Question

In each of Problems 9 through 12, solve the given initial value problem and determine how the interval in which the solution exists depends on the initial value yo.

a. y' = -4t/y, y(0) = yo
b. y' = 2ty², y(0) = yo
c. y'y³ = 0, y(0) = yo
d. y' 12 y(1 + t³) ' = y(0) = yo

Answer 1

In each initial value problem, the interval in which the solution exists depends on the initial value yo.

Step-by-step explanation:

In each of the given initial value problems, the interval in which the solution exists depends on the initial value yo.

a. y' = -4t/y, y(0) = yo: The solution exists as long as y ≠ 0 because division by zero is undefined. So, the interval in which the solution exists is all values of t except where y = 0.

b. y' = 2ty², y(0) = yo: The solution exists as long as y ≠ 0 because division by zero is undefined. So, the interval in which the solution exists is all values of t except where y = 0.

c. y'y³ = 0, y(0) = yo: The equation y'y³ = 0 implies that either y' = 0 or y³ = 0. If y' = 0, the solution exists for all values of t. If y³ = 0, the solution exists only when y = 0. So, the interval in which the solution exists depends on the values of y' and y³.

d. y' 12 y(1 + t³) ' = y(0) = yo: There seems to be a typo in the equation. Please provide the correct equation to determine the interval in which the solution exists.