Question

Show that the differential equation $(x^2-y^2)dx+2xydy=0$
is homogeneous and solve it.

Answer 1

Given differential equation is

$(x^2-y^2)dx+2xydy=0$

$\frac{dy}{dx}=\frac{y^2-x^2}{2xy}$

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

Put

$F(x,y)=\frac{y^2-x^2}{2xy}$,

Now,

$F(\lambda x,\lambda y)=\frac{(\lambda y{)}^2-(\lambda x{)}^2}{2\lambda x.\lambda y}$

$F(\lambda x,\lambda y)=\frac{{\lambda }^2(y^2-x^2)}{{\lambda }^2.2xy}$

$F(\lambda x,\lambda y)=\frac{y^2-x^2}{2xy}=F(x,y)$

Hence, the differential equation is homogeneous.

Given differential equation is

$(x^2-y^2)dx+2xydy=0$

$\frac{dy}{dx}=\frac{y^2-x^2}{2xy}$ ...$\left(i\right)$

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

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

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

Substituting the value of $\frac{dy}{dx}\text{and}y=vx$
in $\left(i\right)$, we get

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

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

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

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

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

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

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

Integrating both sides, we get

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

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

Substituting $t=1+v^2$,

Differentiating both sides $w.r.t.v$, we get
$\frac{d}{dv}(1+v^2)=\frac{dt}{dv}$

$2v=\frac{dt}{dv}$

Substituting $dv=\frac{dt}{2v}$ in $\left(iii\right)$, we get

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

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

$I=\log \left|t\right|+\log c$

Substituting $t=1+v^2$,

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

Substituting in $\left(ii\right)$, we get

$\log |1+v^2|=-\log \left|x\right|+\log c$

$\log |1+v^2|+\log \left|x\right|=\log c$

$\log \left|x\right(v^2+1\left)\right|=\log c$

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

Substituting $v=\frac{y}{x}$

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

$\log \left|\frac{y^2+x^2}{x}\right|=\log c$

$y^2+x^2=cx$

Hence, the required general solution is
$y^2+x^2=cx$