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

If x^(3)dy+xydx=x^(2)dy+2ydx,y(2)=e then y(-1)=

If x^(3)dy+xydx=x^(2)dy+2ydx,y(2)=e then y(-1)=

If x^(3) dy + xy dx = x^(2) dy + 2y dx , y (2) = e and x gt 1 , then y(4) is equal to .

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

Solve x dy + y dx = (ydx - xdy)/(x^2 + y^2)

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

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

ydx - (x + 2y ^(2) ) dy = 0

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

x ^ (2) ydx = (x ^ (3) + y ^ (3)) dy