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

Explanation:

Step1: Rewrite the differential - equation

Given $2yy^{\prime}(t)=3t^{2}$, we can rewrite it as $2y\frac{dy}{dt}=3t^{2}$. Then, separate the variables: $2y;dy = 3t^{2};dt$.

Step2: Integrate both sides

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

Step3: Use the initial condition

Given $y(0) = 3$, substitute $t = 0$ and $y = 3$ into the equation $y^{2}=t^{3}+C$. So, $3^{2}=0^{3}+C$, which gives $C = 9$.

Step4: Solve for $y(t)$

We have $y^{2}=t^{3}+9$. Since $y(0)=3>0$, we take the positive square root: $y(t)=\sqrt{t^{3}+9}$.

Answer:

A. The solution to the initial value problem is $y(t)=\sqrt{t^{3}+9}$