Given
dydx−y2x=y2(y+1)ey
Finding integrating factor I.F.=e∫y−2dx=e−2lny=y21
⇒y2x=∫(y+1)eydy
⇒y2x=y⋅ey+C
0=e+C⇒C=−e
⇒y2x=yey−e
Putting y=e in the equation we get
⇒e2x=e.ee−e
⇒x=e3(ee−1)
If x=x(y) is the solution of the differential equation ydydx=2x+y3(y+1)ey,x(1)=0; then x(e) is equal to
Held on 24 Jun 2022 · Verified 6 Jul 2026.
ee(e3−1)
e3(ee−1)
ee−1
ee(e2−1)
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