The absolute value of the constant term in the solution of the differential equation y2x4-ydydx-1-4xy2x2-if the curve passes through the center of the circle x2-y2-6y-0- is

Y(2x^4+y)dydx=(1 4xy^2)x^2 ⇒ 2x^4ydydx+y^2dydx=x^2 4x^3y^2 ⇒ 2x^4ydydx+4x^3y^2+y^2dydx=x^2 ⇒ddx(x^4y^2)+ddx(y^33)=x^2 ⇒ x^4y^2+y^33=x^33+C The given circle is ...

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Byju's Answer

Standard XII

Mathematics

Normal of a Curve y = f(x)

The absolute ...

Question

The absolute value of the constant term in the solution of the differential equation $y (2 x^{4} + y) \frac{d y}{d x} = (1 - 4 x y^{2}) x^{2}$ ,
if the curve passes through the center of the circle $x^{2} + y^{2} - 6 y = 0$ , is

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Solution

$y (2 x^{4} + y) \frac{d y}{d x} = (1 - 4 x y^{2}) x^{2} \Rightarrow 2 x^{4} y \frac{d y}{d x} + y^{2} \frac{d y}{d x} = x^{2} - 4 x^{3} y^{2} \Rightarrow 2 x^{4} y \frac{d y}{d x} + 4 x^{3} y^{2} + y^{2} \frac{d y}{d x} = x^{2} \Rightarrow \frac{d}{d x} (x^{4} y^{2}) + \frac{d}{d x} (\frac{y^{3}}{3}) = x^{2}$
$\Rightarrow x^{4} y^{2} + \frac{y^{3}}{3} = \frac{x^{3}}{3} + C$

The given circle is $x^{2} + y^{2} - 6 y = 0$
$\Rightarrow x^{2} + (y - 3)^{2} = 9$
So, the centre of the circle is $(0, 3)$
Now, $x^{4} y^{2} + \frac{y^{3}}{3} = \frac{x^{3}}{3} + C$
$\Rightarrow 0 + 3^{2} = C \Rightarrow C = 9$

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The absolute value of the constant term in the solution of the differential equation y(2x4+y)dydx=(1−4xy2)x2, if the curve passes through the center of the circle x2+y2−6y=0, is

y(2x4+y)dydx=(1−4xy2)x2⇒2x4ydydx+y2dydx=x2−4x3y2⇒2x4ydydx+4x3y2+y2dydx=x2⇒ddx(x4y2)+ddx(y33)=x2 ⇒x4y2+y33=x33+C The given circle is x2+y2−6y=0 ⇒x2+(y−3)2=9 So, the centre of the circle is (0,3) Now, x4y2+y33=x33+C ⇒0+32=C⇒C=9

The absolute value of the constant term in the solution of the differential equation $y (2 x^{4} + y) \frac{d y}{d x} = (1 - 4 x y^{2}) x^{2}$ ,
if the curve passes through the center of the circle $x^{2} + y^{2} - 6 y = 0$ , is