Question

The solution of diferential equation $\frac{d y}{d x}=\frac{4 x+6 y+5}{3 y+2 x+4}$ is

• A. $x^2+y^2-x y+x-y=c$
• B. $ y-2 x+\frac{3}{8} \ell n (24 y+16 x+23)=c$
• C. $4 x y+3\left(x^2+y^2\right)-10(x+y)=c$
• D. $y+2 x+\frac{3}{8} \ln (24 y+16 x+23)=c$

Answer 1

Answer: (B)

$3 y+2 x=v \therefore 3 \frac{d y}{d x}+2=\frac{d v}{d x}$
$\left(\frac{d v}{d x}-2\right)=\frac{2 v+5}{v+4}$
$\frac{d v}{d x}=\frac{6 v+15}{v+4}+2=\frac{8 v+23}{v+4}$
$\frac{8 v+32}{8 v+23} d v=8 d x$
$\int\left(1+\frac{9}{8 v+23}\right) d v=\int\left(1+\frac{9}{8 v+23}\right)$
$v+\frac{9}{8} \ln (8 v+23)=8 x+c$
$y-2 x+\frac{3}{8} \ln (24 y+16 x+23)=k$