Question

The solution of the differential equations $\frac{dy}{dx}=\frac{x-2y+1}{2x-4y}$ is?

• A. $(x-2y{)}^2+2x=c$
• B. $(x-2y{)}^2+x=c$
• C. $(x-2y)+2x^2=c$
• D. $(x-2y)+x^2=c$

Answer 1

The correct option is A $(x-2y{)}^2+2x=c$

Given,

$\frac{dy}{dx}=\frac{x-2y+1}{2x-4y}$

Let $x-2yt$

So, $1-\frac{2dy}{dx}=\frac{dt}{dx}$

$=1-\frac{dt}{dx}=2\frac{dy}{dx}$

$=\frac{1-\frac{dt}{dx}}{2}=\frac{dy}{dx}$

substituting,

$\frac{1-\frac{dt}{dx}}{2}=\frac{t+1}{2t}$

$=1-\frac{dt}{dx}=\frac{t+1}{t}$

$=1-\frac{dt}{dx}=1+\frac{1}{t}$

$=-\frac{dt}{dx}=\frac{1}{t}$

$=-tdt=dx$

Integrating,

$-\int tdt=\int dx$

$=-\frac{t^2}{2}=x+C$

$=C=x+\frac{t^2}{2}$

$=C=x+\frac{(x-2y{)}^2}{2}$

$=C=2x+(x-2y{)}^2$

So, the solution of

$\frac{dy}{dx}=\frac{x-2y+1}{2x-4y}$ is

$(x-2y{)}^2+2x=C$