y=vx dxdy=v+xdxdvv+xdxdv=1+v⇒dv=xdx∴v=logx+c⇒xy=logx+c Since, y(1)=1, we have y=xlogx+x
The solution of the differential equation d y d x= x+y x satisfying the condition y(1)=1 is
JEE Main 2008 — Mathematics Calculus
The solution of the differential equation dxdy=xx+y satisfying the condition y(1)=1 is
Held on 30 Apr 2008 · Verified 6 Jul 2026.
Options
- A
y=lnx+x
- B
y=xlnx+x2
- C
y=xe(x−1)
- D
y=xlnx+x
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