Solve the initial value problem: $$ 8 x-3 y \sqrt{x^{2}+1} \frac{d y}{d x}=0, \quad y(0)=-2 $$ $$ y(x)= $$

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We start with the differential equation
$$
8x - 3y \sqrt{x^2+1}\frac{dy}{dx}=0.
$$
Rearrange the equation to solve for $$\frac{dy}{dx}$$:
$$
3y \sqrt{x^2+1}\frac{dy}{dx} = 8x \quad\Longrightarrow\quad \frac{dy}{dx}=\frac{8x}{3y\sqrt{x^2+1}}.
$$

This equation is separable. Multiply both sides by $$y\,dx$$ to obtain:
$$
y\,dy = \frac{8x}{3\sqrt{x^2+1}}\,dx.
$$

Now, integrate both sides:
$$
\int y\,dy = \frac{8}{3}\int \frac{x}{\sqrt{x^2+1}}\,dx.
$$

### Left-hand side integration
$$
\int y\,dy = \frac{y^2}{2} + C_1.
$$

### Right-hand side integration
Let $$u=x^2+1$$. Then, $$du=2x\,dx$$ so that $$x\,dx=\frac{du}{2}$$. Hence, the integral becomes
$$
\frac{8}{3}\int \frac{x}{\sqrt{x^2+1}}\,dx = \frac{8}{3}\int \frac{1}{\sqrt{u}}\cdot \frac{du}{2} = \frac{4}{3}\int u^{-1/2}\,du.
$$
Integrate:
$$
\frac{4}{3}\int u^{-1/2}\,du = \frac{4}{3}\cdot \frac{u^{1/2}}{(1/2)} = \frac{8}{3}\sqrt{u} = \frac{8}{3}\sqrt{x^2+1}.
$$

### Combine the results
Equate the integration results (absorbing the constant into one constant $$C$$):
$$
\frac{y^2}{2}=\frac{8}{3}\sqrt{x^2+1}+C.
$$

### Apply the initial condition
Given $$y(0)=-2$$, substitute $$x=0$$ and $$y=-2$$:
$$
\frac{(-2)^2}{2}=\frac{8}{3}\sqrt{0^2+1}+C \quad\Longrightarrow\quad \frac{4}{2}=\frac{8}{3}+C,
$$
$$
2=\frac{8}{3}+C \quad\Longrightarrow\quad C=2-\frac{8}{3}=\frac{6-8}{3}=-\frac{2}{3}.
$$

Substitute $$C$$ back into the equation:
$$
\frac{y^2}{2}=\frac{8}{3}\sqrt{x^2+1}-\frac{2}{3}.
$$
Multiply both sides by 2:
$$
y^2=\frac{16}{3}\sqrt{x^2+1}-\frac{4}{3}=\frac{4}{3}\Bigl(4\sqrt{x^2+1}-1\Bigr).
$$

Since $$y(0)=-2$$ is negative, we take the negative square root:
$$
y(x)=-\sqrt{\frac{4}{3}\Bigl(4\sqrt{x^2+1}-1\Bigr)} = -\frac{2}{\sqrt{3}}\sqrt{4\sqrt{x^2+1}-1}.
$$

Thus, the solution to the initial value problem is
$$
y(x)=-\frac{2}{\sqrt{3}}\sqrt{4\sqrt{x^2+1}-1}.
$$
The solution to the initial value problem is: $$ y(x) = -\frac{2}{\sqrt{3}}\sqrt{4\sqrt{x^2+1}-1} $$

Solve the initial value problem:

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