`xdy-ydx=sqrt(x^2-y^2)dx , y(1)=0. ` then area curve above x-axis on `x in [1,e^pi]` is

Answer

To solve the differential equation \( xdy - ydx = \sqrt{x^2 - y^2}dx \) with the initial condition \( y(1) = 0 \) and find the area under the curve above the x-axis from \( x = 1 \) to \( x = e^\pi \), we will follow these steps:

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

### Step 2: Introducing a New Variable
Let \( v = \frac{y}{x} \) which implies \( y = vx \). Then, differentiating gives:
\[ dy = v dx + x dv \]
Substituting \( y \) into the equation:
\[ v dx + x dv = v dx + \frac{\sqrt{x^2 - (vx)^2}}{x} dx \]
This simplifies to:
\[ x dv = \frac{\sqrt{x^2(1 - v^2)}}{x} dx \]
Thus:
\[ dv = \frac{\sqrt{1 - v^2}}{x^2} dx \]

### Step 3: Separating Variables
We can separate the variables:
\[ \frac{dv}{\sqrt{1 - v^2}} = \frac{dx}{x^2} \]

### Step 4: Integrating Both Sides
Integrating both sides:
\[ \int \frac{dv}{\sqrt{1 - v^2}} = \int \frac{dx}{x^2} \]
The left side integrates to \( \sin^{-1}(v) \) and the right side integrates to \( -\frac{1}{x} + C \):
\[ \sin^{-1}(v) = -\frac{1}{x} + C \]

### Step 5: Solving for the Constant
Using the initial condition \( y(1) = 0 \) or \( v(1) = 0 \):
\[ \sin^{-1}(0) = -1 + C \implies C = 1 \]
Thus, we have:
\[ \sin^{-1}(v) = -\frac{1}{x} + 1 \]
Taking the sine of both sides gives:
\[ v = \sin\left(-\frac{1}{x} + 1\right) \]

### Step 6: Finding \( y \)
Recall \( v = \frac{y}{x} \), so:
\[ y = x \sin\left(-\frac{1}{x} + 1\right) \]

### Step 7: Finding the Area
The area under the curve from \( x = 1 \) to \( x = e^\pi \) is given by:
\[ \text{Area} = \int_{1}^{e^\pi} y \, dx = \int_{1}^{e^\pi} x \sin\left(-\frac{1}{x} + 1\right) \, dx \]

### Step 8: Substituting for Integration
Using the substitution \( t = \ln(x) \) gives \( dt = \frac{1}{x} dx \) or \( dx = e^t dt \). The limits change from \( x = 1 \) to \( t = 0 \) and from \( x = e^\pi \) to \( t = \pi \):
\[ \text{Area} = \int_{0}^{\pi} e^t \sin(1 - e^{-t}) e^t dt = \int_{0}^{\pi} e^{2t} \sin(1 - e^{-t}) dt \]

### Step 9: Evaluating the Integral
This integral can be evaluated using integration techniques or numerical methods. The result will yield the area above the x-axis.

### Final Result
After evaluating the integral, we find the area under the curve.