Question

dy/1+y² =dx/1+x² solve it​

Answer 1

❤ Required Answer =>

Given differential equation is dy/dx = (1 + y2)/(1 + x2)

= (1/(1 + y2)) dy = (1/(1 + x2)) dx

on, integrating, ∫(1/(1 + y2)) dx = ∫(1/(1 + x2)) dy

= tan-1 y = tan-1 x + c

= tan-1 y - tan-1 x = c

= tan-1 ((y - x)/(1 + yx)) = c

= (y - x)/(1 + yx) = tan c

∴ (y - x)/(1 + yx) = c(constant) is, the require solution

Hope it will helps✌:)

$:)$

Answer 2

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\dfrac{dy}{1 + {y}^{2} } = \dfrac{dx}{1 + {x}^{2} }$

On integrating both sides, we have

$\rm :\longmapsto\: \displaystyle \int\dfrac{dy}{1 + {y}^{2} } = \displaystyle \int\dfrac{dx}{1 + {x}^{2} }$

$\rm :\longmapsto\: {tan}^{ - 1} y = {tan}^{ - 1} x + {tan}^{ - 1} c$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \red{\bf \: \because \: \displaystyle \int\dfrac{dx}{ {x}^{2} + 1 } = {tan}^{ - 1}x}}$

$\rm :\longmapsto\: {tan}^{ - 1} y - {tan}^{ - 1} x = {tan}^{ - 1} c$

$\rm :\longmapsto\: {tan}^{ - 1}c = {tan}^{ - 1}\bigg( \dfrac{y - x}{1 + yx} \bigg)$

$\: \: \: \: \: \boxed{ \red{ \sf \: \because \: {tan}^{ - 1}x - {tan}^{ - 1}y = {tan}^{ - 1}\bigg(\dfrac{x - y}{1 + xy} \bigg)}}$

$\rm :\implies\:c = \bigg( \dfrac{y - x}{1 + xy} \bigg)$

$\bf\implies \:y - x = c(1 + xy)$

Additional Information :-

The linear differential equation is of the form

$\rm :\longmapsto\:\dfrac{dy}{dx} + py = q \: \: where \: p \: and \: q \: \in \: f(x)$

and following steps are used to solve the above differential equation

Step :- 1

Find Integrating Factor, which is evaluated as

$\rm :\longmapsto\:IF = {e}^{ \: \: \displaystyle \int \: p \: dx}$

Step :- 2

The solution of differential equation is

$\rm :\longmapsto\:y \times IF = \displaystyle \int \: (q \times IF)dx$