Question

find the general solution of the given DE (x - 2y)dx - (x - 2y + 1)dy = 0

Answer 1

(x-2y)dx-(x-2y+1)dy=0

(x-2y)dx=(x-2y+1)dy

$\frac{dy}{dx}=\frac{x-2y}{x-2y+1}$ This is equation (i)

Let x-2y=u $\implies1-2\frac{dy}{dx}=\frac{du}{dx}$

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

$\frac{dy}{dx}=\frac{1}{2}-\frac{1}{2}\frac{du}{dx}$ This is equation (ii)

Substituting equation (ii) into equation(i) we get,

$\frac{1}{2}-\frac{1}{2}\frac{du}{dx}=\frac{u}{u+1}$

$1-\frac{du}{dx}=\frac{2u}{2u+2}$

$\frac{du}{dx}=1-\frac{2u}{2u+2}$

$\frac{du}{dx}=\frac{2u+2-2u}{2u+2}=\frac{2}{2u+2}$

$(2u+2)du=2dx$

Integrate both sides$\int (2u+2)du=2\int 1dx$

$u^{2}+2u=2x+C$

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

$x^{2}-4xy+4y^{2}+2x-4y=2x+C$

$4y^{2}-(4x+4)y+(x^{2}+C)=0$

$y=\frac{(4x+4)\pm \sqrt{(-4x-4)^{2}-16(x^{2}+C}}{8}$

$y=\frac{4(x+1)\pm \sqrt{16x^{2}+32x+16-16x^{2}-16C}}{8}$

$y=\frac{4(x+1)\pm \sqrt{32x+16-16C}}{8}$

$y=\frac{4(x+1)\pm 4\sqrt{2x+1-C}}{8}$

$y=\frac{(x+1)\pm \sqrt{2x+1-C}}{2}$

The general solution of the DE is;

$y=\frac{(x+1)\pm \sqrt{2x+1-C}}{2}$