If `(1+x^2)(dy)/(dx)=1+y^2, y(0)=1,` then `y(2)=`

If (dy)/(dx)+(2y)/(x)=0,y(1)=1, then y(2)=

if y=y(x) and (2+sinx)/(y+1)((dy)/(dx))=-cosx ,y(0)=1, then y(pi/2)=

If y=y(x) and (2+sinx)/(y+1)((dy)/(dx))=-cosx ,y(0)=1, then y(pi/2)=

if y=y(x) and (2+sinx)/(y+1)((dy)/(dx))=-cosx ,y(0)=1, then y(pi/2)=

If (dy)/(dx)= y sin 2x, y(0)=1 , then solution is

(1+x)(dy)/(dx)=((1+x)^2+(y-3)) , If y(2)=0 then y(3)=?

(1+x+xy^(2))(dy)/(dx)+(y+y^(3))=0

2 x y+y^2-2 x^2 (dy)/(dx)=0 , y=2 when x=1

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

For each of the following initial value problems verify that the accompanying functions is a solution. (i) x(dy)/(dx)=1, y(1)=0 => y=logx (ii) (dy)/(dx)=y , y(0)=1 => y=e^x (iii) (d^2y)/(dx^2)+y=0, y(0)=0, y^(prime)(0)=1 => y=sinx (iv) (d^2y)/(dx^2)-(dy)/(dx)=0, y(0)=2, y^(prime)(0)=1 => y=e^x+1 (v) (dy)/(dx)+y=2, y(0)=3 => y=e^(-x)+2