JEE MAIN - Mathematics (2021 - 27th August Evening Shift - No. 5)
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 :
y5 $$-$$ 2y $$-$$ 2 = 0
2y5 $$-$$ 2y $$-$$ 1 = 0
2y5 $$-$$ y2 $$-$$ 2 = 0
y5 $$-$$ y2 $$-$$ 1 = 0
Explanation
$$(2x - 10{y^3})dy + ydx = 0$$
$$ \Rightarrow {{dx} \over {dy}} + \left( {{2 \over y}} \right)x = 10{y^2}$$
$$I.F. = {e^{\int {{2 \over y}dy} }} = {e^{2\ln (y)}} = {y^2}$$
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 $$x{y^2} = 2{y^5} - 2$$
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 $${y^5} - {y^2} - 1 = 0$$
$$ \Rightarrow {{dx} \over {dy}} + \left( {{2 \over y}} \right)x = 10{y^2}$$
$$I.F. = {e^{\int {{2 \over y}dy} }} = {e^{2\ln (y)}} = {y^2}$$
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 $$x{y^2} = 2{y^5} - 2$$
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 $${y^5} - {y^2} - 1 = 0$$
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