Solve the given initial value problem and determine how the interval in which the solution exists depends on the initial value y0.

y' + y3 = 0 y (0) = y0

y has a finite solution for any value y_0 ≠ 0.

Step-by-step explanation:

Given the differential equation

y' + y³ = 0

We can rewrite this as

dy/dx + y³ = 0

Multiplying through by dx

dy + y³dx = 0

Divide through by y³, we have

dy/y³ + dx = 0

dy/y³ = - dx

Integrating both sides

-1 / (2y²) = - x + c

Multiplying through by - 1, we have

1 / (2y²) = x + C (Where C = - c)

Applying the initial condition y (0) = y_0, put x = 0, and y = y_0

1 / (2y_0²) = 0 + C

C = 1 / (2y_0²)

So

1 / (2y²) = x + 1 / (2y_0²)

2y² = 1/[x + 1 / (2y_0²) ]

y² = 1/[2x + 1 / (y_0²) ]

y = 1/[2x + 1 / (y_0²) ]½

This is the required solution to the initial value problem.

The interval of the solution depends on the value of y_0. There are infinitely many solutions for y_0 assumes a real number.

For y_0 = 0, the solution has an expression 1/0, which makes the solution infinite.

With this, y has a finite solution for any value y_0 ≠ 0.