Question

The solution of $\frac{dy}{dx}+\frac{2x+3y+1}{3x+4y-1}=0$ is:

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

Answer 1

The correct option is A $x^2+3xy+2y^2+x-y=c$

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

$\Rightarrow 2h+3k+1=\mathrm{0..........1}$

$3h+4k-1=\mathrm{0..........2}$

$\Rightarrow \frac{dY}{dX}+\frac{2X+3Y}{3X+4Y}=0$

put $Y=vX$

$\Rightarrow \frac{dY}{dX}=v+X\frac{dv}{dX}$

$v+X\frac{dv}{dX}+\frac{2X+3Y}{3X+4Y}=0$

$v+X\frac{dv}{dX}+\frac{2+3v}{3+4v}=0$

$\Rightarrow X\frac{dv}{dX}+\frac{2+6v+4v^2}{3+4v}=0$

$\Rightarrow \int \frac{3+4v}{2[2v^2+3v+1]}dv+\int \frac{dX}{X}=constant$

$\Rightarrow \log (2v^2+3v+1)X^2=constant$$\Rightarrow 2Y^2+3YX+X^2=constant$

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

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

$\Rightarrow 2y^2+3xy+x^2+x-y=c$ $\left[from1and2\right]$