Question

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

Answer 1

A first‐order differential equation is said to be homogeneous if $M(x,y)$ and $N(x,y)$ are both homogeneous functions of the same degree.Here degree of $x$ and $y$ are same.Hence the above differential equation is homogeneous.

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

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

Let $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$ where $v$ is a function of $x$

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

$\Rightarrow v+x\frac{dv}{dx}=\frac{1+2v}{1-v}$ by cancelling the common terms in the numerator and denominator

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

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

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

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

$\Rightarrow -\int \frac{(v-1)dv}{v^2+2\times v\times \frac{1}{2}+{\left(\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2+1}=\int \frac{dx}{x}$ by completing square method.

$\Rightarrow -\int \frac{(v-1)dv}{{(v+\frac{1}{2})}^2-\frac{1}{4}+1}=\int \frac{dx}{x}$

$\Rightarrow -\int \frac{(v-1)dv}{{(v+\frac{1}{2})}^2-{\left(\frac{\sqrt{3}}{2}\right)}^2}=\int \frac{dx}{x}$

$\Rightarrow \int \frac{vdv}{{(v+\frac{1}{2})}^2-{\left(\frac{\sqrt{3}}{2}\right)}^2}-\int \frac{dv}{{(v+\frac{1}{2})}^2-{\left(\frac{\sqrt{3}}{2}\right)}^2}=\int \frac{dx}{x}$

Let $t={(v+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2\Rightarrow dt=2(v+\frac{1}{2})$

$\Rightarrow \frac{-1}{2}\int \frac{2(v+\frac{1}{2}-\frac{1}{2})dv}{{(v+\frac{1}{2})}^2-{\left(\frac{\sqrt{3}}{2}\right)}^2}-\int \frac{dv}{{(v+\frac{1}{2})}^2-{\left(\frac{\sqrt{3}}{2}\right)}^2}=\int \frac{dx}{x}$

$\Rightarrow \frac{-1}{2}\int \frac{2(v+\frac{1}{2})dv}{{(v+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}+(\frac{1}{4}-1)\int \frac{dv}{{(v+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}=\int \frac{dx}{x}$

$\Rightarrow \frac{-1}{2}\int \frac{dt}{t}-\frac{\frac{3}{4}}{\frac{1}{2}}{\tan }^{-1}\left(\frac{v+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)=\ln x+c$ where $c$ is the constant of integration

$\Rightarrow -\ln t-3{\tan }^{-1}\left(\frac{2v+1}{\sqrt{3}}\right)=2\ln x+2c$ where $2c$ is the constant of integration

$\Rightarrow \ln t+3{\tan }^{-1}\left(\frac{2v+1}{\sqrt{3}}\right)+2\ln x=-2c$ where $k=-2c$ is the constant of integration

$\Rightarrow \ln ({(v+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2)+3{\tan }^{-1}\left(\frac{2\frac{y}{x}+1}{\sqrt{3}}\right)+2\ln x=k$ where $k=-2c$ is the constant of integration where $t={(v+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2$ and $v=\frac{y}{x}$