Question

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.

Answer 1

`\displaystyle(dy)/(dx) = (x^2)/(y^2)\ \Rightarrow\ y^2 dy = x^2 dx\ \Rightarrow\ \int y^2 dy = \int x^2 dx\ \Rightarrow\textstyle\ (1)/(3)y^3 = (1)/(3)x^3 + C. \\ \\ \text{Now } y(0) = 2\ \Rightarrow\ (1)/(3)(2)^3 = (1)/(3)(0)^3 + C\ \Rightarrow\ (8)/(3) = C,\text{ so } (1)/(3)y^3 = (1)/(3)x^3 + (8)/(3). \\ \\ y^3 = x^3 + 8\ \Rightarrow\ y = \sqrt[3]{x^3 + 8}`