Question

Show that the differential equation
$(1+e^{\frac{x}{y}})dx+e^{\frac{x}{y}}(1-\frac{x}{y})dy=0$
is homogenous and solve it.

Answer 1

Given differential equation:
$(1+e^{\frac{x}{y}})dx+e^{\frac{x}{y}}(1-\frac{x}{y})dy=0$
$\frac{dx}{dy}=\frac{-e^{\frac{x}{y}}(1-\frac{x}{y})}{1+e^{\frac{x}{y}}}$
If $F(\lambda x,\lambda y)=\lambda F(x,y)$, then the differential
equation is homogeneous.
Now,

Substituting $F(x,y)=\frac{-e^{\frac{x}{y}(1-\frac{x}{y})}}{1+e^{\frac{x}{y}}}$,
$F(\lambda x,\lambda y)=\frac{-e^{\frac{\lambda x}{\lambda y}}(1-\frac{\lambda x}{\lambda y})}{1+e^{\frac{\lambda x}{\lambda y}}}$
$F(\lambda x,\lambda y)=\frac{-e^{\left(\frac{x}{y}\right)}(1-\frac{x}{y})}{1+e^{\frac{x}{y}}}=F(x,y)$
Hence, the differential equation is
homogeneous.

Given differential equation :
$(1+e^{\frac{x}{y}})dx+e^{\frac{x}{y}}(1-\frac{x}{y})dy=0$
$\frac{dx}{dy}=\frac{-e^{\left(\frac{x}{y}\right)}(1-\frac{x}{y})}{1+e^{\frac{x}{y}}}$ ...$\left(i\right)$
Let $x=vy$

Differentiate both sides $w.r.t.{}^{\prime }{y}^{\prime }$, we get
$\frac{dx}{dy}=\frac{d\left(vy\right)}{dy}$
Using product rule : ${\left(uv\right)}^{\prime }=u^{\prime }v+u{v}^{\prime }$
$\frac{dx}{dy}=y.\frac{dv}{dy}+v\frac{dy}{dy}$
$\frac{dx}{dy}=y.\frac{dv}{dy}+v$
Substituting the value of $\frac{dx}{dy}$ and $x=vy$
in $\left(i\right)$, we get
$v+y\frac{dv}{dy}=\frac{-e^v(1-v)}{1+e^v}$
$y\frac{dv}{dy}=\frac{-e^v(1-v)}{1+e^v}-v$
$y\frac{dv}{dy}=\frac{-e^v+v{e}^v-v(1+e^v)}{1+e^v}$
$y\frac{dv}{dy}=\frac{-e^v+v{e}^v-v-v{e}^v}{1+e^v}$
$y\frac{dv}{dy}=\frac{-(e^v+v)}{1+e^v}$
$\frac{1+e^v}{v+e^v}dv=\frac{-dy}{y}$

Integrating both sides, we get
$\int \frac{1+e^v}{v+e^v}dv=\int \frac{-dy}{y}$
$\int \frac{1+e^v}{v+e^v}dv=-\log |y|+\log c$ ...$\left(ii\right)$

Substituting $v+e^v=t$,
$(1+e^v)dv=dt$

Substituting in $\left(ii\right)$,
$\int \frac{dt}{t}=-\log |y|+\log c$
$\log |t|=-\log |y|+\log c$
Substituting $t=v+e^v$, we get
$\log |v+e^v|=-\log |y|+\log c$
$\log |v+e^v|+\log |y|=\log c$
$\log \left(\right(v+e^v).y)=\log c$
$(\because \log a+\log b=\log ab)$

Substituting $v=\frac{x}{y}$, we get
$\log (\frac{x}{y}\times y+e^{\frac{x}{y}}y)=\log c$
$x+y{e}^{\frac{x}{y}}=c$

Hence, the required general solution is
$x+y{e}^{\frac{x}{y}}=c$