Form differential equation for `cy+logx=0`

Form differential equation for Ax+By+C=0

Solution of the differential equation ydx-x dy+logx dx=0 is

Solve the differential equation: ydx-xdy+(logx)dx=0

For the differential equation, find the general solution: (x)(logx)(dy)/(dx) + y = (2/x) (logx)

Find the integrating factor of the given differential equation: ydx-xdy+(logx)dx=0

Solve the following differential equation: (dy)/(dx)=logx

If x dy/dx=y(logy-logx+1) , then the solution of the differential equation is (A) log(x/y)=Cy (B) log(y/x)=Cy (C) log(x/y)=Cx (D) log(y/x)=Cx

Integrating factor of the differential equation (x.logx)(dy)/(dx)+y=2logx is

Integrating factor of the differential equation (x.logx)(dy)/(dx)+y=2logx is

Solve the differential equation : xdy/dx+2y=(logx)/x^2 .