If any differentisl equation in the form
`f(f_(1)(x,y)d(f_(1)(x,y)+phi(f_(2)(x,y)d(f_(2)(x,y))+....=0`
then each term can be intergrated separately.
For example,
`intsinxyd(xy)+int((x)/(y))d((x)/(y))=-cos xy+(1)/(2)((x)/(y))^(2)+C`
The solution of the differential equation
`xdy-ydx=sqrt(x^(2)-y^(2))dx` is

Options

A. (a)`Cx=e^(sin^(-1^""(y)/(x)))`
B. (b)`xe^(sin^(-1^""(y)/(x)=c))`
C. (c)`x+e^(sin^(-1^""(y)/(x)=c))`
D. (d)None of these

Answer

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

### Step 1: Rearranging the Equation
We start by rearranging the given equation:
\[ x \, dy - y \, dx = \sqrt{x^2 - y^2} \, dx \]
Dividing both sides by \( dx \):
\[ x \frac{dy}{dx} - y = \sqrt{x^2 - y^2} \]

### Step 2: Homogeneous Equation
This is a homogeneous equation. We can use the substitution \( y = vx \), where \( v \) is a function of \( x \). Thus, we have:
\[ \frac{dy}{dx} = v + x \frac{dv}{dx} \]
Substituting \( y \) and \( \frac{dy}{dx} \) into the equation gives:
\[ x(v + x \frac{dv}{dx}) - vx = \sqrt{x^2 - (vx)^2} \]
This simplifies to:
\[ x^2 \frac{dv}{dx} = \sqrt{x^2(1 - v^2)} \]

### Step 3: Simplifying Further
We can simplify the equation:
\[ \frac{dv}{dx} = \frac{\sqrt{1 - v^2}}{x} \]

### Step 4: Separating Variables
Now, we separate the variables:
\[ \frac{dv}{\sqrt{1 - v^2}} = \frac{dx}{x} \]

### Step 5: Integrating Both Sides
Integrating both sides:
\[ \int \frac{dv}{\sqrt{1 - v^2}} = \int \frac{dx}{x} \]
The left side integrates to \( \sin^{-1}(v) \) and the right side integrates to \( \ln|x| + C \):
\[ \sin^{-1}(v) = \ln|x| + C \]

### Step 6: Back Substitution
Substituting back \( v = \frac{y}{x} \):
\[ \sin^{-1}\left(\frac{y}{x}\right) = \ln|x| + C \]

### Step 7: Exponentiating
Exponentiating both sides gives:
\[ e^{\sin^{-1}\left(\frac{y}{x}\right)} = e^{\ln|x| + C} \]
This simplifies to:
\[ e^{\sin^{-1}\left(\frac{y}{x}\right)} = Cx \]

### Final Result
Thus, the solution of the differential equation is:
\[ e^{\sin^{-1}\left(\frac{y}{x}\right)} = Cx \]