Question

find the solution of
$\frac{dy}{dx}=\frac{4x+6y+5}{3y+2x+4}$

• A. $y+2x+\frac{3}{8}\log (24y+16x+23)=k.$
• B. $y-x+\frac{3}{8}\log (24y+16x+23)=k.$
• C. $y+x+\frac{3}{8}\log (24y+16x+23)=k.$
• D. $y-2x+\frac{3}{8}\log (24y+16x+23)=k.$

Answer 1

The correct option is D $y-2x+\frac{3}{8}\log (24y+16x+23)=k.$

Given differential equation

$\frac{dy}{dx}=\frac{4x+6y+5}{3y+2x+4}$ .....(1)

Put $3y+2x=v$

$\therefore 3\frac{dy}{dx}+2=\frac{dv}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{1}{3}(\frac{dv}{dx}-2)$

So, eqn (1) becomes

$\therefore \frac{1}{3}(\frac{dv}{dx}-2)=\frac{2v+5}{v+4}$

$\frac{dv}{dx}=\frac{6v+15}{v+4}+2=\frac{8v+23}{v+4}$

$\frac{dv}{dx}=\frac{8v+23}{v+4}$

$\frac{8v+32}{8v+23}dv=8dx$

$(1+\frac{9}{8v+23})dv=8dx$

Integrating , we get

$v+\frac{9}{8}\log (8v+23)=8x+c$

$3y+2x+\frac{9}{8}[8(3y+2x)+23]=8x+c$

$y-2x+\frac{3}{8}\log (24y+16x+23)=\frac{c}{3}=k.$