Question

Please answer this Calculus webwork question about differential equations:

Answer 1

a)

`4ydy = xdx`

`2 {y}^(2) = (1)/(2) {x}^(2) + c`

`{y}^(2) = (1)/(4) {x}^(2) + c`

`{y}^(2) - (1)/(4) {x}^(2) = c`

b)

c = 4, so

`{y}^(2) = (1)/(4) {x}^(2) + 4`

`y = - \sqrt{ (1)/(4) {x}^(2) + 4}`

`y = - \frac{ \sqrt{ {x}^(2) + 16} }{2}`