Question

Question #31517

Answer 1

`y= 1/4 x^2 + 3`

Explanation:

`(dy)/(dx) = 1/2 x`

`y = int dy = int 1/2 x dx`

`y = 1/4 x^2 + c`

plug in `y = 3, x = 2` in the above equation to find `c`

`3 =1/4 (2)^2 + c`

`3 = c`

therefore it equation is `y= 1/4 x^2 + 3`