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

Engineering

Mathematics

Linear Differential Equation

Question

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

$(\sqrt{3}, 0)$

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

(3, 0)

(–1, 2)

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$\frac{dy}{dx} = x - \frac{2 y}{x} \Rightarrow \frac{dy}{dx} + \frac{2 y}{x} = x$

$∴ I . F . = e^{\int \frac{2}{x} dx} = x^{2}$

$∴ {yx}^{2} = \int x^{2} \cdot x dx = \frac{x^{4}}{4} + C$

(1, – 2) ⇒ – 2 = $\frac{1}{4}$ + C ⇒ C = – 2 – $\frac{1}{4} = \frac{- 9}{4}$

$∴ {yx}^{2} = \frac{x^{4}}{4} - \frac{9}{4} \Rightarrow$ 4x²y = x⁴ – 9.