express the general solution in explicit form.

y' = (3x^2 -1) / (3+2y)

So, I will skip many steps, because they are easy. However, I am stuck in one of the last ones.
y^2 +3y = x^3- x +C

y(y+3)= x^3 - x +C

I have seen the solution for y, but I don't understand how it is derived. Can someone help?

 

[Chestermiller]

Try completing the square on the left hand side, and then taking the square root of both sides of the equation.

 

[HallsofIvy]

y(y+ 3)= y^2+ 3y= x^3- x+ C

y^2+ 3y+ (x- C)= 0.

Solve that quadratic equation using the quadratic formula.

 

[knowLittle]

y(y+ 3)= y^2+ 3y= x^3- x+ C

y^2+ 3y+ (x- C)= 0.

Solve that quadratic equation using the quadratic formula.

Hello HallsofIvy, How did you get rid of x^3 ?

 

[Chestermiller]

Hello HallsofIvy, How did you get rid of x^3 ?

He appears to inadvertently left it out. The method HallsofIvy suggested and the method I suggested are mathematically equivalent.

Chet

 

[Mark44]

Hello HallsofIvy, How did you get rid of x^3 ?

He appears to inadvertently left it out.

Yes.

The method HallsofIvy suggested and the method I suggested are mathematically equivalent.

Move all of the terms to the left side, and you'll have a quadratic in y. It's probably simpler at this point to just use the Quadratic Formula to solve for y, which will have two parts separated by ±, as is usually the case.

If there is an initial condition, it might lead you to choose one or the other of the two values.

 

[HallsofIvy]

Hello HallsofIvy, How did you get rid of x^3 ?

I waved my magic wand and uttered a spell of "Carelessness"!