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Question Number 91558 by jagoll last updated on 01/May/20

(x^2 +1)y′+y^2 +1 = 0

$$\left({x}^{\mathrm{2}} +\mathrm{1}\right){y}'+{y}^{\mathrm{2}} +\mathrm{1}\:=\:\mathrm{0}\: \\ $$

Commented by jagoll last updated on 01/May/20

it Bernoulli diff eq?

$${it}\:{Bernoulli}\:{diff}\:{eq}? \\ $$

Answered by niroj last updated on 01/May/20

I think its not be a Bernoulli′s equation. It might be First order Veriables Separable Differential Equation. (x^2 +1)y^′ +y^2 +1=0 (x^2 +1)(dy/dx)+y^2 +1=0 (x^2 +1)(dy/dx)= −(y^2 +1) (1/(y^2 +1))dy=− (1/(x^2 +1))dx On Integrating both side, ∫ (1/(y^2 +1))dy =− ∫(1/(x^2 +1))dx tan^(−1) y =− tan^(−1) x +C tan^(−1) x+ tan^(−1) y = C //.

$$\:\:\mathrm{I}\:\mathrm{think}\:\mathrm{its}\:\mathrm{not}\:\mathrm{be}\:\mathrm{a}\:\mathrm{Bernoulli}'\mathrm{s}\:\mathrm{equation}.\:\mathrm{It}\:\mathrm{might}\:\mathrm{be}\: \\ $$$$\:\mathrm{First}\:\mathrm{order}\:\mathrm{Veriables}\:\mathrm{Separable}\:\:\mathrm{Differential}\:\mathrm{Equation}.\: \\ $$$$\:\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{y}^{'} +\mathrm{y}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\:\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\:\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\frac{\mathrm{dy}}{\mathrm{dx}}=\:−\left(\mathrm{y}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\:\:\:\:\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\mathrm{dy}=−\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx} \\ $$$$\:\:\:\mathrm{On}\:\mathrm{Integrating}\:\mathrm{both}\:\mathrm{side}, \\ $$$$\:\:\int\:\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\mathrm{dy}\:=−\:\int\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx} \\ $$$$\:\:\mathrm{tan}^{−\mathrm{1}} \mathrm{y}\:=−\:\mathrm{tan}^{−\mathrm{1}} \mathrm{x}\:+\mathrm{C} \\ $$$$\:\mathrm{tan}^{−\mathrm{1}} \mathrm{x}+\:\mathrm{tan}^{−\mathrm{1}} \mathrm{y}\:=\:\mathrm{C}\://. \\ $$

Commented by jagoll last updated on 01/May/20

yes. agree

$${yes}.\:{agree} \\ $$

Answered by jagoll last updated on 01/May/20

Commented by niroj last updated on 01/May/20

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