Answer:
[tex]\boxed{\boxed{y(t)=\frac{2\sqrt{k}}{3}t^{\frac{3}{2}}+c}}[/tex]
Step-by-step explanation:
Rate of ice forming on the pond is,
[tex]\dfrac{dy}{dt}=\sqrt{kt}[/tex]
where,
y = the thickness of the ice in centimeters
t = time passed in hours
k = positive constant
Then,
[tex]dy=\sqrt{kt}\cdot dt[/tex]
Taking integrals on both sides,
[tex]\int dy=\int \sqrt{kt}\cdot dt[/tex]
[tex]\int dy=\int \sqrt{k}\cdot \sqrt{t}\cdot dt[/tex]
As k is a constant, it can come outside the integral,
[tex]\int dy=\sqrt{k}\int \sqrt{t}\cdot dt[/tex]
[tex]y=\sqrt{k} \dfrac{t^{\frac{3}{2}}}{\frac{3}{2}}+c[/tex]
[tex]y=\frac{2\sqrt{k}}{3}t^{\frac{3}{2}}+c[/tex]
