Question

Show that the given differential equation is homogeneous and then solve it.

$(x^2+xy)dy=(x^2+y^2)dx.$

Answer 1

Given, differential equation is $\frac{dy}{dx}=\frac{x^2+y^2}{x^2+xy}$ ...(i)
Here, the numerator and denominator both have polynomial of degree 2. So, the given differential equation is homogeneous.
Let $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
On putting the values of $\frac{dy}{dx}$ and y in Eq. (i), we get
$v+x\frac{dv}{dx}=\frac{x^2+x^2{v}^2}{x^{2}v{x}^2}\Rightarrow v+x\frac{dv}{dx}=\frac{x^2(1+v^2)}{x^2(1+v)}\Rightarrow x\frac{dv}{dx}=\frac{1+v^2}{1+v}-v\Rightarrow x\frac{dv}{dx}=\frac{1+v^2-v-v^2}{1+v}=\frac{1-v}{1+v}\Rightarrow \frac{1+v}{1-v}dv=\frac{dx}{x}\Rightarrow \frac{2-1+v}{1-v}dv=\frac{dx}{x}$
On integarting both sides, we get
$\int \frac{2}{1-v}dv-\int 1dv=\int \frac{1}{x}dx\Rightarrow -2log(1-v)-v=logx-logC\Rightarrow -v-2log(1-v)-logx=-logC\Rightarrow v+2log(1-v)+logx=logCPutv=\frac{y}{x}\Rightarrow \frac{y}{x}+2log(1-\frac{y}{x})+logx=logC\Rightarrow \frac{y}{x}+log{\left(\frac{x-y}{x}\right)}^2+logx=logC\frac{y}{x}+log(x-y{)}^2-log{x}^2+logx=logC\frac{y}{x}+log(x-y{)}^2-2logx+logx=logC\Rightarrow \frac{y}{x}+log(x-y{)}^2-logx=logC\Rightarrow \frac{y}{x}+log\frac{(x-y{)}^2}{x}-logC=0\Rightarrow log(\frac{(x-y{)}^2}{x}.\frac{1}{C})=-\frac{y}{x}\Rightarrow (x-y{)}^2=Cx{e}^{-\frac{y}{x}}$
This is the required solution of the given differential equation.