1 Answer

Damjan · Answer 1 · 2023-07-09T07:18:18+0000

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.

$\boxed{y(x) = -√(x^2+1)}$

Please log in or register to add a comment.