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

The correct option is D $x-y+C=log(3x-4y+1)$

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