Question

Solve (1+y2)dx = (tan-1y-x)dy

Answer 1

Hope u can understand now

Answer 2

Result is $xe^m = me^m - e^m + c$$xtan^{-1}x = e^{tan^{-1}x}(tan^{-1}x - 1) + c$

Step-by-step explanation:

$(1+y^2)dx = (tan^{-1}y-x)dy$

$\frac {dy}{dx} = \frac {1+y^2}{tan^{-1}y-x}$

$\frac {dx}{dy} = \frac {tan^{-1}y-x}{1+y^2}$

$\frac {dx}{dy}+\frac {x}{1+y^2} = \frac {tan^{-1}y}{1+y^2}$

It proves to be a Linear Differential Equation-

$P = \frac {1}{1+y^2}$

Integrating factor ,

$e^{\int p.dy} = e^{\int \frac {1}{1+y^2}dy} = e^{tan^-^1y}$

Finally the solution becomes,

$xe^{tan^-^1y}=\int e^{tan^{-1}y}\frac {tan^{-1}y}{1+y^2}dy + c$

As $tan^{-1}y = m$

So, $\frac {1}{1+y^2}dy = dm$

$xe^m = \int e^m.mdm + c$

= $m\int e^mdm - \int (\frac {d}{dm}m\int e^m)dm + c$

= $me^m - \int 1.e^mdm + c$

= $xe^m = me^m - e^m + c$$xtan^{-1}x = e^{tan^{-1}x}(tan^{-1}x - 1) + c$