If the solution curve y = y(x) of the differential equation (1 + y^2) (1 + logex)dx + xdy = 0, x > 0

Question

If the solution curve \(y=y(x)\) of the differential equation \(\left(1+y^{2}\right)\left(1+\log _{e} x\right) d x+x d y=0, x>0\) passes through the point \((1,1)\) and \(y(e)=\frac{\alpha-\tan \left(\frac{3}{2}\right)}{\beta+\tan \left(\frac{3}{2}\right)}\), then \(\alpha+2 \beta \) is _____.

Answer 1

Correct answer: 3

\(\mathrm{\int\left(\frac{1}{x}+\frac{\ln x}{x}\right) d x+\int \frac{d y}{1+y^{2}}=0}\)

\(\ln \mathrm{x}+\frac{(\ln \mathrm{x})^{2}}{2}+\tan ^{-1} \mathrm{y}=\mathrm{C}\)

Put \(\mathrm{x}=\mathrm{y}=1\)

\(\therefore \mathrm{C}=\frac{\pi}{4}\)

\(\mathrm{\Rightarrow \ln x+\frac{(\ln x)^{2}}{2}+\tan ^{-1} y=\frac{\pi}{4}}\)

Put \(\mathrm{x}=\mathrm{e}\)

\(\Rightarrow \mathrm{y}=\tan \left(\frac{\pi}{4}-\frac{3}{2}\right)=\frac{1-\tan \frac{3}{2}}{1+\tan \frac{3}{2}}\)

\(\therefore \alpha=1, \beta=1\)

\(\Rightarrow \alpha+2 \beta=3\)