`x+y(dy/dx)=2y`

Solve the following initial value problem: x y(dy)/(dx)=y+2,\ y(2)=0

Solve: 2x y(dy)/(dx)=x^2+y^2

Solve the following initial value problem: x y(dy)/(dx)=(x+2)(y+2),\ y(1)=-1

Solve the following differential equations: x y(dy)/(dx)=x^2+y^2

The differential equation of all circles passing through the origin and having their centres on the x-axis is (1) x^2=""y^2+""x y(dy)/(dx) (2) x^2=""y^2+"3"x y(dy)/(dx) (3) y^2=x^2""+"2"x y(dy)/(dx) (4) y^2=x^2""-"2"x y(dy)/(dx)

For the differential equation x y(dy)/(dx)=(x+2)(y+2) , find the solution curve passing through the point (1, 1) .

For the differential equation x y(dy)/(dx)=(x+2)(y+2) , find the solution curve passing through the point (1, -1) .

((dy)/(dx))^(2) + 2y cot x (dy)/(dx) = y^(2) has the solution

Solve the differential equation x y(dy)/( dx)=(1+y^2)/(1+x^2)(1+x+x^2)

Solve the differential equation x y(dy)/( dx)=(1+y^2)/(1+x^2)(1+x+x^2)