If a curve passes through the point (1, –2) and has slope of the tangent at any point (x, y) on it as x2-2yx, then the curve also passes through the point

dydx=x-2yxdydx+2xy=x ⇒ I.F=e∫2xdx=e2log xy(x2)=∫ x2·x dx=x44+cpasses through (1, -2) ⇒ -2=14+c ⇒ c=-94eq x2y=x44- 94verification of options

If a curve passes through the point (1, –2) and has slope of the tangent at any point (x, y) on it as x2-2yx, then the curve also passes through the point

dydx=x-2yxdydx+2xy=x ⇒ I.F=e∫2xdx=e2log xy(x2)=∫ x2·x dx=x44+cpasses through (1, -2) ⇒ -2=14+c ⇒ c=-94eq x2y=x44- 94verification of options

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

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$(\sqrt{3}, 0)$

(–1, 2)

$(- \sqrt{2}, 1)$

(3, 0)

answer is C.

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$\begin{array}{rcl} \frac{dy}{dx} & = & x - 2 \frac{y}{x} \\ \frac{dy}{dx} + \frac{2}{x} y & = & x \Rightarrow I . F = e^{\int \frac{2}{x} dx} = e^{2 \log x} \\ y (x^{2}) & = & \int x^{2} \cdot x dx = \frac{x^{4}}{4} + c \end{array}$

passes through
$\begin{array}{rcl} (1, - 2) & \Rightarrow & - 2 = \frac{1}{4} + c \\ \Rightarrow & c = \frac{- 9}{4} \end{array}$

eq $x^{2} y = \frac{x^{4}}{4} - \frac{9}{4}$

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If a curve passes through the point (1, –2) and has slope of the tangent at any point (x, y) on it as x2-2yx, then the curve also passes through the point

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