`sqrt(1-x^(2))dy+ydx=0`

Options

A. `y(x+sqrt(1-x^(2)))=c`
B. `logy+sin^(-1)x=c`
C. `ysin^(-1)x=c`
D. `y^(2)=2sin^(-1)x+c`

Answer

To solve the differential equation \( \sqrt{1 - x^2} \, dy + y \, dx = 0 \), we can follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the given equation:
\[ \sqrt{1 - x^2} \, dy = -y \, dx \]
Now, we can separate the variables \( y \) and \( x \):
\[ \frac{dy}{y} = -\frac{dx}{\sqrt{1 - x^2}} \]

### Step 2: Integrating Both Sides
Next, we integrate both sides. The left side integrates to:
\[ \int \frac{dy}{y} = \ln |y| + C_1 \]
The right side requires the integral of \( -\frac{1}{\sqrt{1 - x^2}} \):
\[ \int -\frac{dx}{\sqrt{1 - x^2}} = -\sin^{-1}(x) + C_2 \]
Combining the constants of integration, we can write:
\[ \ln |y| = -\sin^{-1}(x) + C \]

### Step 3: Exponentiating Both Sides
To eliminate the logarithm, we exponentiate both sides:
\[ |y| = e^{-\sin^{-1}(x) + C} = e^C e^{-\sin^{-1}(x)} \]
Let \( K = e^C \) (where \( K > 0 \)), we have:
\[ |y| = K e^{-\sin^{-1}(x)} \]
Thus, we can write:
\[ y = K e^{-\sin^{-1}(x)} \]
or
\[ y = \pm K e^{-\sin^{-1}(x)} \]

### Step 4: Final Form
We can express the solution in a more general form:
\[ y = C e^{-\sin^{-1}(x)} \]
where \( C \) can be any real number (positive or negative).

### Summary of the Solution
The solution to the differential equation \( \sqrt{1 - x^2} \, dy + y \, dx = 0 \) is:
\[ y = C e^{-\sin^{-1}(x)} \]