If `x(dy)/(dx)=y(logy-logx+1)`, then the solutions of the equation is

Solve x(dy)/(dx)=y(logy-logx+1)

Solve: x(dy)/(dx)=y(logy-logx-1)

Solve x((dy)/(dx))=y(logy-logx+1)

Solve x((dy)/(dx))=y(logy-logx+1)

If y_(1)(x) is a solution of the differential equation (dy)/(dx)-f(x)y = 0 , then a solution of the differential equation (dy)/(dx) + f(x) y = r(x) is

x dy/dx+y=y^2 logx

The solution of the equation (dy)/(dx)=cos(x-y) is

The general solution of the equation (dy)/(dx)=1+x y is

Show that y=acos(logx)+bsin(logx) is a solution of the differential equation x^2(d^2y)/(dx^2)+(dy)/(dx)+y=0.

Find dy/dx if x=logy