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Solve the following initial value problem. dxdy​=y2−15y+56,y(0)=5 Problem #1 : Enter your answer as a symbolic function of x as in these examples Problem \# 2: Let y(x) be the solution to the following initial value problem. y2dx−csc2(4x)dy=0,y(0)=6 Find y(π). Problem #2; Enter your answer symbolically, as in these examples

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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.

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