Question

Find the general solution of the following differential equation : 

`(1+y^2)+(x-e^(tan^(-1)y))dy/dx= 0`

Answer 1

Given:

`(1+y^2)+(x-e^(tan^(-1)y))dy/dx= 0`

Let tan⁻¹y=t

⇒y=tant

`=>dy/dx=sec^2tdt/dx`

Therefore, the equation becomes

(1+tan²t)+(x−et)sec²t `dt/dx=0`

`=>sec^2t+(x-e^t)(sec^2t)dt/dx=0`

`=>1+(x-e^t)dt/dx=0`

`=>(x-e^t)dt/dx=-1`

`=>x-e^t=dx/dt`

`=>dx/dt+1.x=e^t`

If =e∫1.dt

= e^t

`:. e^t.(dx/dt+1.x)=e^t.e^t`

 `=>d/dt(xe^t)=e^(2t)`

 Integrating both the sides, we get

`xe^t=inte^(2t)dt`

`=>xe^t=1/2e^(2t)+C " ....(1)"`

Substituting the value of t in (1), we get

`xe^(tan^(1))y=1/2e^(2tan^(-1)y)+C_1`

`=>e^2tan^(-1y)=2xe^(tan^1y)+C`

It is the required general solution.