CALCULUS NEED HELP ASAP PLS! DIFFERENTIAL EQUATIONS

Question

1. Solve the differential equation  .
2. Explain why the initial value problem  with y(0) = 2 does not have a solution.

Answer 1

If y' = 6x √(1-y²)

dy/√(1-y²) = 6x dx

sin^-1 y = 3x² + c

y = sin(3x² + c)

From the solution, we see that y is on the interval[-1, 1]. We also know that √(1-y²) only exists for -1 <= y <= 1.