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

Solve the following differential equation : (1+y^2)dx=(tan^-1y-x)dy

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

Solve the following differential equations.(3) (1+y^2)dx=(tan^-1y-x)dy

Consider (1+y^2)dx=(tan^-1y-x)dy Find the integrating factor.

Consider (1+y^2)dx=(tan^-1y-x)dy Solve the given equation.

Consider (1+y^2)dx=(tan^-1y-x)dy Express the equation in the form dx/dy+Px=Q

Solve the differential equation: (i) (1+y^(2))+(x-e^( tan ^(-1)y))(dy)/(dx)=0 (ii) x(dy)/(dx)+cos^(2)y=tan y(dy)/(dx)

If log (x^(2)+y^(2))=tan^(-1)((y)/(x)), then show that (dy)/(dx)=(x+y)/(x-y)

The solution of the differential equation (1+y^(2)) tan^(-1) x dx + y(1+x^(2)) dy = 0 is

If logsqrt(x^(2)+y^(2))=tan^(-1)((x)/(y)) , then show that (dy)/(dx)=(y-x)/(y+x) .