We are given the separable differential equation:

$4x - 6y\sqrt{x^2 + 1}\frac{dy}{dx} = 0$

with the initial condition $y(0) = 10$.

Step 1: Rearrange the equation

First, let's isolate $\frac{dy}{dx}$ by moving the terms to one side:

$6y\sqrt{x^2 + 1} \frac{dy}{dx} = 4x$

Now divide both sides by $6y\sqrt{x^2 + 1}$ to separate the variables:

$\frac{dy}{dx} = \frac{4x}{6y\sqrt{x^2 + 1}} = \frac{2x}{3y\sqrt{x^2 + 1}}$

Step 2: Separate variables

Rearranging the equation to separate $y$ and $x$:

$y \, dy = \frac{2x}{3\sqrt{x^2 + 1}} \, dx$

Step 3: Integrate both sides

Now integrate both sides.

For the left-hand side:

$\int y \, dy = \frac{y^2}{2}$

For the right-hand side:

$\int \frac{2x}{3\sqrt{x^2 + 1}} \, dx$

Let’s perform a substitution for the right-hand side:

• Set $u = x^2 + 1$, so $du = 2x \, dx$.

Thus, the integral becomes:

$\int \frac{2x}{3\sqrt{x^2 + 1}} \, dx = \frac{1}{3} \int \frac{du}{\sqrt{u}} = \frac{1}{3} \cdot 2\sqrt{u} = \frac{2}{3}\sqrt{x^2 + 1}$

Step 4: Combine the results

Now we combine the two integrals:

$\frac{y^2}{2} = \frac{2}{3}\sqrt{x^2 + 1} + C$

Multiply both sides by 2:

$y^2 = \frac{4}{3}\sqrt{x^2 + 1} + 2C$

Let $C_1 = 2C$, so:

$y^2 = \frac{4}{3}\sqrt{x^2 + 1} + C_1$

Step 5: Apply the initial condition

We know $y(0) = 10$, so substitute $x = 0$ and $y = 10$ into the equation:

$10^2 = \frac{4}{3}\sqrt{0^2 + 1} + C_1$

This simplifies to:

$100 = \frac{4}{3} \times 1 + C_1$

$100 = \frac{4}{3} + C_1$

$C_1 = 100 - \frac{4}{3} = \frac{300}{3} - \frac{4}{3} = \frac{296}{3}$

Step 6: Final solution

Substitute $C_1$ back into the equation:

$y^2 = \frac{4}{3}\sqrt{x^2 + 1} + \frac{296}{3}$

Thus, the solution is:

$y(x) = \sqrt{\frac{4}{3}\sqrt{x^2 + 1} + \frac{296}{3}}$

Do you have any questions or want more details?

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Here are 5 related questions you might explore:

1. How do you perform separation of variables in differential equations?
2. What is the method of substitution in integrals?
3. How does the initial condition affect the solution of a differential equation?
4. What are separable and non-separable differential equations?
5. Can all first-order differential equations be solved by separation of variables?

Tip: Always check the initial conditions after solving a differential equation to ensure the solution is consistent.