If the solution curve of the differential equation $\left(2 x-10 y^{3}\right) d y+y d x=0$, passes through the points $(0,1)$ and $(2, \beta)$, then $\beta$ is a root of the equation:

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Differential Equations

If the solution curve of the differential equation

(2 x - 10 y^{3}) d y + y d x = 0

, passes through the points

(0, 1)

and

(2, β)

, then

β

is a root of the equation:

Answer: (A)

Treating

x

as a function of

y

, the equation

(2 x - 10 y^{3}) d y + y d x = 0

becomes

\frac{d x}{d y} + \frac{2}{y} x = 10 y^{2}

, linear with integrating factor

y^{2}

. Then

\frac{d}{d y} (x y^{2}) = 10 y^{4}

, so

x y^{2} = 2 y^{5} + C

. The point

(0, 1)

gives

C = - 2

, hence

x y^{2} = 2 y^{5} - 2

. At

(2, β)

2 β^{2} = 2 β^{5} - 2

, i.e.

β^{5} - β^{2} - 1 = 0

. So

β

is a root of

y^{5} - y^{2} - 1 = 0

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Single correct option

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