# Discuss the existence and uniquness of IVP $y=g(x)\dfrac{dy}{dx}, y(0)=1$

Discuss the existence and uniquness of IVP $$y=g(x)\dfrac{dy}{dx}, y(0)=1$$ where

$$g(x)= \dfrac{\sin(x)}{x}; x\ne 0$$

and

$$g(x)=1; x=0$$

I calculated $$dy/dx=f(x)$$ first:

$$f(x)= \dfrac{xy}{\sin(x)}; x \ne 0$$ and $$f(x)=y; x=0$$

For existence, $$f(x)$$ must be continuous, since the limit $$\dfrac{xy}{\sin(x)}$$ for $$(0,1)$$ tends to $$1$$, the function $$f(x)$$ is continuous. Hence, a solution to the IVP exists.

Now, for uniqueness $$\dfrac{\partial f(x)}{\partial y}$$ must be continuous.

$$\dfrac{\partial f(x)}{\partial y} = \dfrac{x}{\sin x}; x \ne 0$$ and $$\dfrac{\partial f(x)}{\partial y} = 1; x=0$$ Hence, $$\dfrac{\partial f(x)}{\partial y}$$ is continuous.

Therefore, the solution to the given IVP exists and is unique.

My question is, are there any errors in what I did? Or, if I write this solution in my math exam, due to what missing conditions/calculations/writing style may I lose credits?