# Solve differential equation \frac{dy}{dx}= \frac{x}{y}, \ y(0)= -8

Question
Solve differential equation $$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{x}}}{{{y}}}},\ {y}{\left({0}\right)}=-{8}$$

2021-03-09
$$\displaystyle{y}{\left.{d}{y}\right.}={x}{\left.{d}{x}\right.}$$
$$\displaystyle\int{y}{\left.{d}{y}\right.}=\int{x}{\left.{d}{x}\right.}$$
$$\displaystyle{\frac{{{y}^{{2}}}}{{{2}}}}={\frac{{{x}^{{2}}}}{{{2}}}}+{c}$$
Now we use initial condition
$$\displaystyle{y}{\left({0}\right)}=-{8}$$
Substitute the value we get
$$\displaystyle{\frac{{{\left(-{8}\right)}^{{2}}}}{{{2}}}}={\frac{{{0}^{{2}}}}{{{2}}}}+{c}$$
$$\displaystyle{\frac{{{64}}}{{{2}}}}={c}$$
$$\displaystyle{c}={32}$$
$$\displaystyle{\frac{{{y}^{{2}}}}{{{2}}}}={\frac{{{x}^{{2}}}}{{{2}}}}+{32}$$
$$\displaystyle{y}^{{2}}={x}^{{2}}+{64}$$

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