dentify the type of the differential equation (as linear, homogeneous,exact, separable or Bernoulli) and solve the initial value problemdydt = 2(ty)2, with y(0) = 3.Describe the maximal interval where the solution of the initial valueproblem is defined

The given differential equation is dy/dt = 2(ty)^2. This is a nonlinear differential equation. It is not linear, homogeneous, exact, or Bernoulli. However, it is separable.

To solve this separable differential equation, we first rewrite it in the form that separates the variables y and t:

1/y^2 dy = 2t dt

We then integrate both sides:

∫1/y^2 dy = ∫2t dt

This gives us:

-1/y = t^2 + C

We can solve for y to get:

y = -1/(t^2 + C)

Given the initial condition y(0) = 3, we can solve for C:

3 = -1/C

This gives C = -1/3. So the solution to the differential equation is:

y = -1/(t^2 - 1/3)

The maximal interval where the solution is defined is (-∞, √(1/3)) U (√(1/3), ∞). This is because the denominator of the solution, t^2 - 1/3, cannot be zero, so t cannot be ±√(1/3).