`dy/dx+xy=y^3*e^(x^2)" ";" "y(0)=1/2`

xy-dy/dx=y^3e^(-x^2)

If (dy)/dx = (xy)/(x^2 + y^2), y(1) = 1 and y(x) = e then x =

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

Solution of xy-(dy)/(dx)=y^(3)e^(-x^(2)) is

dy/dx + 2xy - x = 0, y(0) = 3.

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

Solve the differential equation (dy)/(dx)=1+x+y^(2)+xy^(2) , when y=0 and x=0.