Q.

If the solution curve f(x,y)=0 of the differential equation (1+logex)dxdy-xlogex=ey, x>0, passes through the points (1, 0) and (α,2), then αα is equal to     [2023]

1 ee2  
2 e2e2  
3 e2e2  
4 e2e2  

Ans.

(4)

The given differential equation is,

(1+logx)dxdy-xlogx=ey

Put xlogx=t⇒(x1x+logx)dx=dt

⇒(1+logx)dx=dt

∴ Given equation becomes:  dtdy-t=ey, which is a linear differential equation.

∴   I.F.=e∫-dy=e-y

So, required solution is given by

       e-y·t=∫ey·e-y dy=y+C

⇒t=(y+C)ey

⇒xlogx=(y+C)ey    ⋯(i)

Now, (i) passes through (1, 0) and (α,2)

∴ C=0 and αlogα=2e2

⇒logαα=2e2  ⇒  αα=e2e2