We have, ∫e−ydy=∫eαxdx
⇒−e−y=αeαx+c...(1)
Put (x,y)=(ℓn2,ℓn2)
−21=α2α+c...(2)
Put (x,y)≡(0,−ℓn2)
−2=α1+c
From equation (1)−(2), we get
α2α−1=23
⇒α=2( as α∈N).
JEE Main 2021 — Mathematics Calculus
Let y=y(x) be the solution of the differential equation dy=eαx+ydx;α∈N. If y(loge2)=loge2 and y(0)=loge(21), then the value of α is equal to ___.
Held on 27 Jul 2021 · Verified 6 Jul 2026.
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