$$\left\{y' = \frac{x}{y}, y(2) = 10\right\} $$
I am asked what $y(1)$ is. In order to solve this Cauchy's problem, this is what I've tried.
$$\frac{dy}{dx} = \frac{x}{y} \rightarrow ydy = xdx$$ $$\int{ydy} = \int{xdx} \rightarrow \frac{y^2}{2} = \frac{x^2}{2}+c$$ $$y = x + \sqrt{2c}$$
Then I have that $$y(2) = 10 \rightarrow 10 = 2 + \sqrt{2c} \rightarrow 8 = \sqrt{2c} \rightarrow c = 32$$
So $y(1)$ would be: $$y(1) = 1 + 8 = 9$$
but, the answer is actually $y(1) = \sqrt{97}$.
This is demotivating. What am I doing wrong?