Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So, I am failing to understand some potentially simple algebra here. I have a separable equation:

$ \frac{dy}{dx} = \frac{e^{-x} - e^x}{3+4y} $

and after the easy integration $=>$ $3y + 2y^2 = -e^{-x} - e^x + c$

Now, to the algebra.. how do I solve for y? The book has a fairly long answer involving a square root... It could come down to, I did the separable/integration part incorrectly or I've lost my mind but I'm kind of shaking my head over my lack of algebra skills.

share|cite|improve this question
up vote 2 down vote accepted

This is a quadratic equation for $y$, so you can feed it to the quadratic formula. The result is $$y=\frac {-3 \pm \sqrt{9-4\cdot 2(c-e^{-x}-e^x)}}{4\cdot 2}$$

share|cite|improve this answer
So important, yet I fill my head with other things! – intervade Sep 10 '12 at 23:50
The $a$ term is $2$ right? Shouldn't it be $$y=\displaystyle\frac{-3 \pm \sqrt{9-4(2)(c-e^{-x}-e^{x})}}{4(2)}$$? – Joseph Skelton Sep 11 '12 at 1:30
@JosephSkelton: correct. fixed. – Ross Millikan Sep 11 '12 at 1:55

The equation

$$2y^2+3y +e^{-x} + e^x - c=0$$

is a quadratic in terms of $y$, right?

So, use Bhaskara's formula (also called Quadratic Equation) to solve for $y$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.