Solve: $x d y - y d x - (1 - x^{2}) d x = 0$

y^{2} + x^{2} + 1 = C y

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y + x^{2} + 1 = C x

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y^{2} - x^{2} + 1 = C x

No worries! We‘ve got your back. Try BYJU‘S free classes today!

y - x^{2} + 1 = C y

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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Solution

The correct option is A $y + x^{2} + 1 = C x$
$x^{2} - y + x \frac{d y}{d x} - 1 = 0 \Rightarrow \frac{d y}{d x} - \frac{y}{x} = - \frac{x^{2} - 1}{x}$

Let $μ = e^{\int - \frac{1}{x} d x} = \frac{1}{x}$
$\frac{\frac{d y}{d x}}{x} - \frac{y}{x^{2}} = - \frac{x^{2} - 1}{x^{2}}$

Substitute $- \frac{1}{x^{2}} = \frac{d}{d x} (\frac{1}{x})$

$\frac{\frac{d y}{d x}}{x} + \frac{d}{d x} (\frac{1}{x}) = - \frac{x^{2} - 1}{x^{2}}$

Using $g \frac{d f}{d x} + f \frac{d g}{d x} = \frac{d}{d x} (f g)$

$\frac{d}{d x} (\frac{y}{x}) = - \frac{x^{2} - 1}{x^{2}} \Rightarrow \int \frac{d}{d x} (\frac{y}{x}) d x = \int - \frac{x^{2} - 1}{x^{2}} d x \Rightarrow \frac{y}{x} = - \frac{1}{x} - x + c \Rightarrow y = {- x}^{2} + c x - 1$

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