`xdy/dx + 2y = log x `

Solve the following differential equation : xdy/dx + y = x log x, x!=0 .

Find the general solution of the differential equation xdy/dx+2y=x^2 log x

If y=x sin(log x)+x log x, prove that x^(2)(d^(2)y)/(dx^(2))-x(dy)/(dx)+2y=x log x

y dx - xdy + log x dx = 0

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

The integrating factor of the differential equation xdy/dx - y = x^(2) is -

Express the following differential equations in the form (dy)/(dx) = F((y)/(x)) . (i) xdy - ydx = sqrt(x^(2) + y^(2))dx (ii) [x - y Tan^(-1)((y)/(x))] dx + x Tan^(-1)((y)/(x)) dy = 0 (iii) xdy = y(log y - log x+1)dx

Find (d^2y)/(dx^2) if y= log(log x)