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?

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- Dec 31st 2017, 11:31 PM #1

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- Dec 31st 2017, 11:57 PM #2

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- Jan 1st 2018, 03:39 AM #3
## Re: DE help

$\displaystyle \begin{align*} \left( 1 + x^2 \right) \,\frac{\mathrm{d}y}{\mathrm{d}x} + y^2 + 1 &= 0 \\ \left( 1 + x^2 \right) \,\frac{\mathrm{d}y}{\mathrm{d}x} &= - \left( 1 + y^2 \right) \\ \frac{1}{1 + y^2 } \,\frac{\mathrm{d}y}{\mathrm{d}x} &= -\frac{1}{1 + x^2} \\ \int{ \frac{1}{1 + y^2} \,\frac{\mathrm{d}y}{\mathrm{d}x}\,\mathrm{d}x} &= \int{ -\frac{1}{1 + x^2}\,\mathrm{d}x} \\ \int{ \frac{1}{1 + y^2}\,\mathrm{d}y} &= -\arctan{ \left( x \right) } + C_1 \\ \arctan{ \left( y \right) } + C_2 &= -\arctan{ \left( x \right) } +C_1 \\ \arctan{ \left( y \right) } &= -\arctan{ \left( x \right) } + C \textrm{ where } C = C_1 - C_2 \end{align*}$

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

- Jan 6th 2018, 06:00 AM #4

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