Solve The Differential Equation. Dy Dx 6x2y2 -

The general solution (y = \frac{1}{K - 2x^3}) is defined wherever the denominator is not zero, i.e., (x \neq \sqrt[3]{\frac{K}{2}}) (the real cube root). At that point, the solution has a vertical asymptote. For example, if (K = 0), then (y = -\frac{1}{2x^3}), which blows up at (x = 0).

To isolate $y$, we take the reciprocal of both sides (raise both sides to the power of -1). solve the differential equation. dy dx 6x2y2

[ y = -\frac{1}{2x^3 + C} ]

When we divided by (y^2) in Step 2, we assumed (y \neq 0). What if (y = 0) everywhere? Let's test it: if (y = 0), then (\frac{dy}{dx} = 0). The right-hand side is (6x^2 (0)^2 = 0). So (y = 0) satisfies the equation. However, our general solution (y = \frac{1}{K - 2x^3}) cannot equal (0) for any finite (K) (since the numerator is 1). Therefore, (y = 0) is a not covered by the general formula. The general solution (y = \frac{1}{K - 2x^3})

[ y = \frac{1}{-2x^3 - C} ]