Question

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

Answer 1

Given: $(x-y)\frac{dy}{dx}=x+2y$
$\frac{dy}{dx}=\left(\frac{x-2y}{x-y}\right)$

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

$\text{Substituting}F(x,y)=\left(\frac{x+2y}{x-y}\right)$

Now,
$F(\lambda x,\lambda y)=\frac{\lambda x+2\left(\lambda y\right)}{\lambda x-\lambda y}$

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

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

Hence,the differential equation is homogeneous.
Given differential equation is

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

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

$\text{Let}y=vx$

Differentiating both sides $W.r.t^{\prime }{x}^{\prime }$
$\frac{dy}{dx}=\frac{d\left(vx\right)}{dx}$

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 $\frac{dy}{dx}andy=vx$
in $\left(i\right)$, we get

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

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

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

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

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

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

Integrating both sides, we get

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

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

$\int \frac{(v+\frac{1}{2})-\frac{3}{2}}{(v+\frac{1}{2}{)}^2+\frac{3}{4}}dv=-\log \left|x\right|+c$

$\int \frac{(v+\frac{1}{2})dv}{{(v+\frac{1}{2})}^2+\frac{3}{4}}-\frac{3}{2}\int \frac{dv}{{(v+\frac{1}{2})}^2+\frac{3}{4}}$

$=-\log \left|x\right|+c$
$I_1-\frac{3}{2}{I}_2=-\log \left|x\right|+c...\left(ii\right)$

$I_1=\int \frac{(v+\frac{1}{2})dv}{{(v+\frac{1}{2})}^2+\frac{3}{4}}$

Substituting ${(v+\frac{1}{2})}^2+\frac{3}{4}=t$

Differetiating $w.r.t.v,$ we get

$\frac{d({\left(\right(v+\frac{1}{2})}^2+\frac{3}{4})}{dv}=\frac{dt}{dv}$

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

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

Substituting value of $v$ and $dv$ in $I_1,$

$I_1=\int \frac{(v+\frac{1}{2})}{t}\times \frac{dt}{2(v+\frac{1}{2})}$

$I_1=\frac{1}{2}\int \frac{dt}{t}$

$I_1=\frac{1}{2}log|t|$

$\text{Substituting}t={(v+\frac{1}{2})}^2+\frac{3}{4}\text{in}{l}_1,$

$l_1=\frac{1}{2}\log |{(v+\frac{1}{2})}^2+\frac{3}{4}|$

$l_1=\frac{1}{2}\log |v^2+2v\times \frac{1}{2}+\frac{1}{4}+\frac{3}{4}|$

$l_1=\frac{1}{2}\log |v^2+v+1|$

$I_2=\int \frac{dv}{{(v+\frac{1}{1})}^2}+\frac{3}{4}$

$l_2=\int \frac{dv}{{(v+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$

$l_2=\frac{1}{\frac{\sqrt{3}}{2}}{\tan }^{-1}\frac{(v+\frac{1}{2})}{\frac{\sqrt{3}}{2}}$

$I_2=\frac{2}{\sqrt{3}}{\tan }^{-1}\frac{2(v+\frac{1}{2})}{\sqrt{3}}$

$I_2=\frac{2}{\sqrt{3}}ta{n}^{-1}\left(\frac{2v+1}{\sqrt{3}}\right)$

$\text{Substituting value of}{l}_1\text{and}{I}_2\text{in~(ii),}$

$I_1-\frac{3}{2}{I}_2=-\log |x|+c$

$\frac{1}{2}\log |v^2+v+1|-\frac{3}{2}\times \frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2v+1}{\sqrt{3}}\right)$

$=-\log |x|+c$

$\text{Since},\int \frac{dy}{1+y^2}={\tan }^{-1}y+C)$

$\frac{1}{2}\log |v^2+v+1|-\sqrt{3}{\tan }^{-1}\left(\frac{2v+1}{\sqrt{3}}\right)$

$=-\log |x|+c$

$\text{Replacing}v\text{by}\frac{y}{x},\text{We get}$

$=-\log |x|+c$

$\frac{1}{2}\log |\frac{y^2}{x^2}+\frac{y}{x}+1|+\log |x|$

$=\sqrt{3}{\tan }^{-1}\left(\frac{2y+x}{\sqrt{3}x}\right)+c$

$\log |\frac{y^2}{x^2}+\frac{y}{x}+1|+2\log |x|$

$=2\sqrt{3}{\tan }^{-1}\left(\frac{2y+x}{\sqrt{3}x}\right)+2c$

$\text{Substituting 2c = C and}$

$\text{using}(\log ab=\log a+\log b)$

$\log [|\frac{y^2}{x^2}+\frac{y}{x}+|\times |x^2|]$

$=2\sqrt{3}{\tan }^{-1}\left(\frac{2y+x}{\sqrt{3}x}\right)+C$

$\log |\frac{x^2{y}^2}{x^2}+\frac{x^{2}y}{x}+x^2|$

$=2\sqrt{3}{\tan }^{-1}\left(\frac{x+2y}{\sqrt{3}x}\right)+C$

$\log |x^2+xy+y^2|=2\sqrt{3}{\tan }^{-1}\left(\frac{x+2y}{\sqrt{3}x}\right)+C$

$\therefore$ The required general solution is

$\log |x^2+xy+y^2|=2\sqrt{3}{\tan }^{-1}\left(\frac{x+2y}{\sqrt{3}x}\right)+C$