Given the initial value problem y′=3x2y2(x3+1)𝑦′=3𝑥2𝑦2(𝑥3+1) with y(0)=1𝑦(0)=1. It follows that y(2)

The given differential equation is y′=3x²y²/(x³+1) with the initial condition y(0)=1.

To solve this, we can use the method of separation of variables.

First, we separate the variables by dividing both sides by y² and multiplying both sides by dx:

1/y² dy = 3x²/(x³+1) dx

Next, we integrate both sides:

∫1/y² dy = ∫3x²/(x³+1) dx

The left side of the equation becomes -1/y after integration.

The right side of the equation is a bit more complex. We can use the substitution method to solve it. Let u = x³ + 1, then du = 3x² dx. The integral becomes:

∫1/u du = ln|u| + C

Substituting back for u, we get:

ln|x³ + 1| + C

So, the solution to the differential equation is:

-1/y = ln|x³ + 1| + C

To find the constant C, we use the initial condition y(0) = 1. Substituting these values in, we get:

-1/1 = ln|0 + 1| + C C = -1

So, the solution to the differential equation is:

-1/y = ln|x³ + 1| - 1

Finally, we want to find y(2). Substituting x = 2 into the equation, we get:

-1/y = ln|8 + 1| - 1 -1/y = ln|9| - 1 -1/y = 2 - 1 -1/y = 1

Solving for y, we get y = -1.

So, y(2) = -1.