Solve the following differential equation: 6−x2−−−−−√dydx−x(y+2)=06−x2dydx−x(y+2)=0 Use the letter c to denote the arbitrary constant occurring in the general solution.

The differential equation given is:

\[
\sqrt{6 - x^2} \frac{dy}{dx} - x(y + 2) = 0
\]

We can solve this by separating the variables.

### Step 1: Rearrange the Equation
First, let's rearrange the equation to separate the variables \( x \) and \( y \):

\[
\sqrt{6 - x^2} \frac{dy}{dx} = x(y + 2)
\]

Next, divide both sides by \( \sqrt{6 - x^2} \) and \( y + 2 \):

\[
\frac{dy}{y + 2} = \frac{x}{\sqrt{6 - x^2}} dx
\]

### Step 2: Integrate Both Sides
Now, integrate both sides:

\[
\int \frac{dy}{y + 2} = \int \frac{x}{\sqrt{6 - x^2}} dx
\]

#### Left Side:
The integral on the left side is straightforward:

\[
\int \frac{dy}{y + 2} = \ln |y + 2|
\]

#### Right Side:
For the right side, use the substitution \( u = 6 - x^2 \), so \( du = -2x dx \):

\[
\int \frac{x}{\sqrt{6 - x^2}} dx = -\frac{1}{2} \int \frac{-2x dx}{\sqrt{6 - x^2}} = -\frac{1}{2} \int \frac{du}{\sqrt{u}}
\]

The integral of \( \frac{1}{\sqrt{u}} \) is \( 2\sqrt{u} \):

\[
-\frac{1}{2} \times 2\sqrt{u} = -\sqrt{u} = -\sqrt{6 - x^2}
\]

So, the integral on the right side is:

\[
\int \frac{x}{\sqrt{6 - x^2}} dx = -\sqrt{6 - x^2}
\]

### Step 3: Combine and Solve for \( y \)

Combining the results from both sides:

\[
\ln |y + 2| = -\sqrt{6 - x^2} + C
\]

Where \( C \) is the constant of integration. Exponentiate both sides to solve for \( y \):

\[
|y + 2| = e^{-\sqrt{6 - x^2} + C} = e^{-\sqrt{6 - x^2}} \cdot e^C
\]

Let \( e^C = c \), where \( c \) is a new arbitrary constant:

\[
|y + 2| = c e^{-\sqrt{6 - x^2}}
\]

So, the general solution is:

\[
y + 2 = \pm c e^{-\sqrt{6 - x^2}}
\]

Finally, subtract 2 from both sides:

\[
y(x) = -2 \pm c e^{-\sqrt{6 - x^2}}
\]

### Final Answer:

The general solution to the differential equation is:

\[
y(x) = -2 + c e^{-\sqrt{6 - x^2}} \quad \text{or} \quad y(x) = -2 - c e^{-\sqrt{6 - x^2}}
\]

where \( c \) is an arbitrary constant.

Do you need further explanation on any part of this solution, or do you have additional questions?

### Relative Questions:
1. How do you determine the appropriate substitution for solving the integral?
2. Why do we need to include the arbitrary constant \( c \) when solving differential equations?
3. How would the solution change if the initial condition \( y(x_0) = y_0 \) were provided?
4. What other methods can be used to solve this type of differential equation?
5. Can you explain the role of absolute value in the solution process?

### Tip:
When solving differential equations, remember to check whether the equation can be separated into variables before applying more complex methods. This often simplifies the solution process.

Learn how to solve the differential equation 6 - sqrt(x^2) dy/dx - x(y + 2) = 0 using separation of variables and integration techniques. This problem covers key concepts in differential equations and calculus. Suitable for advanced high school students and college-level calculus enthusiasts.

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