Question

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

• 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)

Put $3x-4y=X$
$\Rightarrow 3-4\frac{dy}{dx}=\frac{dX}{dx}\Rightarrow \frac{dy}{dx}=\frac{1}{4}\left(3-\frac{dX}{dx}\right)$
$\Rightarrow \frac{3}{4}-\frac{1}{4}\frac{dX}{dx}=\frac{X-2}{X-3}$
$\Rightarrow -\frac{1}{4}\frac{dX}{dx}=\frac{4X-8-3\left(X-3\right)}{4\left(X-3\right)}$
$=-\frac{1}{4}\frac{dX}{dx}=\frac{X+1}{4\left(X-3\right)}$
Now, solve it.
Required solution is x - y + c = log (3x - 4y + 1)