Question

Show that the given differential equation is homogeneous and solve:$(x-y)dy-(x+y)dx=0$

Answer 1

The given differential equation is:
$(x-y)dy-(x+y)dx=0$
$⇒\frac{dy}{dx}=\frac{x+y}{x-y}...(1)$
Let $F(x,y)=\frac{x+y}{x-y}$
$∴F(λx,λy)=\frac{λx+λy}{λx-λy}=λ.F(x,y)$
Thus,the given differential equation is a homogenous equation.
To solve it,we make the substitution as:
$y=vx$
$⇒\frac{d}{dx}(y)=\frac{d}{dx}(vx)$
$⇒\frac{dy}{dx}=v+x \frac{dy}{dx}$
Substituting the values of y and $\frac{dy}{dx}$ in equation(1),we get:
$v+x \frac{dv}{dx}=\frac{x+vx}{x-vx}=\frac{1+v}{1-v}$
$x\frac{dv}{dx}=\frac{1+v}{1-v-v}=\frac{1+v-v(1-v)}{1-v}$
$⇒x\frac{dv}{dx}=\frac{1+v^2}{1-v}$
$⇒\frac{1-v}{(1+v)^2}dv=\frac{dx}{x}$
$⇒(\frac{1}{1+v^2}-\frac{1}{1-v^2})dv=\frac{dx}{x}$
Integrating both sides,we get:
$tan^{-1}v-\frac{1}{2}log(1+v^2)=logx+C$
$⇒tan^{-1}(\frac{y}{x})-\frac{1}{2}log[1+(\frac{y}{x})^2]logx+C$
$⇒tan^{-1}(\frac{y}{x})-\frac{1}{2}log(x^2+\frac{y^2}{x^2})=logx+C$
$⇒tan^{-1}(\frac{y}{x})-\frac{1}{2}[log(x^2+y^2)-logx^2]=logx+C$
$⇒tan{-1}(\frac{y}{x})=\frac{1}{2}log(x^2+y^2)+C$
This is the required solution of the given differential equation.