Question

Find the solution of the differential equation that satisfies the given initial condition. dy/dx=x/y , y(0)=-1

Answer 1

Separating variables, we have

`(dy)/(dx) = \frac xy \implies y\,dy = x\,dx`

Integrate both sides.

`\displaystyle \int y\,dy = \int x\,dx`

`\frac12 y^2 = \frac12 x^2 + C`

Given that
`y(0)=-1`, we find

`\frac12 (-1)^2 = \frac12 0^2 + C \implies C = \frac12`

Then the particular solution is

`\frac12 y^2 = \frac12 x^2 + \frac12`

`y^2 = x^2 + 1`

`y = \pm√(x^2 + 1)`

and because
`y(0)=-1`, we take the negative solution to accommodate this initial value.

`\boxed{y(x) = -√(x^2+1)}`