The curve satisfying the differential equatio | vedclass

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(B) The given differential equation is $y dx - (x + 3y^2) dy = 0$.Rearranging the terms,we get $y dx = (x + 3y^2) dy$.Dividing by $y dy$,we get $\frac

Vedclass · Accepted Answer

$\left( -\frac{1}{3}, \frac{1}{3} \right)$

Vedclass · Answer

(B) The given differential equation is $y dx - (x + 3y^2) dy = 0$.Rearranging the terms,we get $y dx = (x + 3y^2) dy$.Dividing by $y dy$,we get $\frac{dx}{dy} = \frac{x}{y} + 3y$.This is a linear differential equation in $x$ of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = 3y$.The integrating factor is $IF = e^{\int P(y) dy} = e^{-\int \frac{1}{y} dy} = e^{-\ln y} = \frac{1}{y}$.The solution is given by $x \cdot IF = \int Q(y) \cdot IF dy + C$.Substituting the values,$x \cdot \frac{1}{y} = \int 3y \cdot \frac{1}{y} dy + C$.$\frac{x}{y} = \int 3 dy + C = 3y + C$.Thus,$x = 3y^2 + Cy$.Since the curve passes through $(1, 1)$,we have $1 = 3(1)^2 + C(1)$,which gives $C = -2$.The equation of the curve is $x = 3y^2 - 2y$.Checking the options,for $y = \frac{1}{3}$,$x = 3(\frac{1}{9}) - 2(\frac{1}{3}) = \frac{1}{3} - \frac{2}{3} = -\frac{1}{3}$.Thus,the curve passes through $\left( -\frac{1}{3}, \frac{1}{3} \right)$.

$X=x$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$	$0$

The curve satisfying the differential equation,$ydx - (x + 3y^2) dy = 0$ and passing through the point $(1, 1)$,also passes through the point

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