Discussion Forum

The solution  of the differential equation (2x-3y+5)dx+(9y+6x-7) dy =0 , is 

Answer:

Explanation:

We have

 (2x-3y+5)dx+(9y+6x-7) dy =0 

$\frac{dy}{dx}=\frac{2x-3y+5}{3(2x-3y)+7}$

Put 2x-3y=z

$\Rightarrow$    $2-\frac{3dy}{dx}=\frac{dz}{dx}$

$\Rightarrow$      $\frac{dy }{dx}=\frac{1}{3}$  $\left(2-\frac{dz}{dx}\right)$

$\therefore$  $\frac{1}{3}\left(2-\frac{dz}{dx}\right)=\frac{z+5}{3z+7}$

$\Rightarrow$    $2-\frac{dz}{dx}=\frac{3z+15}{3z+7}$

$\Rightarrow$    $\frac{dz}{dx}=2-\frac{3z+15}{3z+7}$

$\Rightarrow$     $\frac{dz}{dx}=\frac{3z+1}{3z+7}$

$\Rightarrow$    $\frac{3z+1}{3z+7}dz= dx$

$\Rightarrow$     $\int\left( \frac{3z-1}{3z-1}+\frac{8}{3z-1}\right)dz= dx$

 $=z+\frac{8}{3}log| 3z-1|=x+c$

 $=3z+8log( 3z-1)=3x+c$

 =3x-9y+8 log (6x-9y-1)=c