Question

Solve for y:

y^2/2-2y=x^3+x^2+x+C

Answer 1

I multiplied the whole equation by 2, I got y^2-4y=2(x^3+x^2+x+C)

Answer 2

\[\frac{y^2}{2}-2y=\cdots~~?\]
Try completing the square.

Answer 3

Sorry, I'm not good at that.

Answer 4

Okay, well we don't really have to worry about the right hand side, so I'll just focus on the left and you can take it from there.
\[\frac{1}{2}y^2-2y=\frac{1}{2}\left(y^2-4y\right)\]
Add and subtract 4:
\[\frac{1}{2}\left(y^2-4y+4-4\right)\]
Factor the first three terms:
\[\frac{1}{2}\left((y-2)^2-4\right)\]
Distribute the 1/2:
\[\frac{1}{2}(y-2)^2-2=x^3+\cdots\]

Answer 5

Now you can isolate the \(y\) to get an explicit solution. Don't forget the \(\pm\sqrt{\cdots}\).

Answer 6

HAHA, didn't learn the completing square method fully in algebra before. Sorry. :)

Answer 7

Got it! But I don't know why the answers in the book are slightly different. I guess they're all wrong?

Answer 8

Two possibilities that I can see. (1) When you add 2 to the right side, then multiply by 2, it could be that the 2 you add is absorbed into \(C\). Or, (2) it's not absorbed into \(C\)... I think (1) is more likely. What does your book say?

Answer 9

My book says y=2+/-sqrt(2(x^3+x^2+x+c)) but I got y=2+/-sqrt(2(2+x^3+x^2+x+c)).

Answer 10

So what's the correct answer then?

Answer 11

Looks like it does get absorbed. I'd go with the book's answer.

Answer 12

Okay. :)

Answer 13

From Mathematica 9:\[y=2\pm \sqrt{2} \sqrt{c+x^3+x^2+x} \]