If the solution curve f(x, y)=0 of the differential equation (1+\log _e x) \fracd xd y-x \log _e x=e^y, x \gt 0 , passes through the points (1,0) and (\alpha, 2) , then \alpha^\alpha is equal to :

Differential Equations PYQ — JEE Main 2023 · 6 Apr · Shift 2 · Q30 with Solution

View written solutionFree

Correct answer: D

We have

(1+\ln x)\frac{dx}{dy}-x\ln x=e^y, \qquad x>0

with solution curve passing through $(1,0)$ and $(\alpha,2)$ .

We need to find $\alpha^\alpha$ .

Notice that

\frac{d}{dy}(x\ln x)=\frac{dx}{dy}(\ln x+1)

since

\frac{d}{dx}(x\ln x)=\ln x+1.

So the given equation becomes

\frac{d}{dy}(x\ln x)-x\ln x=e^y.

Let

z=x\ln x.

Then the differential equation reduces to

\frac{dz}{dy}-z=e^y.

We solve

\frac{dz}{dy}-z=e^y.

This is a first-order linear ODE.

Its integrating factor is

IF=e^{\int -1\,dy}=e^{-y}.

Multiplying throughout by $e^{-y}$ ,

e^{-y}\frac{dz}{dy}-e^{-y}z=1.

Thus,

\frac{d}{dy}\left(ze^{-y}\right)=1.

Integrating,

ze^{-y}=y+C.

Hence,

z=e^y(y+C).

That is,

x\ln x=e^y(y+C).

At $(x,y)=(1,0)$ ,

1\cdot \ln 1= e^0(0+C).

Since $\ln 1=0$ , left side is $0$ . Therefore,

0=C.

So the solution becomes

x\ln x=ye^y.

Substitute $(x,y)=(\alpha,2)$ :

\alpha\ln \alpha=2e^2.

Now,

\alpha^\alpha=e^{\alpha\ln \alpha}=e^{2e^2}.

We obtained

\alpha^\alpha=e^{2e^2}.

This matches Option D.

Stored correct answer: D

Our derived answer: D

So they agree.