`y=Ax : xy' = y (x ne 0)`

verify that the given functions(explicit or implicit) is a solution of the corresponding differential equation : y = Bx : xy' = y (x ne 0)

For the differential equation, find the general solution: (x)(dy)/(dx) + y - x + xy cotx = 0 (x ne 0)

Solve the following differential equations x (dy)/(dx) + y - x + xy cot x = 0 (x ne 0)

x(dy)/(dx) + y - x + xy cot x = 0 (x ne 0)

Solve x(dy)/(dx) + y + xy cot x = 0 (x ne 0)

Verify that the given function is a solution of the corresponding differential equation: y=Ax: xy\'=y (x!=0)

x + y = 5 xy , 3x + 2y = 13 xy ( x ne 0, y ne 0 )

4 x + 6y = 3 xy , 8x + 9 y = 5xy ( x ne 0 , y ne 0 )

Solve for x and y : 6x + 3y = 7xy, 3x + 9y = 11 xy (x ne 0 , y ne 0 )