Find all functions (over real numbers) such that $$f(x^2+y)=xf(x)+f(y).$$
My idea: Put $x=1$. Therefore the given equation becomes $f(y+1)=f(y)+f(1)$. It is in the Cauchy's first form, so we get $f(x)=cx$. It also satisfies that given functional equation.
Is my answer correct?? If not then what is the complete solution?? Please help me!