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  • Let y = y(x) be the solution of the differential equation , with y(1) =0. If th

Let y = y(x) be the solution of the differential equation xdy -ydx = \sqrt{(x^2 +y^2)} dx, x\geqslant 1, with y(1) =0. If the area bounded by the line x=1, x=e^{\pi} , y=0 and y=y(x) is \alpha e^{2 \pi } + \beta, then the va;ue of 10(\alpha +\beta ) is equal to_______
Option: 1 2
Option: 2 4
Option: 3 6
Option: 4 8

Answers (1)

best_answer

\\x d y-y d x=\sqrt{x^{2}-y^{2}} d x \\ \Rightarrow \int \frac{x d y-y d x}{x^{2}}=\frac{1}{x} \sqrt{1-\frac{y^{2}}{x^{2}}} d x \\ \Rightarrow \int \frac{d\left(\frac{y}{x}\right)}{\sqrt{1-\left(\frac{y}{x}\right)^{2}}}=\int \frac{d x}{x} \\ \Rightarrow \sin ^{-1}\left(\frac{y}{x}\right)=\ell n|x|+c

\\\text {at } x=1, y=0 \Rightarrow c=0 \\ y=x \sin (\ell n x)

\\\mathrm{A}=\int_{1}^{\mathrm{e}^{\pi}} \mathrm{x} \sin (\ell \mathrm{n} \mathrm{x}) \mathrm{d} \mathrm{x} \\ \mathrm{x}=\mathrm{e}^{\mathrm{t}}, \mathrm{d} \mathrm{x}=\mathrm{e}^{\mathrm{t}} \mathrm{dt}\\\Rightarrow \int_{0}^{\pi} \mathrm{e}^{2 \mathrm{t}} \sin (\mathrm{t}) \mathrm{dt}=\text{A}

\\\alpha \mathrm{e}^{2 \pi}+\beta=\left(\frac{\mathrm{e}^{2 \mathrm{t}}}{5}(2 \sin \mathrm{t}-\cos \mathrm{t})\right)_{0}^{\pi}=\frac{1+\mathrm{e}^{2 \pi}}{5} \\ \alpha=\frac{1}{5}, \beta=\frac{1}{5} \text { so } 10(\alpha+\beta)=4

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