Question

Solve:

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

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

Answer 1

The correct option is A $x+y-2={k(x-y)}^3$

Substitute $x\to X+h$ and $y=Y+k$

$\therefore h+2k=3,2h+k=3$

$\Rightarrow h=1,k=1$

Now $\frac{dy}{dx}=\frac{X+2Y}{2X+Y}$

Substitute $Y=vx$

Now $\frac{xdv}{dx}=\frac{1+2v}{2+v}-v$

$\Rightarrow \frac{2+v}{(1+v)(1-v)}dv=\frac{dx}{x}$

$=\int (\frac{1}{1-v}+\frac{1}{1-v^2})dv=\ln \left|X\right|+c$

$=-\ln |1-v|+\frac{1}{2}\ln \left|\frac{v+1}{1-v}\right|=\ln \left|X\right|+c\Rightarrow \ln |X-Y|+\ln {\left|\frac{Y+X}{X-Y}\right|}^{\frac{1}{2}}=c$

$\Rightarrow \ln \left|\frac{\sqrt{\frac{X+Y}{X-Y}}}{X-Y}\right|\Rightarrow \frac{X+Y}{{(X-Y)}^3}=k=(x+y-2)=k{(x-y)}^3$