Question

The solution of the equation $\frac{dy}{dx}=\frac{3x-4y-2}{3x-4y-3}$ is:

• A. ${(x-y)}^2+C=\log (3x-4y+1)$
• B. $x-y+C=\log (3x-4y+4)$
• C. $x-y+C=\log (3x-4y-3)$
• D. $x-y+C=\log (3x-4y+1)$

Answer 1

Answer: (D)

$x-y+C=\log (3x-4y+1)$

Substitute $3x-4y=X\Rightarrow 3-4\frac{dy}{dx}=\frac{dX}{dx}$
$\Rightarrow \frac{dy}{dx}=\frac{1}{4}(3-\frac{dX}{dx})$
Therefore the given equation is reduced to
$\frac{3}{4}-\frac{1}{4}\frac{dX}{dx}=\frac{X-2}{X-3}\Rightarrow -\frac{1}{4}\frac{dX}{dx}=\frac{4X-8-3(X-3)}{4(X-3)}$
$\Rightarrow -\frac{X-3}{X+1}dx=dx\Rightarrow -(1-\frac{4}{X+1})dX=dx$
$\Rightarrow -X+4\log (X+1)=x$+ constant
$\Rightarrow 4\log (3x-4y+1)=x+3x-4y$+ constant
$\Rightarrow \log (3x-4y+1)=x-y+C$