Find the general solution of the following differential equations :
`sqrt(1+x^(2))dy+sqrt(1+y^(2))dx=0`.

Answer

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

### Step 1: Rearranging the equation
We start by rearranging the equation to separate the variables \( y \) and \( x \):
\[ \sqrt{1+x^2} \, dy = -\sqrt{1+y^2} \, dx \]
This can be rewritten as:
\[ \frac{dy}{\sqrt{1+y^2}} = -\frac{dx}{\sqrt{1+x^2}} \]

### Step 2: Integrating both sides
Now, we integrate both sides:
\[ \int \frac{dy}{\sqrt{1+y^2}} = -\int \frac{dx}{\sqrt{1+x^2}} \]

### Step 3: Applying the integration formula
The integral \( \int \frac{dy}{\sqrt{1+y^2}} \) is known to be \( \ln(y + \sqrt{1+y^2}) + C_1 \), and similarly, \( \int \frac{dx}{\sqrt{1+x^2}} \) is \( \ln(x + \sqrt{1+x^2}) + C_2 \). Therefore, we have:
\[ \ln(y + \sqrt{1+y^2}) = -\ln(x + \sqrt{1+x^2}) + C \]
where \( C = C_1 - C_2 \).

### Step 4: Simplifying the equation
We can express the equation as:
\[ \ln(y + \sqrt{1+y^2}) + \ln(x + \sqrt{1+x^2}) = C \]
Using the property of logarithms \( \ln(a) + \ln(b) = \ln(ab) \), we can rewrite this as:
\[ \ln\left((y + \sqrt{1+y^2})(x + \sqrt{1+x^2})\right) = C \]

### Step 5: Exponentiating both sides
Exponentiating both sides gives:
\[ (y + \sqrt{1+y^2})(x + \sqrt{1+x^2}) = e^C \]
Letting \( k = e^C \), we can write:
\[ (y + \sqrt{1+y^2})(x + \sqrt{1+x^2}) = k \]

### Final General Solution
Thus, the general solution of the differential equation is:
\[ (y + \sqrt{1+y^2})(x + \sqrt{1+x^2}) = C \]
where \( C \) is a constant.