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

\(\displaystyle{y}=-\sqrt{{{x}^{{2}}+{9}}}\)

\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{x}}}{{{y}}}}\)

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

\(\displaystyle{y}=-\sqrt{{{x}^{{2}}+{9}}}\)