How do you solve this differential equation: ((e^x)*sin(y)+2x)dx + ((e^x)*cos(y)+(1/(1+y^2)))dy = 0?

( (e^x)*sin(y) + 2x )dx + ( (e^x)*cos(y) + (1 / (1 + y^2)) )dy = 0

The solution is of the form

M(x,y)dx + N(x,y)dy = 0

Now checking for exactness we will take the partial derivative

∂/∂y [M(x,y)] = e^x cos(y) = ∂/∂x [N(x,y)]

Above implies the equation is EXACT. Thus there is a φ(x,y) such that

Please note: Underscore represnts the subscript implying its partial

φ_x = (e^x)*sin(y) + 2x

φ_y = (e^x)*cos(y) + ( 1 / (1 + y^2) )

Integrating the first of these equations we get

φ = (e^x)*sin(y) + x^2 + ψ(y)..........(1)

and hence again the partial derivative will be

φ_y = (e^x)*cos(y) + dψ/dy

comparing both ψ(y) we will get,

dψ/dy = ( 1 / (1 + y^2) )

dψ = dy / (1 + y^2)

Integrate,

∫dψ = ∫dy / (1 + y^2)

=>

ψ = arctan(y)

Substitute ψ in equation (1)

φ = (e^x)*sin(y) + x^2 + arctan(y)

e^x sin y + e^x cos y dy/dx + 2x + 1/(1+y^2) dy/dx = 0

d/dx(e^x sin y) + 2x + d/dx( tan^-1 y) = 0

e^x sin y + x^2 + tan^-1 y = C