Answer:
Explanation:
Problem #1:
To solve the initial value problem dx/dy = y^2 - 15y + 56, y(0) = 5, we can use separation of variables.
Separating the variables, we have:
dx = (y^2 - 15y + 56) dy
Integrating both sides, we get:
∫ dx = ∫ (y^2 - 15y + 56) dy
Integrating the right side, we have:
x = (1/3)y^3 - (15/2)y^2 + 56y + C
Now we can use the initial condition y(0) = 5 to find the value of C:
0 = (1/3)(5^3) - (15/2)(5^2) + 56(5) + C
Simplifying, we have:
0 = 125/3 - 375/2 + 280 + C
0 = -625/6 + 280 + C
C = 625/6 - 280
C = 625/6 - 1680/6
C = -1055/6
Therefore, the solution to the initial value problem is:
x = (1/3)y^3 - (15/2)y^2 + 56y - 1055/6
Problem #2:
To solve the initial value problem y^2 dx - csc^2(4x) dy = 0, y(0) = 6, we can also use separation of variables.
Separating the variables, we have:
y^2 dx = csc^2(4x) dy
Integrating both sides, we get:
∫ y^2 dx = ∫ csc^2(4x) dy
Integrating the left side, we have:
x = -cot(4x) + C
Now we can use the initial condition y(0) = 6 to find the value of C:
0 = -cot(4(0)) + C
0 = -cot(0) + C
0 = -∞ + C
C = ∞
Therefore, the solution to the initial value problem is:
x = -cot(4x) + ∞
To find y(π), substitute x = π into the equation:
π = -cot(4π) + ∞
Since cot(4π) = cot(0) = ∞, we have:
π = -∞ + ∞
The equation is undefined since ∞ - ∞ is an indeterminate form.
Hence, the value of y(π) cannot be determined from the given initial value problem.