The differential equation of all straight lin | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The equation of a straight line passing through the point $(1, -1)$ with slope $m$ is given by:$y - y_1 = m(x - x_1)$Substituting $(x_1, y_1) = (1, -1)$,we get:$y - (-1) = m(x - 1)$$y + 1 = m(x - 1)$Since $m$ represents the slope of the line,we can write $m = \frac{dy}{dx}$.Substituting this into the equation,we get:$y + 1 = \frac{dy}{dx}(x - 1)$Rearranging the terms to solve for $y$:$y = (x - 1)\frac{dy}{dx} - 1$Thus,the correct option is $D$.

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

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