(a) Find the explicit solution of the initial value problem: $$ \frac{d y}{d x}=\frac{2 x}{-8 y-10}, \quad y(0)=-8 $$ $$ y(x)= $$ (b) Determine the interval where the solution is defined. Interval:

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We start with the differential equation

$$
\frac{dy}{dx} = \frac{2x}{-8y-10}
$$

with the initial condition

$$
y(0)=-8.
$$

### (a) Finding the Explicit Solution

1. **Separate the Variables.**  
   Multiply both sides of the differential equation by $$-8y-10$$ and $$dx$$:

$$
   (-8y-10)\,dy = 2x\,dx.
   $$

2. **Integrate Both Sides.**

Integrate the left side with respect to $$y$$ and the right side with respect to $$x$$:

$$
   \int (-8y-10)\,dy = \int 2x\,dx.
   $$

Compute the integrals:

- For the left side:
   
     $$
     \int (-8y-10)\,dy = -8\int y\,dy - 10\int dy = -8\left(\frac{y^2}{2}\right) - 10y = -4y^2 - 10y.
     $$

- For the right side:
   
     $$
     \int 2x\,dx = x^2.
     $$

Thus, we have

$$
   -4y^2 - 10y = x^2 + C.
   $$

3. **Determine the Constant of Integration.**

Use the initial condition $$y(0) = -8$$:

$$
   -4(-8)^2 - 10(-8) = 0^2 + C.
   $$

Compute step by step:
   
   $$
   -4(64) + 80 = -256 + 80 = -176.
   $$
   
   So, we have

$$
   -176 = C.
   $$

Thus, the equation becomes

$$
   -4y^2 - 10y = x^2 - 176.
   $$

4. **Solve for $$y$$.**

Multiply the equation by $$-1$$ to make it easier to handle:

$$
   4y^2 + 10y = -x^2 + 176.
   $$

Bringing all terms to one side:

$$
   4y^2 + 10y + x^2 - 176 = 0.
   $$

This is a quadratic in $$y$$. Write it in the standard form:

$$
   4y^2 + 10y + (x^2 - 176) = 0.
   $$

Use the quadratic formula where $$a=4$$, $$b=10$$, and $$c=x^2-176$$:

$$
   y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-10 \pm \sqrt{10^2 - 4(4)(x^2-176)}}{8}.
   $$

Compute the discriminant:

$$
   10^2 - 4(4)(x^2-176) = 100 - 16x^2 + 16\cdot176.
   $$

Notice that

$$
   16\cdot176 = 2816,
   $$

so the discriminant becomes

$$
   100 - 16x^2 + 2816 = 2916 - 16x^2.
   $$

Thus, the solution for $$y$$ is

$$
   y(x)=\frac{-10 \pm \sqrt{2916 - 16x^2}}{8}.
   $$

5. **Select the Correct Sign.**

Apply the initial condition $$y(0)=-8$$:

$$
   y(0) = \frac{-10 \pm \sqrt{2916}}{8} = \frac{-10 \pm 54}{8}.
   $$

We have two possibilities:
   
   - $$\frac{-10 + 54}{8} = \frac{44}{8} = 5.5,$$
   - $$\frac{-10 - 54}{8} = \frac{-64}{8} = -8.$$
   
   The initial condition $$y(0) = -8$$ tells us that we must choose the negative sign.

Therefore, the explicit solution is

$$
   y(x)=\frac{-10-\sqrt{2916-16x^2}}{8}.
   $$

We can simplify the square root by factoring:

$$
   \sqrt{2916-16x^2} = \sqrt{4\cdot(729-4x^2)} = 2\sqrt{729-4x^2},
   $$

so that

$$
   y(x)=\frac{-10-2\sqrt{729-4x^2}}{8}=\frac{-5-\sqrt{729-4x^2}}{4}.
   $$

### (b) Determining the Interval of Definition

The expression under the square root must be nonnegative:

$$
729-4x^2\ge0.
$$

Solve for $$x$$:

1. Divide by 4:

$$
   x^2\le \frac{729}{4}.
   $$

2. Take square roots:

$$
   |x|\le \frac{27}{2}.
   $$

Thus, the solution is real and defined for

$$
x\in\left[-\frac{27}{2},\frac{27}{2}\right].
$$

However, also note that the original differential equation has a singularity when the denominator $$-8y-10=0$$ (i.e., when $$y=-\frac{10}{8}=-\frac{5}{4}$$). Our solution evaluates to

$$
y(x)=\frac{-5-\sqrt{729-4x^2}}{4}.
$$

At the endpoints $$x=\pm\frac{27}{2}$$, we have

$$
\sqrt{729-4\left(\frac{27}{2}\right)^2} = \sqrt{729-4\left(\frac{729}{4}\right)} = \sqrt{729-729} = 0,
$$

so

$$
y\left(\pm\frac{27}{2}\right)=\frac{-5}{4}=-\frac{5}{4}.
$$

Since the differential equation is not defined at $$y=-\frac{5}{4}$$, the maximal interval of existence (the open interval where the solution remains away from the singularity) is

$$
x\in\left(-\frac{27}{2},\frac{27}{2}\right).
$$

### Final Answers

$$ y(x)=\frac{-5-\sqrt{729-4x^2}}{4} $$

Interval: $$\left(-\frac{27}{2},\frac{27}{2}\right)$$
$$ y(x) = \frac{-5 - \sqrt{729 - 4x^2}}{4} $$ Interval: $$ \left( -\frac{27}{2}, \frac{27}{2} \right) $$

(a) Find the explicit solution of the initial value problem: \[ \frac{d y}{d x}=\frac{2 x}{-8 y-10}, \quad y(0)=-8 \] \( y(x)= \) (b) Determine the interval where the solution is defined. Interval:

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