Question-103664

Question Number 103664 by byaw last updated on 16/Jul/20

Answered by bramlex last updated on 16/Jul/20

(x^{2} + y^{2}) dx = 2 xy dy

\frac{dy}{dx} = \frac{x^{2} + y^{2}}{2 xy}; set y = zx

\Rightarrow \frac{dy}{dx} = z + x \frac{dz}{dx}

\Leftrightarrow z + x \frac{dz}{dx} = \frac{x^{2} + z^{2} x^{2}}{{2 x}^{2} z}

x \frac{dz}{dx} = \frac{1 + z^{2}}{2 z} - z = \frac{1 - z^{2}}{2 z}

\frac{2 z}{1 - z^{2}} = \frac{dx}{x} \Rightarrow \int \frac{d (1 - z^{2})}{1 - z^{2}} = - \ln ∣ x ∣ + c

\ln ∣ 1 - z^{2} ∣ = \ln ∣ \frac{C}{x} ∣ \Rightarrow 1 - z^{2} = \frac{C}{x}

1 - \frac{C}{x} = \frac{y^{2}}{x^{2}} \Rightarrow y^{2} = x^{2} - Cx ⊛

Commented by byaw last updated on 16/Jul/20

Thank you . I am greatful .

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