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$$x=\pm\frac{y'}{\sqrt{(y')^2+1}}$$

$$x=\pm\frac{y'}{\sqrt{(y')^2+1}}$$

$$x^2=\frac{(y')^2}{{(y')^2+1}}$$

$$x^2(y')^2+x^2=(y')^2$$

$$(y')^2[x^2-1]=-x^2$$

$$(y')^2=\frac{-x^2}{x^2-1}$$

$$y'=\pm \sqrt{\frac{-x^2}{x^2-1}}$$

Have I got it wrong? as there is no ODE

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    it looks correct to me,why didn't you proceed? – 2017-01-23
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    It meant to be solve as ODE and I just got a simple integral – 2017-01-23

2 Answers 2

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It looks alright.

You're mistaken in saying there's no ODE. Every line in your posting is just an ODE.

From where you left off I'd go on to say $$ \int\pm\sqrt{\frac{-x^2}{x^2-1}}\,dx = \int\pm\sqrt{\frac{x^2}{1-x^2}} = \int\pm \frac x {\sqrt{1-x^2}} \, dx = \int\pm \frac {du}{2\sqrt{u}} = \pm \sqrt u + C $$ $$ = \pm \sqrt{1-x^2} + C. $$

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Here is a solution

$$y'=\pm \sqrt {\frac{-x^2}{x^2-1}} \\ \frac{dy}{dx} =\pm \frac x {\sqrt {1-x^2}} \\ \int dy = \pm \int \frac x {\sqrt {1-x^2}} \, dx =\pm \frac 1 2 \int \frac {d(1-x^2)}{\sqrt{ 1-x^2}}=\pm \sqrt {1-x^2} \\ y=\pm \sqrt{1-x^2} +C$$

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    What's wrong with the answer? please show my mistake.It is second time somebody downvotes without any explanation. – 2017-01-23
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    Sometimes down-votes here are completely irrational and sometimes they are dishonest. – 2017-01-23