- \(\displaystyle{y' + \frac{4}{x} y = x^3 y^2, \quad x > 0}\) View Solution
- \(\displaystyle{x \frac{dy}{dx} + y = x^2 y^2}\) View Solution
- \(\displaystyle{\frac{dy}{dx} = y(xy^3 - 1)}\) View Solution
- \(\displaystyle{x y' + y = \frac{1}{y^2}, \quad y(1) = 3}\) View Solution