Question

solve the differential equation (1+y^2)+(x-e^tan^-1y) dy/dx=0

Answer 1

(1 + y^2) * dx + (x - e^(arctan(y))) * dy = 0 
(1 + y^2) * dx - e^(arctan(y)) * dy + x * dy = 0 

e^(arctan(y)) = u 
arctan(y) = ln(u) 
y = tan(ln(u)) 
dy = (1/u) * sec(ln(u))^2 * du 
dy = (1/u) * (1 + tan(ln(u))^2) * du 
dy = (1/u) * (1 + y^2) * du 
u * dy = (1 + y^2) * du 
dy = (1 + y^2) * du / u 

(1 + y^2) + (x - e^(arctan(y)) * dy/dx = 0 
(1 + y^2) * dx + (x - e^(arctan(y)) * dy = 0 
(1 + y^2) * dx + (x - u) * (1 + y^2) * du / u = 0 
dx + (x - u) * du / u = 0 
dx + x * du/u - u * du/u = 0 
dx + x * du/u - du = 0 
u * dx + x * du - u * du = 0 
ux - (1/2) * u^2 + C = 0 
2ux - u^2 + C = 0 
2 * e^(arctan(y)) * x - (e^(arctan(y)))^2 + C = 0 
2x * e^(arctan(y)) + C = e^(2 * arctan(y))