bradhunt7966 bradhunt7966
  • 20-04-2018
  • Mathematics
contestada

Solve the differential equation dy dx equals the quotient of x squared and y squared for y = f(x) with the initial condition y(0) = 2.

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JцstinBieber
JцstinBieber JцstinBieber
  • 23-04-2018
[tex]\displaystyle\frac{dy}{dx} = \frac{x^2}{y^2}\ \Rightarrow\ y^2 dy = x^2 dx\ \Rightarrow\ \int y^2 dy = \int x^2 dx\ \Rightarrow\textstyle\ \frac{1}{3}y^3 = \frac{1}{3}x^3 + C. \\ \\ \text{Now } y(0) = 2\ \Rightarrow\ \frac{1}{3}(2)^3 = \frac{1}{3}(0)^3 + C\ \Rightarrow\ \frac{8}{3} = C,\text{ so } \frac{1}{3}y^3 = \frac{1}{3}x^3 + \frac{8}{3}. \\ \\ y^3 = x^3 + 8\ \Rightarrow\ y = \sqrt[3]{x^3 + 8}[/tex]
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