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.