Question

Find the solution of $\frac{dy}{dx}=\frac{8x+6y+12}{4x+3y+2}$.

• A. $5(3y+6x)-12\log (20x+15y+22)=k$
• B. $5(3y-6x)-12\log (20x+15y+22)=k$
• C. $5(3y-5x)-12\log (20x+15y+22)=k$
• D. $5(3y+5x)-12\log (20x+15y+22)=k$

Answer 1

The correct option is B $5(3y-6x)-12\log (20x+15y+22)=k$

Given differential equation
$\frac{dy}{dx}=\frac{8x+6y+12}{4x+3y+2}$ ....(1)
Substitute $4x+3y=v$
$\Rightarrow 4+3\frac{dy}{dx}=\frac{dv}{dx}$
$\Rightarrow \frac{dy}{dx}=\frac{1}{3}(\frac{dv}{dx}-4)$
So, equation (1) becomes
$\frac{1}{3}(\frac{dv}{dx}-4)=\frac{2v+12}{v+2}$
$\frac{dv}{dx}-4=\frac{6v+36}{v+2}$
$\Rightarrow \frac{v+2}{10v+44}dv=dx$
$\Rightarrow \frac{10v+20}{10v+44}dv=10dx$
$(1-\frac{24}{10v+44})dv=10dx.$
$(1-\frac{12}{(5v+22})dv=10dx.$
Integrating both sides, we get
$v-\frac{12}{5}\log (5v+22)=10x+k$
or $4x+3y-\frac{12}{5}\log (20x+15y+22)=10x+k$
or $5(3y-6x)-12\log (20x+15y+22)=c$