If the solution curve y=y(x) of the different | vedclass

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(B) The given differential equation is $(1+y^2)(1+\ln x) dx + x dy = 0$.Rearranging the terms,we get $\frac{1+\ln x}{x} dx + \frac{dy}{1+y^2} = 0$.Int

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$3$

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(B) The given differential equation is $(1+y^2)(1+\ln x) dx + x dy = 0$.Rearranging the terms,we get $\frac{1+\ln x}{x} dx + \frac{dy}{1+y^2} = 0$.Integrating both sides,we have $\int \frac{1}{x} dx + \int \frac{\ln x}{x} dx + \int \frac{dy}{1+y^2} = C$.This simplifies to $\ln x + \frac{(\ln x)^2}{2} + 	an^{-1} y = C$.Since the curve passes through $(1,1)$,we substitute $x=1$ and $y=1$: $\ln(1) + \frac{(\ln 1)^2}{2} + 	an^{-1}(1) = C$,which gives $0 + 0 + \frac{\pi}{4} = C$,so $C = \frac{\pi}{4}$.The equation of the curve is $\ln x + \frac{(\ln x)^2}{2} + 	an^{-1} y = \frac{\pi}{4}$.For $x=e$,we have $\ln(e) + \frac{(\ln e)^2}{2} + 	an^{-1} y = \frac{\pi}{4}$,which implies $1 + \frac{1}{2} + 	an^{-1} y = \frac{\pi}{4}$.Thus,$	an^{-1} y = \frac{\pi}{4} - \frac{3}{2}$,so $y = 	an(\frac{\pi}{4} - \frac{3}{2}) = \frac{	an(\pi/4) - 	an(3/2)}{1 + 	an(\pi/4)	an(3/2)} = \frac{1 - 	an(3/2)}{1 + 	an(3/2)}$.Comparing this with $y(e) = \frac{\alpha - 	an(3/2)}{\beta + 	an(3/2)}$,we get $\alpha = 1$ and $\beta = 1$.Therefore,$\alpha + 2\beta = 1 + 2(1) = 3$.

If the solution curve $y=y(x)$ of the differential equation $(1+y^2)(1+\log_e x) dx + x dy = 0, x>0$ passes through the point $(1,1)$ and $y(e) = \frac{\alpha-\tan(3/2)}{\beta+\tan(3/2)}$,then $\alpha+2\beta$ is equal to:

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