Question

Show that the differential equation
$x.\cos \left(\frac{y}{x}\right)\frac{dy}{dx}=y.\cos \left(\frac{y}{x}\right)+x$
is homogeneous and solve it.

Answer 1

$\text{Given}:x.\cos \left(\frac{y}{x}\right)\frac{dy}{dx}=y.\cos \left(\frac{y}{x}\right)+x$

$\frac{dy}{dx}=\frac{y\cos \left(\frac{y}{x}\right)+x}{x\cos \left(\frac{y}{x}\right)}$

If $F(\lambda x,\lambda y)=\lambda F(x,y),$
then the differential equation is homogeneous .

Substituting $F(x,y)=\frac{y\cos \left(\frac{y}{x}\right)+x}{x\cos \left(\frac{y}{x}\right)}$
Now,

$F(\lambda x,\lambda y)=\frac{\left(\lambda y\right)\cos \left(\frac{\lambda y}{\lambda x}\right)+\lambda x}{\left(\lambda y\right).\cos \left(\frac{\lambda y}{\lambda x}\right)}$

$F(\lambda x,\lambda y)=\frac{\lambda y\cos \left(\frac{y}{x}\right)+\lambda x}{\lambda y\cos \left(\frac{y}{x}\right)}$

$F(\lambda x,\lambda y=\frac{\lambda (y\cos \left(\frac{y}{x}\right)+x)}{\lambda x\cos \left(\frac{y}{x}\right)}$

$F(\lambda x,\lambda y)=\frac{y\cos \left(\frac{y}{x}\right)+x}{x\cos \left(\frac{y}{x}\right)}=F(x,y)$


Hence,the differential equation is homogeneous.

Given differential equation is

$x.\cos \left(\frac{y}{x}\right)\frac{dy}{dx}=y.\cos \left(\frac{y}{x}\right)+x$

$\frac{dy}{dx}=\frac{y\cos \left(\frac{y}{x}\right)+x}{x\cos \left(\frac{y}{x}\right)}...\left(i\right)$

$\text{Let}y=vx$

Differentiating both sides w.r.t. 'x'.

$\frac{dy}{dx}=\frac{d\left(vx\right)}{dx}$

Using product rule : $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$

$\frac{dy}{dx}=\frac{dv}{dx}x+v\frac{dx}{dx}$

$\frac{dy}{dx}=\frac{dv}{dx}x+v$

Substituting value of $\frac{dy}{dx}\text{and}y=\text{vx in (i),}$

$\frac{dv}{dx}x+v=\frac{\left(vx\right)\cos \left(\frac{vx}{x}\right)+x}{x\cos \left(\frac{vx}{x}\right)}$

$\frac{dv}{dx}x+v=\frac{x(v\cos v+1)}{x\cos v}$

$\frac{dv}{dx}x+v=\frac{v\cos v+1}{x\cos v}$

$\frac{dv}{dx}x=\frac{v\cos v+1}{\cos v}-v$

$\frac{dv}{dx}x=\frac{v\cos v+1-v\cos v}{\cos v}$

$\frac{dv}{dx}x=\frac{1}{\cos v}$

$\cos vdv=\frac{dx}{x}$

Integrating both sides, we get

$\int \cos vdv=\int \frac{dx}{x}$

$\sin v=\log |x|+\log |c|$

$\text{Substituting}v=\frac{y}{x},\text{we get}$

$\sin \frac{y}{x}=\log |x|+\log |c|$

$\sin \frac{y}{x}=\log |cx|(\because \log +\log b=\log ab)$

$\therefore$ The required general solution is

$\sin \frac{y}{x}=\log |cx|$