Question

$\frac{dy}{dx}=\frac{2y+x+1}{2x+4y+1}$ is solution is $4(x+2y)-\log (ax+by+3)=8x+k.$.
Find $\frac{b}{a}$?

Answer 1

$\frac{dy}{dx}=\frac{2y+x+1}{2x+4y+1}$
Here, $\frac{a_1}{a_2}=\frac{1}{2}$
$\frac{b_1}{b_2}=\frac{2}{4}=\frac{1}{2}$
$\Rightarrow \frac{a_1}{a_2}=\frac{b_1}{b_2}$
Put $x+2y=v$
$1+2\frac{dy}{dx}=\frac{dv}{dx}$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2}(\frac{dv}{dx}-1)$
So, eqn (1) becomes
$\Rightarrow \frac{1}{2}(\frac{dv}{dx}-1)=\frac{v+1}{2v+1}$
$\Rightarrow \frac{dv}{dx}=\frac{4v+3}{2v+1}$
$\Rightarrow \int \frac{(2v+1)dv}{4v+3}=\int dx$
$\Rightarrow \frac{1}{2}\int \frac{(4v+3-1)dv}{4v+3}=x+c$
$\Rightarrow \frac{1}{2}\int dv-\frac{1}{2}\int \frac{dv}{4v+3}=x+c$
$\Rightarrow \frac{1}{2}v-\frac{1}{8}\log (4v+3)=x+c$
$\Rightarrow 4(x+2y)-\log (4x+8y+3)=8x+k$
On comparing with given solution
$a=2,b=8$
$\Rightarrow \frac{b}{a}=2$