dxdy+xy=bx3
I.F.=e∫x1dx=x
So, solution of D.E. is given by
y⋅x=∫b⋅x3⋅xdx+c
y=xc+5bx4
Passes through (1,2)
2=c+5b…(1)
∫12f(x)dx=562
[clnx+25bx5]12=562
cln2+2531b=562…(2)
By equation (1) and (2)
c=0 and b=10
JEE Main 2021 — Mathematics Calculus
If a curve y=f(x) passes through the point (1,2) and satisfies xdxdy+y=bx4, then for what value of b,∫12f(x)dx=562?
Held on 24 Feb 2021 · Verified 6 Jul 2026.
531
10
5
562
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