Question

Show that the given differential equation is homogeneous and solve:$(x^2+xy)dy=(x^2+y^2)dx$

Answer 1

The given differential equation i.e.,$(x^2+xy)dy=(x^2+y^2)dx$ can be written as:
$\frac{dy}{dx}=\frac{x^2+y^2}{x^2+xy}...(1)$
Let $F(x,y)=\frac{x^2+y^2}{x^2+xy}$
Now,$F(λx,λy)=\frac{(λx)^2+(λy)^2}{(λx)^2(λx)(λy)}=\frac{x^2+y^2}{x^2+xy}=λ.F(x,y)$
This shows that equation(1)is a homogeneous equation.
To solve it we make the substitution as:
$y=vx$
Differentiating both sides with respect to x,we get:
$\frac{dy}{dx}=v+\frac{xdv}{dx}$
Substituting the value of v and $\frac{dy}{dx}$ in equation(1),we get:
$v+\frac{dv}{dx}=\frac{x^2+(vx)^2}{x^2+x(vx)}$
$⇒v+x \frac{dv}{dx}=\frac{1+v^2}{1+v}$
$⇒x\frac{dv}{dx}=\frac{1+v^2}{1+v-v}=\frac{(1+v)^2-v(1+v)}{1+v}$
$⇒x\frac{dv}{dx}=\frac{1-v}{1+v}$
$⇒(\frac{1+v}{1-v})=dv=\frac{dx}{x}$
$⇒(\frac{2-1+v}{1-v})dv=\frac{dx}{x}$
$⇒(\frac{2}{1-v}-1)dv=\frac{dx}{x}$
Integrating,both sides,we get:
$-2log(1-v)-v=logx-logk$
$⇒v=-2log(1-v)-logx+logk$
$⇒v=log[\frac{k}{x(1-v)^2}]$
$⇒\frac{y}{x}=log[\frac{k}{x(1-\frac{y}{x})^2}]$
$⇒\frac{y}{x}=log[\frac{kx}{(x-y)^2}]$
$⇒\frac{kx}{(x-y)^2}=\frac{ey}{x}$
$⇒(x-y)^2=kxe^ {-\frac{y}{x}}$
This is the required solution of the given differential equation.