Question

The solution $\frac{dy}{dx}+\frac{3x+2y-5}{2x+3y-5}=0$

• A. $x^2+xy-y^2-4x+6y=0$
• B. $3x^2+4xy-3y^2+10x+10y=c$
• C. $(x+2y{)}^2+3x+3y=c$
• D. $3x^2+4xy+3y^2-10x-10y=c$

Answer 1

The correct option is D $3x^2+4xy+3y^2-10x-10y=c$

put $y=y+k;x=x+h$

$\Rightarrow 3h+2k-5=0$ .......... $\left(1\right)$

$2h+3k-5=0$ ......... $\left(2\right)$

$\Rightarrow \frac{dy}{dx}+\frac{3x+2y}{2x+3y}=0$

put $y=vx$

$\Rightarrow v+x\frac{dv}{dx}+\frac{3+2v}{2+3v}=0$

$\Rightarrow x\frac{dv}{dx}+\frac{3+4v+3v^2}{2+3v}=0$

$\Rightarrow \int 2\frac{(2+3v)}{3v^2+4v+3}dv+\int 2\frac{dx}{x}=c$

$\Rightarrow \log x^2(3v^2+4v+3)=c$

$\Rightarrow x^2(3v^2+4v+3)=c$

$\Rightarrow 3y^2+4yx+3x^2=c$

$\Rightarrow 3{(y-k)}^2+4(y-k)(x-h)+3{(x-h)}^2=c$

$\Rightarrow 3x^2+3y^2+4xy+x(-6h-4k)+y(-6k-4h)=c$

$\Rightarrow 3x^2+3y^2+4xy-10x-10y=c$ ........ From $\left(1\right)$ and $\left(2\right)$