Question

By replacing $x=X+h,$ $y=Y+K$ where $h=1,K=0$.The transformed equation of $\frac{dy}{dx}+\frac{3y-7x+7}{7y-3x+3}=0$ is

• A. $\frac{dY}{dX}=\frac{7X+3Y}{3X-7Y}$
• B. $\frac{dY}{dX}=\frac{-7X+3Y}{3X-7Y}$
• C. $\frac{dY}{dX}=\frac{7X-3Y}{3X-7Y}$
• D. $\frac{dY}{dX}=\frac{7X+3Y}{3X+7Y}$

Answer 1

The correct option is B $\frac{dY}{dX}=\frac{-7X+3Y}{3X-7Y}$

$\Rightarrow \frac{dy}{dx}=-\left[\frac{3y-7x+y}{7y-3x+3}\right]$

$\frac{dy}{dx}=-\left[\frac{3\left(Y\right)-7(X+1)+7}{7Y-3(X+1)+3}\right]$

$\Rightarrow \frac{dy}{dx}=-\left[\frac{3Y-7X}{7Y-3X}\right]$

$=\frac{3Y-7X}{3X-7Y}$