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

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

Solve (1+x^(2))dy = xydx given that y (0) = 1

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

(1+e^(2x))dy+e^(x)(1+y^(2))dx=0 it being given that y=1 when x=0

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

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

If 2x^2dy + (e^y-2x)dx=0 and y(e)=1 then find the value of y(1)