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Step 1: Write the Given Differential Equation
The differential equation provided is:
x\,dy - y\,dx = \sqrt{x^2 - y^2}\,dx,\quad x \ge 1,\quad \text{with}\quad y(1)=0.
Step 2: Convert to a More Manageable Form
Divide both sides of the equation by x^2 (assuming x \neq 0 ), to get:
\frac{x\,dy - y\,dx}{x^2} = \frac{\sqrt{x^2 - y^2}}{x^2}\,dx.
This can be rewritten as:
d\!\Bigl(\frac{y}{x}\Bigr) = \frac{1}{x}\sqrt{1 - \Bigl(\frac{y}{x}\Bigr)^2}\,dx.
Step 3: Separate Variables and Integrate
Rearrange to separate variables:
\frac{d\bigl(y/x\bigr)}{\sqrt{1 - \bigl(y/x\bigr)^2}} = \frac{dx}{x}.
Now integrate both sides:
\int \frac{d\bigl(y/x\bigr)}{\sqrt{1 - \bigl(y/x\bigr)^2}} = \int \frac{dx}{x}.
The left integral is recognized as the standard form of the inverse sine function, while the right integral is the natural logarithm:
\sin^{-1}\!\Bigl(\frac{y}{x}\Bigr) = \ln(x) + C.
Step 4: Apply the Initial Condition
We are given y(1) = 0. Substitute x=1 and y=0 into the equation:
\sin^{-1}\!\bigl(0\bigr) = \ln(1) + C\quad\Longrightarrow\quad 0 = 0 + C\quad\Longrightarrow\quad C=0.
Hence, the particular solution is:
\sin^{-1}\!\Bigl(\frac{y}{x}\Bigr) = \ln(x).
This implies:
\frac{y}{x} = \sin\bigl(\ln(x)\bigr)\quad\Longrightarrow\quad y = x\,\sin\bigl(\ln(x)\bigr).
Step 5: Express the Required Area
We need the area bounded by the lines x=1 , x=e^{\pi} , y=0 , and y=y(x) . For this region, the area A is given by the definite integral:
A = \int_{1}^{e^{\pi}} \bigl[\text{(top function)} - \text{(bottom function)}\bigr] \,dx
= \int_{1}^{e^{\pi}} \bigl[x \,\sin(\ln(x)) - 0\bigr]\;dx.
Step 6: Make a Suitable Substitution
Let t = \ln(x) . Then x = e^{t} and dx = e^{t}\,dt.
When x=1 , t=\ln(1)=0. When x=e^{\pi} , t = \ln\bigl(e^{\pi}\bigr)=\pi.
Rewriting the integral in terms of t :
A = \int_{0}^{\pi} e^{t}\,\sin(t)\,\bigl(e^{t}\,dt\bigr)
= \int_{0}^{\pi} e^{2t}\,\sin(t)\,dt.
Step 7: Evaluate the Integral
We need to compute:
\int_{0}^{\pi} e^{2t}\,\sin(t)\,dt.
This can be done by integration by parts or by a known standard result. A common approach is to use the formula:
\int e^{at}\sin(bt)\,dt = \frac{e^{at}\bigl(a\sin(bt) - b\cos(bt)\bigr)}{a^2 + b^2},
specialized to a=2 and b=1. Hence:
\int e^{2t}\sin(t)\,dt
= \frac{e^{2t}(2\sin(t) - \cos(t))}{2^2 + 1^2}
= \frac{e^{2t}(2\sin(t) - \cos(t))}{5}.
Evaluating from t=0 to t=\pi :
\left[\frac{e^{2t}(2\sin(t) - \cos(t))}{5}\right]_{0}^{\pi}
= \frac{1}{5}\Bigl(e^{2\pi}(2\sin(\pi) - \cos(\pi)) - e^{0}(2\sin(0) - \cos(0))\Bigr).
Recall \sin(\pi) = 0 , \cos(\pi) = -1 , and \sin(0) = 0 , \cos(0)=1. Therefore, this becomes:
\frac{1}{5}\Bigl(e^{2\pi}(0 - (-1)) - (0 - 1)\Bigr)
= \frac{1}{5}\bigl(e^{2\pi}\cdot 1 + 1\bigr)
= \frac{e^{2\pi}}{5} + \frac{1}{5}.
Thus, we can identify \alpha = \frac{1}{5} and \beta = \frac{1}{5}.
Step 8: Find the Requested Value
The problem states the area is \alpha\,e^{2\pi} + \beta, and we need 10(\alpha + \beta). Since \alpha = \frac{1}{5} and \beta = \frac{1}{5},
\alpha + \beta = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}.
10(\alpha + \beta) = 10\cdot\frac{2}{5} = 4.
The final required value is therefore \boxed{4}.
