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 $y(e)=\frac{\alpha-\tan\left(\frac{3}{2}\right)}{\beta+\tan\left(\frac{3}{2}\right)}$, then $\alpha+2\beta$ is

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Differential Equations

If the solution curve

y = y (x)

of the differential equation

(1 + y^{2}) (1 + lo g_{e} x) d x + x d y = 0, x > 0

passes through the point

(1, 1)

and

y (e) = \frac{α - t a n ( \frac{3}{2} )}{β + t a n ( \frac{3}{2} )}

, then

α + 2 β

Answer: 3

(1 + y^{2}) (1 + ln x) d x + x d y = 0

separates as

\frac{d y}{1 + y ^{2}} = - \frac{1 + ln x}{x} d x

Integrating:

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

(1, 1)

tan^{- 1} (1) = \frac{π}{4}

, and the right side is

C

(since

ln 1 = 0

), so

C = \frac{π}{4}

tan^{- 1} y = \frac{π}{4} - ln x - \frac{( ln x ) ^{2}}{2}

x = e

(

ln x = 1

tan^{- 1} y (e) = \frac{π}{4} - 1 - \frac{1}{2} = \frac{π}{4} - \frac{3}{2}

y (e) = tan (\frac{π}{4} - \frac{3}{2}) = \frac{1 - tan \frac{3}{2}}{1 + tan \frac{3}{2}}

Comparing with

y (e) = \frac{α - tan ( 3/2 )}{β + tan ( 3/2 )}

gives

α = 1, β = 1

α + 2 β = 1 + 2 = 3

Key confirmed: 3.

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