# (y')^2+y^2=-2 why this equation has no general solution ?

1. Oct 15, 2010

### young_eng

hi

(y')^2+y^2=-2
why this differential equation has no general solution ?

2. Oct 15, 2010

### Pengwuino

It's non-linear. Having a general solution for these types of equations is the exception, not the rule.

Not that there isn't a general solution because I don't know. It's just that non-linear equations rarely have general solutions

3. Oct 15, 2010

### Dickfore

Because the lhs is necessarily non-negative (sum of squares), whereas the lhs is negative. You can have a solution in complex numbers.

4. Oct 16, 2010

### jackmell

Why not just solve it the regular way:

$$\frac{dy}{\sqrt{-2-y^2}}=\pm dx$$

or $y=\pm i\sqrt{2}[/tex] are solutions, maybe singular ones. Not sure. Otherwise: $$\frac{y\sqrt{-2-y^2}}{2+y^2}=\tan(c\pm x)$$ $$y(x)=\pm \frac{\sqrt{2}\tan(c\pm x)}{\sqrt{-\sec^2(c\pm x)}}$$ so that the solution is in the form of y(z)=u+iv 5. Oct 17, 2010 ### HallsofIvy Staff Emeritus There is no general solution in terms of real valued functions because if y' and y are both real numbers (for a given x) then [itex](y')^2+ y^2$ cannot be negative!

Oops! Dickfore had already said that, hadn't he?

Last edited: Oct 18, 2010
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