`dy/dx=(2xy)/(x^2-y^2)`

(dy)/(dx) = (2xy)/(x^(2)-1-2y)

STATEMENT -1 : The differential equation (dy)/(dx) = (2xy)/(x^(2) + y^(2)) Can't be solved by the substitution x = vy. and STATEMENT-2 : When the differential equation is homogeneous of first order and first degree, then the substitution that solves the equation is y = vx.

If (dy)/dx = (xy)/(x^2 + y^2), y(1) = 1 and y(x) = e then x =

Solve: x+y dy/dx=(a^2((x dy/dx-y))/(x^2+y^2))

dy/dx+2xy=e^(-x^2)

The straight line y=2x meets y=f(x) at P, where f(x) is a solution of the differential equation (dy)/(dx)=(x^(2)+xy)/(x^(2)+y^(2)) such that f(1)=3 , then f'(x) at point P is

If the straight line y=x meets y=f(x) at P, where f(x) is a solution of the differential equation (dy)/(dx)=(x^(2)+xy)/(x^(2)+y^(2)) such that f(1)=3 , then the value of f'(x) at the point P is

Solve the differential equation : (dy)/(dx)=(x^(2)-y^(2))/(xy) .

On putting (y)/(x)=v the differential equation (dy)/(dx)=(2xy-y^(2))/(2xy-x^(2)) is transferred to

The differential equations of all circle touching the x-axis at orgin is (a) (y^(2)-x^(2))=2xy((dy)/(dx)) (b) (x^(2)-y^(2))(dy)/(dx)=2xy ( c ) (x^(2)-y^(2))=2xy((dy)/(dx)) (d) None of these