Question

Solution of diff. equation $(6x+2y-10)\frac{dy}{dx}=2x+9y-20$ is

• A. $(y+2x{)}^2=C(x-2y+5)$
• B. $(y+2x{)}^3=C(x+2y-5)$
• C. $(y-2x{)}^2=C(x+2y-5)$
• D. none of these

Answer 1

Answer: (C)

$\frac{dy}{dx}=\frac{2x+9y-20}{6x+2y-10}$.
Put $x=X+h$ ; $y=Y+k$
$\frac{dY}{dX}=\frac{2X+9Y}{6X+2Y}$
Where $2h+9k-20=0$
$6h+2k-10=0$
$\frac{h}{-50}=\frac{k}{-100}=\frac{1}{-50}$
$\therefore h=1$, $k=2$
Put $Y=vX$
$v+X\frac{dv}{dX}=\frac{2+9v}{6+2v}$
$\therefore X\frac{dv}{dX}=\frac{2+9v-6v-2v^2}{6+2v}$
$=\frac{2+3v-2v^2}{6+2v}$
$\frac{6+2v}{2+3v-2v^2}dv=\frac{dX}{X}$
$\Rightarrow \int \left[\frac{2}{1+2v}+\frac{2}{2-v}\right]dv$
$=\int \frac{dX}{X}+log{C}_1$
$\Rightarrow log\left(1+2v\right)-2log\left(2-v\right)=log{C}_{1}X$
$\Rightarrow log\frac{1+2v}{{\left(2-v\right)}^2}=log{C}_{1}X$
$\Rightarrow \frac{1+2\frac{Y}{X}}{\left(2-\frac{Y}{X}\right)}=C_{1}X$
$\Rightarrow \frac{\left(X+2Y\right)X}{{\left(2X-Y\right)}^2}=C_{1}X$
$\Rightarrow \left(X+2Y\right)={\left(2X-Y\right)}^2{C}_1$
$\Rightarrow C\left(x+2y-5\right)-{\left(2x-y\right)}^2$
$\Rightarrow {\left(y-2x\right)}^2=C\left(x+2y-5\right)$