$$ y(x) = \int_{0}^{x} (y(t))^{\frac{1}{2}} dt + 1 $$
What will be a general solution? I couldn't find any related problems in stackexchange. I tried
$$u'(x) = \frac{1}{2}\int_{0}^{x} (y(t))^{\frac{1}{2}} (y ′ (t))dt $$
and
$$u''(x) = \frac{1}{4}\int_{0}^{x} y(t)^(1/2) (y ′(t))^2dt$$
But, I couldn't go further. Thanks in advance.
On
You made a mistake when you took the derivative $$y(x) = \int_{0}^{x} (y(t))^{\frac{1}{2}} dt + 1$$ $$y'(x) =\frac {dy}{dx}=\frac {d}{dx} (\int_{0}^{x} (y(t))^{\frac{1}{2}} dt + 1)$$ And not ( which is what you did) : $$y'(x) = \frac {d}{dt}(\int_{0}^{x} (y(t))^{\frac{1}{2}} dt + 1)$$ Use the Fondamental Theorem of Calculus $$y'=(y(x))^{\frac{1}{2}}.1$$ It's an easy ode to solve...since it's separable
You will have a constant C to evaluate it use the fact that
$$y(0) = \int_{0}^{0} (y(t))^{\frac{1}{2}} dt + 1=1$$
The integral is zero..
From $y(x) = \int_{0}^{x} (y(t))^{\frac{1}{2}} dt + 1$ you get, by the FTC, the initial value problem
$y'(x)=(y(x))^{\frac{1}{2}}, $
$y(0)=1$.
Can you proceed ?