`(1+y^(2))dx=(tan^(-1)y-x)dy` को हल कीजिए।

Solve :(1+y^(2))dx=(tan^(-1)y-x)dy

Solve : (1+y^(2))dx=(tan^(-1)y-x)dy , given that y=0 when x = -1.

The solution of (1 + y^(2)) dx = (Tan^(-1) y -x) dy is

Solve the following differential equations (i) (1+y^(2))dx = (tan^(-1)y - x)dy (ii) (x+2y^(3))(dy)/(dx) = y (x-(1)/(y))(dy)/(dx) + y^(2) = 0 (iv) (dy)/(dx)(x^(2)y^(3)+xy) = 1

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

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

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

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

SolveL (1+y)^2dx=(tan^(-1)y-x)dy , given that y= 0 when x= -1.