Find the solution of the differential equation that satisfies the given initial

Find the solution of the differential equation that satisfies the given initial condition. $$dy/dx=x/y, y(0)=-3$$

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Nathaniel Kramer
Step 1
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{x}}}{{{y}}}}$$
we can rewrite it in differential notation and integrate:
xdx=ydy
$$\displaystyle\int{x}{\left.{d}{x}\right.}=\int{y}{\left.{d}{y}\right.}$$
$$\displaystyle{\frac{{{1}}}{{{2}}}}{x}^{{2}}+{c}={\frac{{{1}}}{{{2}}}}{y}^{{2}}$$
$$\displaystyle{y}^{{2}}={x}^{{2}}+{C}$$
$$\displaystyle{y}=\pm\sqrt{{{x}^{{2}}+{c}}}$$
square root always positive and initial value of y is negative
$$\displaystyle{y}=-\sqrt{{{x}^{{2}}+{C}}}$$
Since y(0)=-3
$$\displaystyle{y}{\left({0}\right)}=-\sqrt{{{0}^{{2}}+{C}}}=-{3}$$
C=9
$$\displaystyle{y}=-\sqrt{{{x}^{{2}}+{9}}}$$
Result
$$\displaystyle{y}=-\sqrt{{{x}^{{2}}+{9}}}$$