`dy/dx+y f\'(x)-f(x)f\'(x)=0`

The solution of (dy)/(dx)+yf'(x)-f(x).f'(x)=0,y!=f(x) is

Solve: (dy)/(dx)+y*f'(x)=f(x)*f'(x), where f(x) is a given function.

Solve: dy/dx=(y f\'(x)-y^2)/f(x) , where f(x) is a given function of x

Solve (dy)/(dx) = yf^(')(x) = f(x) f^(')(x) , where f(x) is a given integrable function of x .

The solution of the D.E (dy)/(dx)=(y*f'(x)-y^(2))/(f(x)) equal to

If f(x) is differentiable, then the solution of dy+f\'(x)(y-f(x))dx=0 is (A) yf(x)=Ce^(-f(f(x))^2) (B) y+1=f(x)+Ce^(-f(x)) (C) f(x)=Cye^(-y^2/2) (D) none of these

General solution of differential equation of f (x) (dy)/(dx) =f ^(2) (x)+yf(x) +f'(x)y is: (c being arbitary constant.)

Solve: (dy)/(dx) = (yf^(')(x)-y^(2))/(f(x))

Let y=f(x) satisfies (dy)/(dx)=(x+y)/(x) and f(e)=e then the value of f(1) is

Let y=f(x) is a solution of differential equation e^(y)((dy)/(dx)-1)=e^(x) and f(0)=0 then f(1) is equal to