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

$\dfrac{dy}{dx} = \dfrac xy \implies y\,dy = x\,dx$

Integrate both sides.

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

$\dfrac12 y^2 = \dfrac12 x^2 + C$

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

$\dfrac12 (-1)^2 = \dfrac12 0^2 + C \implies C = \dfrac12$

Then the particular solution is

$\dfrac12 y^2 = \dfrac12 x^2 + \dfrac12$

$y^2 = x^2 + 1$

$y = \pm\sqrt{x^2 + 1}$

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

$\boxed{y(x) = -\sqrt{x^2+1}}$