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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?

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