Question

If $\frac{dy}{dx}=\frac{x-y}{x+y}$ then:

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

Answer 1

The correct option is D $x^2-y^2-2xy=c$

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

Let $y=vx$ since the above function $\frac{x-y}{x+y}$ is homogeneous function.

Therefore, $\frac{dy}{dx}=v+\frac{xdv}{dx}$.

Hence

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

$\to v+\frac{xdv}{dx}=\frac{x-vx}{x+vx}$.

$\to v+\frac{xdv}{dx}=\frac{1-v}{1+v}$.

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

$\frac{xdv}{dx}=\frac{1-v-v-v^2}{(1+v)}$

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

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

$\frac{-xdv}{dx}=\frac{(v+1{)}^2-2}{1+v}$

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

$\int \frac{1+v}{(v+1{)}^2-2}dv=-lnx+lnC$

Consider

$\int \frac{1+v}{(v+1{)}^2-2}dv=I$

Then let $(v+1{)}^2-2=t$

$2(v+1)dv=dt$

Hence

$\int \frac{1+v}{(v+1{)}^2-2}dv$

$=\int \frac{dt}{2t}$

$=\frac{1}{2}lnt$

$=\ln \left(\sqrt{t}\right)$

$=\ln \left(\sqrt{(v+1{)}^2-2}\right)$

$=\ln \left(\sqrt{(x^2+y^2+2xy-2x^2}\right)-\ln x$

$=\ln \left(\sqrt{(y^2+2xy-x^2}\right)-\ln x$

Hence, the integral becomes

$\ln \left(\sqrt{(y^2+2xy-x^2}\right)-\ln x=-\ln x+\ln C$

or

$\ln \left(\sqrt{(y^2+2xy-x^2}\right)=\ln c$

$\sqrt{(y^2+2xy-x^2}=C$

$y^2+2xy-x^2=C$

Or

$x^2-y^2-2xy=$ Constant