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JцstinBieber
[tex]\displaystyle \frac{dy}{dx} = \frac{x}{y} \ \Rightarrow\ y \,dy = x\, dx\ \Rightarrow\textstyle \ \int y \,dy = \int x\, dx\ \Rightarrow\ \frac{1}{2}y^2 = \frac{1}{2}x^2 + C \Rightarrow \\ \\ y(0) = -3\ \Rightarrow\ \frac{1}{2}(-3)^2 = \frac{1}{2}(0)^2 + C\ \Rightarrow\ C = \frac{9}{2},\\ \\ \text{so } \frac{1}{2}y^2 = \frac{1}{2}x^2 + \frac{9}{2}\ \Rightarrow\ y^2 = x^2 + 9 \ \Rightarrow \ \\ \\ y = -\sqrt{x^2 + 9}\ \text{ since $y(0) = -3 \ \textless \ 0$}[/tex]

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