Solve the differential equation. $$ \frac{d y}{d x}=x \sqrt{y} $$ for $$ y \neq 0 $$

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We start with the differential equation
$$
\frac{dy}{dx} = x \sqrt{y}.
$$
Since $$ y \neq 0 $$, we can separate the variables. Dividing both sides by $$\sqrt{y}$$, we have
$$
\frac{1}{\sqrt{y}}\,dy = x\,dx.
$$
Now, we integrate both sides:
$$
\int \frac{1}{\sqrt{y}}\,dy = \int x\,dx.
$$
The left-hand side integrates as follows:
$$
\int y^{-\frac{1}{2}}\,dy = 2\sqrt{y} + C_1,
$$
and the right-hand side is
$$
\int x\,dx = \frac{x^2}{2} + C_2.
$$
We can combine the constants $$ C_1 $$ and $$ C_2 $$ into a single constant $$ C $$ (since both are arbitrary). Thus, we write:
$$
2\sqrt{y} = \frac{x^2}{2} + C.
$$
To solve for $$\sqrt{y}$$, divide both sides by 2:
$$
\sqrt{y} = \frac{x^2}{4} + C_0,
$$
where $$ C_0 = \frac{C}{2} $$ is an arbitrary constant.

Finally, to express $$ y $$ explicitly, square both sides:
$$
y = \left(\frac{x^2}{4} + C_0\right)^2.
$$

Thus, the general solution to the differential equation is
$$
\boxed{y = \left(\frac{x^2}{4} + C_0\right)^2, \quad \text{where } C_0 \text{ is an arbitrary constant.}}
$$
The solution to the differential equation $$ \frac{dy}{dx} = x \sqrt{y} $$ is: $$ y = \left(\frac{x^2}{4} + C\right)^2, $$ where $$ C $$ is an arbitrary constant.

Solve the differential equation. \[ \frac{d y}{d x}=x \sqrt{y} \] for \( y \neq 0 \)

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