Answer:
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.
