determine whether the following equation is separable. if so, solve the given initial value problem…

Explanation:

Step1: Rewrite the equation

Rewrite $2yy^{\prime}(t)=9t^{2}$ as $2y\frac{dy}{dt}=9t^{2}$. Then separate variables: $2y;dy = 9t^{2};dt$.

Step2: Integrate both sides

Integrate $\int 2y;dy=\int 9t^{2};dt$. Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we get $y^{2}=3t^{3}+C$.

Step3: Use the initial condition

Given $y(0) = 8$, substitute $t = 0$ and $y=8$ into $y^{2}=3t^{3}+C$. Then $8^{2}=3\times0^{3}+C$, so $C = 64$.

Step4: Solve for y

We have $y^{2}=3t^{3}+64$, so $y(t)=\sqrt{3t^{3}+64}$ (we take the positive square root because when $t = 0$, $y=8>0$).

Answer:

A. The solution to the initial value problem is $y(t)=\sqrt{3t^{3}+64}$