Question

Show that the differential equation $x^2\frac{dy}{dx}=x^2-2y^2+xy$ is homogeneous and solve it.

Answer 1

Given differential equation:

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

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

$\frac{dy}{dx}=1-\frac{2y^2}{x^2}+\frac{y}{x}$

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

Substituting $F(x,y)=1-\frac{2y^2}{x^2}+\frac{y}{x}$,

Now,

$F(\lambda x,\lambda y)=1-\frac{2(\lambda y{)}^2}{(\lambda x{)}^2}+\frac{\lambda y}{\lambda x}$

$F(\lambda x,\lambda y)=1-\frac{2{\lambda }^2{y}^2}{{\lambda }^2{x}^2}+\frac{y}{x}$

$F(\lambda x,\lambda y)=1-\frac{2y^2}{x^2}+\frac{y}{x}=F(x,y)$

Hence, the differential equation is homogeneous.

Given differential equation:

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

$\frac{dy}{dx}=1-\frac{2y^2}{x^2}+\frac{y}{x}$ ...$\left(i\right)$

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,

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

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

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

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

Integrating both sides, we get

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

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

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

$\frac{1}{2}\times \frac{1}{2\left(\frac{1}{\sqrt{2}}\right)}\log \left|\frac{\frac{1}{\sqrt{2}}+v}{\frac{1}{\sqrt{2}}-v}\right|=\log \left|x\right|+c$

$\frac{\sqrt{2}}{4}\log \left|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}\right|=\log \left|x\right|+c$

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

$\frac{1}{2\sqrt{2}}\log \left|\frac{1+\sqrt{2}\frac{y}{x}}{1-\sqrt{2}\frac{y}{x}}\right|=\log \left|x\right|+c$

$\frac{1}{2\sqrt{2}}\log \left|\frac{x+\sqrt{2}y}{x-\sqrt{2}y}\right|=\log \left|x\right|+c$

Hence, the required general solution is

$\frac{1}{2\sqrt{2}}\log \left|\frac{x+\sqrt{2}y}{x-\sqrt{2}y}\right|=\log \left|x\right|+c$