1. ## DE help

how to solve (1+x^2)y'+y^2+1=0?

what I come up with is arctany=-arctanx-C
then how can i proceed?

2. ## Re: DE help

could you try tan both side? then use tan(A+B) formula solve it?

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3. ## Re: DE help

Originally Posted by elmomleo
how to solve (1+x^2)y'+y^2+1=0?

what I come up with is arctany=-arctanx-C
then how can i proceed?
\displaystyle \begin{align*} \left( 1 + x^2 \right) \,\frac{\mathrm{d}y}{\mathrm{d}x} + y^2 + 1 &= 0 \\ \left( 1 + x^2 \right) \,\frac{\mathrm{d}y}{\mathrm{d}x} &= - \left( 1 + y^2 \right) \\ \frac{1}{1 + y^2 } \,\frac{\mathrm{d}y}{\mathrm{d}x} &= -\frac{1}{1 + x^2} \\ \int{ \frac{1}{1 + y^2} \,\frac{\mathrm{d}y}{\mathrm{d}x}\,\mathrm{d}x} &= \int{ -\frac{1}{1 + x^2}\,\mathrm{d}x} \\ \int{ \frac{1}{1 + y^2}\,\mathrm{d}y} &= -\arctan{ \left( x \right) } + C_1 \\ \arctan{ \left( y \right) } + C_2 &= -\arctan{ \left( x \right) } +C_1 \\ \arctan{ \left( y \right) } &= -\arctan{ \left( x \right) } + C \textrm{ where } C = C_1 - C_2 \end{align*}

I agree with your answer, why do you believe that it's not right?

4. ## Re: DE help

I don't believe that elmomleo thinks that is incorrect. He apparently thinks that he has to solve for y in order to present the answer as "y= something". In that case, Hailiang96 is correct- just take tangent of both sides.