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oksian1 [2.3K]
4 years ago
15

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

Mathematics
1 answer:
andriy [413]4 years ago
5 0
\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 $y(0) = -3 \ \textless \  0$}
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Mazyrski [523]
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Answer:

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Step-by-step explanation:

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