Question

Solve the differential equation $(2x+y+1)dx+(4x+2y-1)dy=0$.

Answer 1

Given that:

$(2x+y+1)dx+(4x+2y-1dy)=0$

Solution:

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

Let, $v=2x+y$

$\frac{dv}{dx}=2+\frac{dy}{dx}$

$\frac{dv}{dx}=2-\frac{v+1}{2v-1}$

$=\frac{4v-2-v-1}{2v-1}$

$\frac{dv}{dx}=\frac{3v-3}{2v-1}$

$\frac{2v-1}{3v-3}dv=dx$ $-----\left(i\right)$

$\frac{2v-1}{3v-3}=\frac{2}{3}+\frac{1}{3v-3}$

Integrating both sides,

or, $\frac{2}{3}\int dv+\int \frac{1}{3v-3}=\int dx$

or, $\frac{2}{3}v+\frac{1}{3}\log (3v-3)=x+c$

or, $2v+\log (3v-3)=3x+3c$

or, $2(2x+y)+\log \left(3\right(2x+y)-3)=3x+c^{\prime }$

or, $4x+2y+\log (6x+3y-3)=3x+c^{\prime }$

or, $x+2y+\log (6x+3y-3)=k$

Hence, the solution is $x+2y+\log (6x+3y-3)=k.$