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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.

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2 Answers 2

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 $$y=\frac {-3 \pm \sqrt{9-4\cdot 2(c-e^{-x}-e^x)}}{4\cdot 2}$$

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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 $y$.

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