Question

Solution of $\frac{dy}{dx}=\frac{x-2y+3}{2x+y-5}$ is:

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

Answer 1

The correct option is A $x^2-y^2-4xy+6x+10y=c$

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

$\Rightarrow h-2k+3=0$

$2h+k-5=0$

$\Rightarrow h=\frac{7}{5};k_1=\frac{11}{5}$

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

put $y=vx$

$\Rightarrow \frac{dy}{dy}=v+x\frac{dy}{dx}=\frac{1-2v}{2+v}$

$\Rightarrow -\frac{1}{2}\left[\frac{4+2v}{v^2+4v-1}\right]dv=\frac{dx}{x}$

$\Rightarrow -\log (v^2+4v-1)=\log x^2-\log c$

$\log c=\log x^2(v^2+4v-1)$

$\Rightarrow c=x^2(v^2+4v-1)$

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

$\Rightarrow c=y^2+k^2-2yk+4(xy-kx-hy+hk)-x^2-h^2-2xh$

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

$\Rightarrow constant=y^2-x^2+4xy-6x-10y$

$\Rightarrow x^2-y^2-4xy+6x+10y=constant$