If the solution curve of the differential equation (2x - 10y3)dy + ydx = 0, passes through the points (0, 1) and (2, \[\beta\]), then \[\beta\] is a root of the equation :
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Official solution
(2x−10y3)dy+ydx=0
\[ \Rightarrow {{dx} \over {dy}} + \left( {{2 \over y}} \right)x = 10{y^2}\]
\[I.F. = {e^{\int {{2 \over y}dy} }} = {e^{2\ln (y)}} = {y^2}\]
Solution of D.E. is
\[\therefore\] \[x\,.\,y = \int {(10{y^2}){y^2}.\,dy} \]
\[x{y^2} = {{10{y^5}} \over 5} + C \Rightarrow x{y^2} = 2{y^5} + C\]
It passes through (0, 1) \[\to\] 0 = 2 + C \[\Rightarrow\] C = -2
\[\therefore\] Curve is xy2=2y5−2
Now, it passes through (2, \[\beta\])
\[2{\beta ^2} = 2{\beta ^5} - 2 \Rightarrow {\beta ^5} - {\beta ^2} - 1 = 0\]
\[\therefore\] \[\beta\] is root of an equation y5−y2−1=0