Let y+1=Y
∴dxdY=Y2e2x2−xY
Put −Y1=k
⇒dxdk+k(−x)=e2x2
I.F.=e−2x2
∴k=(x+c)ex2/2
Put k=−y+11
∴y+1=−(x+c)ex2/21...(i)
when x=2,y=0, then c=−2−e21
Differentiate equation (i)& put x=1 we get,
(dxdy)x=1=−(1+e2)2e3/2
Let y=y(x) be the solution of the differential equation dxdy=(y+1)((y+1)ex2/2−x),0<x<2.1, with y(2)=0. Then the value of dxdy at x=1 is equal to
Held on 18 Mar 2021 · Verified 6 Jul 2026.
(e2+1)2−e3/2
−(1+e2)22e2
(1+e2)2e5/2
(e2+1)25e1/2
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