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.