4(c)

The equation is not linear, separable, or exact: $$\frac{\partial M}{\partial y} = x^2 + 2 y$$while $$\frac{\partial N}{\partial x} = 3 x^2 - 2 y$$. But

$$\frac{\partial M/\partial y - \partial N/\partial x}{N} = \frac{-2 x^2 + 4 y}{x^3 - 2 x y} = \frac{-2}{x}$$

is a function of x alone. Therefore we will have an integrating factor that is a function of x. We solve $\mu' = -2\mu/x$ to obtain $\mu = 1/x^2$. Multiplying the equation by $\mu$, y + y²/x² + (x - 2 y/x) y' = 0 is exact.

$$F(x,y) = \int \left(y + \frac{y^2}{x^2}\right)\, dx = x y - \frac{y^2}{x} + h(y)$$

$$x - \frac{2 y}{x^2} = \frac{\partial F}{\partial y} = x - \frac{2 y}{x} + h'(y)$$

so h'(y) = 0 and h(y) = 0. The general solution is x y - y²/x =C. However, it is impossible to satisfy the initial condition because this is not defined for x=0. In fact, we might have saved ourselves a lot of work by being clever enough to notice at the start that substituting x=0, y=1 into the differential equation yields 1 = 0.