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.

she played for 18.75 hours.

step-by-step explanation:

if she played 2/4 of a hour on tuesday. and there is 60 minutes in an hour and 60 divided by 4 is 7.5. and she played 2/4 of an hour on thursday then again 60 divided by 4 is 7.5, you add 7.5+7.5=56.25. now you have to do 1/4 of an hour, it equals 3.75. now you have to add them all up. 3.75+7.5+7.5=18.75

i hope this and i hope you get it right. it took me a while to get this done for you. : )

step-by-step explanation: