The differential equation of all straight lin | vedclass

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(C) The equation of a straight line passing through the point $(1, -1)$ with slope $m$ is given by the point-slope form: $y - y_1 = m(x - x_1)$ Substi

Vedclass · Accepted Answer

$y + 1 = (x - 1) \frac{dy}{dx}$

Vedclass · Answer

(C) The equation of a straight line passing through the point $(1, -1)$ with slope $m$ is given by the point-slope form: $y - y_1 = m(x - x_1)$ Substituting the point $(1, -1)$,we get: $y - (-1) = m(x - 1)$ $y + 1 = m(x - 1)$ Since $m = \frac{dy}{dx}$,we substitute this into the equation: $y + 1 = \frac{dy}{dx}(x - 1)$ Rearranging the terms,we get: $y + 1 = (x - 1) \frac{dy}{dx}$ Comparing this with the given options,the correct form is $y + 1 = (x - 1) \frac{dy}{dx}$.

The differential equation of all straight lines passing through the point $(1, -1)$ is

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