If `y(x)` is solution of `x(dy)/(dx)+2y=x^2, y(1)=1` then value of `y(1/2)=`

" Let "y=y(x)" be the solution of "(y+2x^(3))(dx)/(dy)=x*y(1)=2" The value of "y(2)" is "

x^2dy+(y-1/x)dx=0 at y(1)=1 then find the value of y(1/2)

If y(x) is a solution of (dy)/(dx)-(xy)/(1+x)=(1)/(1+x) and y(0)=-1 , then the value of y(2) is

If y(x) is a solution of (dy)/(dx)-(xy)/(1+x)=(1)/(1+x) and y(0)=-1 , then the value of y(2) is

The solution of (dy)/(dx)=(x^2+y^2+1)/(2x y) satisfying y(1)=1 is

If y=y(x) is the solution of the differential equation x(dy)/(dx)+2y=x^(2) satisfying y(1)=1 , then y(1/2) is equal to (7k)/16 . The value of k is___________________.

If y_(1)(x) and y_(2)(x) are two solutions of (dy)/(dx)+f(x)y=r(x), then y_(1)(x)+y_(2)(x) is solution of : (A) (dy)/(dx)+f(x)y=0 (B) (dy)/(dx)+2f(x)y=r(x)(C)(dy)/(dx)+f(x)y=2r(x)(D)(dy)/(dx)+2f(x)y=2r(x)

The solution of (dy)/(dx) = (1+y^(2))(1+x^(2))^(-1) is