Question

Show that the differential equation $(x-y)dy-(x+y)dx=0$ is homogeneous and solve it.

Answer 1

Given differential equation:
$(x-y)dy-(x+y)dx=0$

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

If $F(\lambda x,\lambda y)=\lambda F(x,y)$, then the
differential equation is homogeneous.

Put
$F(x,y)=\frac{x+y}{x-y}$

Now,

$F(\lambda x,\lambda y)=\frac{\lambda x+\lambda y}{\lambda x-\lambda y}$

$F(\lambda x,\lambda y)=\frac{\lambda (x+y)}{\lambda (x-y)}$

$F(\lambda x,\lambda y)=\frac{x+y}{(x-y)}=F(x,y)$

Hence, the differential equation is homogeneous.

Given differential equation is

$(x-y)dy-(x+y)dx=0$

$\frac{dy}{dx}=\frac{x+y}{x-y}...\left(i\right)$

Let $y=vx$

Differentiating both sides $w.r.t.{}^{\prime }{x}^{\prime }$, we get

$\frac{dy}{dx}=\frac{d\left(vx\right)}{dx}$

Using product rule : $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$

$\frac{dy}{dx}=\frac{dv}{dx}x+v\frac{dx}{dx}$

$\frac{dy}{dx}=\frac{dv}{dx}x+v$

Putting the value of
$\frac{dy}{dx}$

and $y=vx$ in $\left(i\right)$,then we get

$x\frac{dv}{dx}+v=\frac{x+xv}{x-xv}$

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

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

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

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

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

Integrating both sides, we get

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

$\int \square \frac{dv}{1+v^2}-\int \frac{vdv}{1+v^2}=\int \frac{dx}{x}$

${\tan }^{-1}v-\int \frac{vdv}{1+v^2}=\log \left|x\right|+c$

${\tan }^{-1}v-I=\log \left|x\right|+c$ ...$\left(i\right)$

$I=\int \frac{vdv}{1+v^2}$ ...$\left(ii\right)$

Putting $t=1+v^2$,

Differentiating both side $w.r.t.v$, we get,

$\frac{d}{dv}(1+v^2)=\frac{dt}{dv}$

$vdv=\frac{dt}{2}$

putting $vdv=\frac{dt}{2}$ in $\left(ii\right)$, then we get

$I=\int \frac{vdv}{1+v^2}=\int \frac{dt}{2t}$

$I=\frac{1}{2}\log \left|t\right|+c$

Putting $t=1+v^2$,

$I=\frac{1}{2}\log |1+v^2|+c$

putting in $\left(i\right)$, then we get,

${\tan }^{-1}v-\frac{1}{2}\log |1+v^2|=\log \left|x\right|+c$

${\tan }^{-1}v=\frac{1}{2}\log |1+v^2|+\frac{2}{2}\log \left|x\right|+c$

${\tan }^{-1}v=\frac{1}{2}\log [|1+v^2|.{\left|x\right|}^2]+c$

$(\log a+\log b=\log ab)$

Putting $v=\frac{y}{x}$,

${\tan }^{-1}\frac{y}{x}=\frac{1}{2}\log [(1+{\left(\frac{y}{x}\right)}^2)\times x^2]+c$

${\tan }^{-1}\frac{y}{x}=\frac{1}{2}\log [x^2+y^2]+c$

Final answer:

Hence, the required general solution is
${\tan }^{-1}\frac{y}{x}=\frac{1}{2}\log [x^2+y^2]+c$