Given that: (2x−10y3)dy+ydx=0
⇒dydx=10y2−y2x
⇒dydx+y2x=10y2
⇒ I.F. =e2∫y1dx=y2
⇒x⋅y2=∫10y4dy+C
⇒x⋅y2=2y5+c
\mathrm{Put}x=0&y=1\Rightarrow c=-2
⇒x⋅y2=2y5−2
Passing Through (2,β)
⇒2⋅β2=2β5−2
⇒β5−β2−1=0
root of the equations
y5−y2−1=0
If the solution curve of the differential equation (2x−10y3)dy+ydx=0, passes through the points (0,1) and (2,β), then β is a root of the equation?
Held on 27 Aug 2021 · Verified 6 Jul 2026.
y5−2y−2=0
y5−y2−1=0
2y5−y2−2=0
2y5−2y−1=0
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