[" 19."(1+y^(2))(1+log x)dx+x*dy=0],[" काकि "quad x=1,y=1]

Solve the differential equation : (1 + y^2) (1 + log x)dx + x dy = 0 given that y = 1 when x = 1

Solve the differential equation (1+y^(2))(1+log x) dx +x dy =0, it being given that y = 1 when x = 1.

The solution of the equation x^(2)(d^(2)y)/(dx^(2))=log x when x=1,y=0 and (dy)/(dx)=-1 is

A: If y = x ^(y) then (dy)/(dx) = (y ^(2))/(x(1- log y )) If y = f (x) ^(y), then (dy)/(dx) = (y ^(2) f '(x))/(f (x) [1- ylog f (x)])= (y ^(2) f'(x))/(f (x) [1- log y])

If y=log((1-x^(2))/(1+x^(2)))," then "(dy)/(dx)=

If y=log((1-x^(2))/(1+x^(2)))," then "(dy)/(dx)=

solve x((dy)/(dx))=y(log y-log x+1)