If dy/dx = x^2 + 2, then find the value of y in terms of x. It is given that initially, x = 0 and y = 3.

Hi

dy/dx = x^2 + 2
integrating both sides w.r.t x, we get each term as follows:

Integration of dy/dx = y + a
                      x^2 = (x^3)/3  + b
                      2 = 2x + c                     (a,b and c are constants of integration)

Now, combining the terms and taking a,b and c on one side of equation and replacing the resulting constant as Z we get:

y= (x^3)/3 + 2x + Z

But we are given that y=3 when x=0
putting these values in above equation:

3=0 + 0 + Z
Z = 3

Hence, we get our answer as : y= (x^3)/3 + 2x + 3