Question

Let y=y(x) be the solution of the differential equation x dy - y dx = √(x² - y²) dx, x ≥ 1, with y(1) = 0. If the area bounded by the line x=1, x=e^{π}, y=0 and y=y(x) is α e^{2π} + β, then the value of 10(α + β) is equal to ________.

Hint:

The form $x dy - y dx$ suggests using the substitution $y = vx$ or looking for the derivative of $y/x$.

Answer 1

Correct Answer: 4

Step 1: Rearrange: $\frac{x dy - y dx}{x^2} = \frac{\sqrt{x^2 - y^2}}{x^2} dx = \frac{\sqrt{1 - (y/x)^2}}{x} dx$. Let $y/x = \sin \theta \implies d(y/x) = \frac{\sqrt{1 - (y/x)^2}}{x} dx \implies \frac{d(y/x)}{\sqrt{1 - (y/x)^2}} = \frac{dx}{x}$.
Step 2: Integrate: $\sin^{-1}(y/x) = \ln x + C$. At $x=1, y=0 \implies C=0$. So $y = x \sin(\ln x)$.
Step 3: Area $= \int_1^{e^\pi} x \sin(\ln x) dx$. Let $\ln x = t \implies x = e^t, dx = e^t dt$. Area $= \int_0^\pi e^{2t} \sin t dt$.
Step 4: Using $\int e^{at} \sin bt dt = \frac{e^{at}}{a^2+b^2}(a \sin bt - b \cos bt)$: $= [\frac{e^{2t}}{5}(2 \sin t - \cos t)]_0^\pi = \frac{e^{2\pi}}{5}(0 - (-1)) - \frac{1}{5}(0 - 1) = \frac{1}{5}e^{2\pi} + \frac{1}{5}$.
Step 5: $\alpha = 1/5, \beta = 1/5$. $10(\alpha + \beta) = 10(2/5) = 4$.