The curve satisfying the differential equatio | vedclass

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Question

(D) Given the differential equation: $y \, dx - (x + 3y^2) \, dy = 0$ Rearranging the terms: $y \, dx = (x + 3y^2) \, dy$ Dividing by $y \, dy$: $\fra

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given the differential equation: $y \, dx - (x + 3y^2) \, dy = 0$ Rearranging the terms: $y \, dx = (x + 3y^2) \, dy$ Dividing by $y \, dy$: $\frac{dx}{dy} = \frac{x + 3y^2}{y} = \frac{x}{y} + 3y$ This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = 3y$. The Integrating Factor $(IF)$ is given by: $IF = e^{\int P(y) \, dy} = e^{\int -\frac{1}{y} \, dy} = e^{-\ln y} = \frac{1}{y}$. The general solution is $x \cdot IF = \int Q(y) \cdot IF \, dy + c$. Substituting the values: $x \cdot \frac{1}{y} = \int 3y \cdot \frac{1}{y} \, dy + c$ $\frac{x}{y} = \int 3 \, dy + c = 3y + c$ So,$x = 3y^2 + cy$. Since the curve passes through $(1, 1)$,we substitute $x=1$ and $y=1$: $1 = 3(1)^2 + c(1) \Rightarrow 1 = 3 + c \Rightarrow c = -2$. The equation of the curve is $x = 3y^2 - 2y$. Checking option $(D)$ $(-\frac{1}{3}, \frac{1}{3})$: $x = 3(\frac{1}{3})^2 - 2(\frac{1}{3}) = 3(\frac{1}{9}) - \frac{2}{3} = \frac{1}{3} - \frac{2}{3} = -\frac{1}{3}$. Since the point satisfies the equation,the correct option is $(D)$.

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

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