Answer
$1.5275$
Work Step by Step
Here, $\int y dy=\int \sqrt x dx$ ...(1)
and $(\dfrac{y^2}{2})=\dfrac{2}{3}x^{(3/2)}+c$ ...(2)
Further, apply the initial conditions.
$\dfrac{1}{2}=\dfrac{2}{3}(0)^{3/2}+c$
or, $c=\dfrac{1}{2}$
Now, the particular solution as follows:
$(\dfrac{y^2}{2})=\dfrac{2}{3}x^{(3/2)}+c \implies \dfrac{y^2}{2}=(\dfrac{2}{3})x^{(3/2)}+(\dfrac{1}{2})$
Thus, $y=\sqrt {\dfrac{4}{3}x^{(3/2)}+1}$
and $y(1)=\sqrt {\dfrac{4}{3}+1} \approx 1.5275$
