If `ye^(y) dx = (y^(3) + 2xe^(y))dy, y(0) = 1`, then the value of x when y = 0 is

Let (dy)/(dx)=(6)/(x+y) where y(0)=0 then the value of y when x+y=6 is

If (dy)/(dx)=e^(-2y) and y=0 when x=5, then the value of x for y=3 is

Solve 2ye^(x/y)dx+(y-2xe^(x/y))dy=0

if ( dy )/(dx) =1 + x +y +xy and y ( -1) =0 , then the value of ( y (0) +3 - e^(1//2))

Let y^(2)(dy)/(dx)=x and "y(0)=2" ,then value of y(1) is

If (dy)/(dx) = (2^(x)y + 2^(y).2^(x))/(2^(x) + 2^(x+y) log_(e)2), y(0) = 0 , then for y = 1, the value of x lies in the interval :

If (dy)/(dx)=(2^(x+y)-2^(x))/(2^(y)),y(0)=1 then y(1) is equal to

If y satisfies (dy)/(dx)=(e^(y))/(x^(2))-(1)/(x) and y(1)=0 then the value of e^(y(2)) is

Solve the differential equation (dy)/(dx) = y + int_(0)^(1)y(x)dx , given that the value of y is 1, when x = 0.