Let y = y(x) be the solution of the different | vedclass

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(C) Given the differential equation: $x dy - y dx = \sqrt{x^2 - y^2} dx$.Dividing by $x^2$ (for $x \geq 1$): $\frac{x dy - y dx}{x^2} = \frac{1}{x} \s

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Given the differential equation: $x dy - y dx = \sqrt{x^2 - y^2} dx$.Dividing by $x^2$ (for $x \geq 1$): $\frac{x dy - y dx}{x^2} = \frac{1}{x} \sqrt{1 - (\frac{y}{x})^2} dx$.This simplifies to: $d(\frac{y}{x}) = \frac{1}{x} \sqrt{1 - (\frac{y}{x})^2} dx$.Integrating both sides: $\int \frac{d(\frac{y}{x})}{\sqrt{1 - (\frac{y}{x})^2}} = \int \frac{dx}{x}$.$\sin^{-1}(\frac{y}{x}) = \ln|x| + C$.Using the initial condition $y(1) = 0$: $\sin^{-1}(0) = \ln(1) + C \Rightarrow C = 0$.Thus,$y = x \sin(\ln x)$.The area $A$ is given by: $A = \int_{1}^{e^{\pi}} x \sin(\ln x) dx$.Let $x = e^t$,then $dx = e^t dt$. When $x=1, t=0$; when $x=e^{\pi}, t=\pi$.$A = \int_{0}^{\pi} e^t \sin(t) e^t dt = \int_{0}^{\pi} e^{2t} \sin(t) dt$.Using the formula $\int e^{at} \sin(bt) dt = \frac{e^{at}}{a^2 + b^2} (a \sin(bt) - b \cos(bt)) + C$:$A = [\frac{e^{2t}}{5} (2 \sin t - \cos t)]_{0}^{\pi} = \frac{e^{2\pi}}{5} (2(0) - (-1)) - \frac{1}{5} (2(0) - 1) = \frac{e^{2\pi}}{5} + \frac{1}{5}$.Comparing with $\alpha e^{2\pi} + \beta$,we get $\alpha = \frac{1}{5}$ and $\beta = \frac{1}{5}$.Therefore,$10(\alpha + \beta) = 10(\frac{1}{5} + \frac{1}{5}) = 10(\frac{2}{5}) = 4$.

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