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19 September, 01:59

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

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  1. 19 September, 02:46
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    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.
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