Given differential equation can be written as
dxdy+x2y=x , which is a linear differential equation.
If =e∫x2dx=e2lnx=x2
Solution is
y.x2=∫x.x2dx
⇒y.x2=4x4+c
Since, given curve passes through (1,1)
⇒c=43
Hence, y(x)=4x2+4x23.
So, y(21)=1649.
JEE Main 2019 — Mathematics Calculus
If y=y(x) is the solution of the differential equation, xdxdy+2y=x2 satisfying y(1)=1, then y(21) is equal to
Held on 9 Jan 2019 · Verified 6 Jul 2026.
647
41
1613
1649
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