If the curve y=f(x) passes through the point (1,2) and satisfies xdydx+y=bx4 then for what value of b,  when ∫12f(x)dx=625

The given differential equation is xdydx+y=bx4⇒dydx+y1x=x3.bThe integrating factor is =e∫1xdx=xThe solution for the differential equation is xy=∫bx4dx⇒xy=bx55+cpassing  through  1,  2 Hence, 2=b5+c⇒c=2−b5⇒fx=bx45+cx   Given  ∫12f(x)dx=625  ∫12bx45+cxdx=625 b2531+ c ln2=625 b2531+ b5-2 ln2=625 Equate the coefficient of ln2 to zero b5-2=0 b=10