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I have solved a differential equation (1) for x, where i think is a real variable.

(1) [tex] \frac{dy}{dx} = \frac{1}{x^{\frac{3}{2}} (1+x)^2} [/tex]

And get a solution

[tex] y(x) = \frac{2}{\sqrt{x}} - \frac{\sqrt{x}}{(1+x)} - 3 \arctan{\sqrt{x}} [/tex]

Now, i want to find solutions for x<0, and i have said that the above is a solution for x>0.

So far i have simply done a coord transformation x→-x and then solved this equation

[tex] \frac{dy}{dx} = \frac{1}{x^{\frac{3}{2}} (1-x)^2} [/tex]

and then substituted in "-x" under my square roots.

Is this a correct way of finding x<0?

Konig.

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# Homework Help: Solutions for x<0

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