If a curve passes through the point (1, -2) a | vedclass

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(B) The slope of the tangent is given by $\frac{dy}{dx} = \frac{x^2 - 2y}{x}$.This can be rewritten as a linear differential equation: $\frac{dy}{dx}

Vedclass · Accepted Answer

$(\sqrt{3}, 0)$

Vedclass · Answer

(B) The slope of the tangent is given by $\frac{dy}{dx} = \frac{x^2 - 2y}{x}$.This can be rewritten as a linear differential equation: $\frac{dy}{dx} + \frac{2}{x}y = x$.This is of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{2}{x}$ and $Q = x$.The integrating factor $IF = e^{\int P dx} = e^{\int \frac{2}{x} dx} = e^{2 \ln|x|} = x^2$.The general solution is $y \cdot IF = \int Q \cdot IF dx + C$.$y \cdot x^2 = \int x \cdot x^2 dx + C = \int x^3 dx + C = \frac{x^4}{4} + C$.Since the curve passes through $(1, -2)$,we substitute $x = 1$ and $y = -2$:$-2(1)^2 = \frac{1^4}{4} + C \Rightarrow -2 = \frac{1}{4} + C \Rightarrow C = -2 - \frac{1}{4} = -\frac{9}{4}$.Thus,the equation of the curve is $y x^2 = \frac{x^4}{4} - \frac{9}{4}$,or $4yx^2 = x^4 - 9$.Checking option $(B)$: For $x = \sqrt{3}$,$4y(3) = (\sqrt{3})^4 - 9 = 9 - 9 = 0$,so $y = 0$. Thus,the curve passes through $(\sqrt{3}, 0)$.

If a curve passes through the point $(1, -2)$ and has the slope of the tangent at any point $(x, y)$ on it as $\frac{x^2 - 2y}{x}$,then the curve also passes through the point:

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