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I need to solve an equation of this type

$$\left(\frac{dy}{dx}\right)^2+\frac{y^2}{b^2}=a^2$$

but I don't know how to start Any help would be welcome, Thanks !

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1 Answer

$$\left(\dfrac{dy}{dx} \right)^2 = a^2 - \dfrac{y^2}{b^2}$$ This gives us $$\dfrac{dy}{dx} = \pm\sqrt{a^2 - \dfrac{y^2}{b^2}}$$ $$\dfrac{dy}{\sqrt{a^2b^2 - y^2}} = \pm\dfrac{dx}b$$ Setting $y = ab \cos(t)$, gives us $$\dfrac{-ab \sin(t) dt}{ab \sin(t)} = \pm \dfrac{dx}b$$ This gives us $$dt = \mp \dfrac{dx}b \implies t = \mp \dfrac{x}b + c$$ Hence, $$y = ab \cos(t) = ab \cos \left(\dfrac{x}b + k\right)$$

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Well, is this the only solution ? Because the paper I'm reading now produces $ab\cos(x/b)$ –  simon Nov 24 '12 at 20:44
    
Thank you very much –  simon Nov 24 '12 at 20:59
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