Question

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

dy/dx = x/y, y(0) = -3

Answer 1

`\displaystyle (dy)/(dx) = (x)/(y) \ \Rightarrow\ y \,dy = x\, dx\ \Rightarrow\textstyle \ \int y \,dy = \int x\, dx\ \Rightarrow\ (1)/(2)y^2 = (1)/(2)x^2 + C \Rightarrow \\ \\ y(0) = -3\ \Rightarrow\ (1)/(2)(-3)^2 = (1)/(2)(0)^2 + C\ \Rightarrow\ C = (9)/(2),\\ \\ \text{so } (1)/(2)y^2 = (1)/(2)x^2 + (9)/(2)\ \Rightarrow\ y^2 = x^2 + 9 \ \Rightarrow \ \\ \\ y = -√(x^2 + 9)\ \text{ since $y(0) = -3 \ \textless \ 0$}`