Let y=y(x) be the solution of the differential equation xdy-ydx=√(x^2-y^2)dx,x≥ 1, with y(1)=0. If the area bounded by the line x=1,x=e^π ,y=0 and…

Given $xdy-ydx=\sqrt{{x}^{2}-{y}^{2}}dx$xdy−ydx=x2−y2​dx

$$\Rightarrow \frac{xdy-ydx}{{x}^{2}}=\frac{1}{x}\sqrt{1-\frac{{y}^{2}}{{x}^{2}}}dx$$⇒x2xdy−ydx​=x1​1−x2y2​​dx

$$\Rightarrow \int \frac{d(\frac{y}{x})}{\sqrt{1-{(\frac{y}{x})}^{2}}}=\int \frac{dx}{x}$$⇒∫1−(xy​)2​d(xy​)​=∫xdx​

$$\Rightarrow {\mathrm{sin}}^{-1}(\frac{y}{x})=ln|x|+c$$⇒sin−1(xy​)=ln∣x∣+c

at $x=1,y=0\Rightarrow c=0$x=1,y=0⇒c=0

$$y=x\mathrm{sin}(lnx)$$y=xsin(lnx)

$$A={\int }_{1}^{{e}^{\pi }}x\mathrm{sin}(lnx)dx$$A=∫1eπ​xsin(lnx)dx

$$x={e}^{t},dx={e}^{t}dt\Rightarrow {\int }_{0}^{\pi }{e}^{2t}\mathrm{sin}(t)dt=A$$x=et,dx=etdt⇒∫0π​e2tsin(t)dt=A

$$\alpha {e}^{2\pi }+\beta ={(\frac{{e}^{2t}}{5}(2\mathrm{sin}t-\mathrm{cos}t))}_{0}^{\pi }=\frac{1+{e}^{2\pi }}{5}$$αe2π+β=(5e2t​(2sint−cost))0π​=51+e2π​

$$\alpha =\frac{1}{5},\beta =\frac{1}{5}$$α=51​,β=51​

So, $10(\alpha +\beta )=4$10(α+β)=4