If the solution curve y=y(x) of the differential equation (1+y^2)(1+\log _e x) d x+x d y=0, x \gt 0 passes through the point (1,1) and y(e)=\frac\alpha-\tan (\frac32)\beta+\tan (\frac32) , then \alpha+2 \beta is \underline\hspace2cm .

Differential Equations PYQ — JEE Main 2024 · 29 Jan · Shift 1 · Q60 with Solution

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Correct answer: 3

(1+y^2)(1+\ln x)\,dx + x\,dy = 0, \qquad x>0

We rewrite it in derivative form:

x\frac{dy}{dx}=-(1+y^2)(1+\ln x)

So,

\frac{dy}{1+y^2}=-\frac{1+\ln x}{x}\,dx

Using

\int \frac{dy}{1+y^2}=\tan^{-1}y

and

\int \frac{1+\ln x}{x}\,dx = \ln x + \frac{(\ln x)^2}{2}

we get

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

At $x=1$ , we have $\ln 1=0$ , and $y=1$ . Hence

\tan^{-1}(1)=C

Since

\tan^{-1}(1)=\frac{\pi}{4}

therefore

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

At $x=e$ , we have $\ln e=1$ . So

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

Thus

y(e)=\tan\left(\frac{\pi}{4}-\frac{3}{2}\right)

Now use the identity

\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}

with $A=\frac{\pi}{4}$ and $B=\frac{3}{2}$ . Since $\tan\frac{\pi}{4}=1$ ,

y(e)=\frac{1-\tan\left(\frac{3}{2}\right)}{1+\tan\left(\frac{3}{2}\right)}

Comparing with

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

we get

\alpha=1, \qquad \beta=1

\alpha+2\beta = 1+2(1)=3