(b) Find a solution of the initial-value problem. (Hint: First verify that all members of the family

y = 4/x + C are solutions of the given equation.) y' = −1/4y^2
y(0) = 0.25

To find a solution to the given initial-value problem, we need to first verify that all members of the family of functions y = 4/x + C are solutions of the given equation y' = −1/4y^2.

Let's differentiate the function y = 4/x + C with respect to x using the derivative rules:

dy/dx = d(4/x + C)/dx
= d(4/x)/dx + dC/dx
= (-4/x^2) + 0 (dC/dx = 0 since C is a constant)
= -4/x^2

Now we can substitute this derivative into the given equation y' = −1/4y^2:

-4/x^2 = -1/4(4/x + C)^2

Next, we can simplify the equation:

-4/x^2 = -1/4(16/x^2 + 8C/x + C^2)

Multiplying both sides by -4:

16/x^2 = 16/x^2 + 8C/x + C^2

Canceling out the common terms on both sides:

0 = 8C/x + C^2

Now, let's substitute the initial condition y(0) = 0.25 into the equation y = 4/x + C:

0.25 = 4/0 + C
C = 0.25

So, the value of C is 0.25.

Now we can substitute this value of C back into the equation 0 = 8C/x + C^2:

0 = 8(0.25)/x + (0.25)^2
0 = 2/x + 0.0625

Now, we can solve for x:

2/x = -0.0625
x = -2/0.0625
x = -32

Therefore, the solution to the initial-value problem y' = −1/4y^2, y(0) = 0.25 is y = 4/x + C, where C = 0.25.