If `(x^2+y^2)dy=xydx` and y(1)=1 and `y(x_o)=e`, then `x_o=`

If (x^2+y^2)dy=xydx and y(1)=1 . If y(x_0)=e then x_0 is equal to (A) sqrt(2)e (B) sqrt(3)e (C) 2e (D) e

The solution of the primitive integral equation (x^(2)+y^(2))dy=xydx is y=y(x)* If y(1)=1 and y(x_(0))=e, then x_(0) is

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

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

(x^(2)+1)dy=2xydx;y(1)=2

Solve: (x^2+y^2)dy-xydx=0

(x^(2)+y^(2))dy=xydx . It is given that y(1)=1 and y(x_(0))=e . Find the vale of x_(0) .

If x^3dy+ xydx= 2ydx +x^2dy and y(2) =e then y(4) = ?