Given, (x−1)dxdy+2xy=x−11
⇒dxdy+x−12x⋅y=(x−1)21
I.F=e∫x−12x=e2x×(x−1)2
Solution will be,
y×I.F=∫(x−1)21×I.F
Solving above with y(2)=2e41+e4 we get,
y=(x−1)21[2e2xe2x+1]
So, y(3)=8e6e6+1, now on comparing with y(3)=βeαeα+1 we get, α+β=14
JEE Main 2022 — Mathematics Calculus
Let y=y(x),x>1, be the solution of the differential equation (x−1)dxdy+2xy=x−11, with y(2)=2e41+e4. If y(3)=βeαeα+1. then the value of α+β is equal to ______.
Held on 29 Jun 2022 · Verified 6 Jul 2026.
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