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

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

Find the solution of (y dy/dx+2x)^2=(y^2+2x^2)[1+(dy/dx)^2] and hence find the curve passing through (1,0)

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)

Solve each of the following initial value problems: (dy)/(dx)+(2x)/(x^2+1)y=1/((x^2+1)^2),y(0)=0 (ii) (x^2+1)(dy)/(dx) -2x y=(x^4+2x^2+1)cosx ,y(0)=0

dy/dx = (2x-y)/(x-2y)

dy/dx = (2x-y)/(x-2y)

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

Find dy/dx if 2x-4y=sinx-x^2

dy/dx - y/x = 2x^2

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