# Show that the differential equation is homogeneous and solve$y'=\frac{x+y}{x}$

 Solution: $y = x. logx + C.x$

Tool Box:

• A differential equation of the form dy/d = F(x,y) is said to be homogenous if F(x,y) is a homogenous function of degree zero.
• To solve equations of this type we substitute $y = vx and dy/dx = v + xdv/dx$

Step1.

$F(x,y) = (x+y)/x$
$F(kx,ky) = k(x+y)/kx = k^0.f(x,y)$
Hence $F(x,y)$ is a homogenous degree of zero.

Step2.

Using the information in the tool box, let us substitute $y = vx and dy/dx = v+xdv/dx.$
$v + x.dv/dx = (x+vx)/x$
Taking x as the common factor and cancelling we get
$v + x.dv/dx = 1 + v$
Bringing v from LHS to the RHS and cancelling wwe get $x.dv/dx = 1$

Step3.

Seperating the variables we get, $dv = dx/x$

integrating on both sides we get, integration of v = integration of dx/x
i.e, $v= logx + C$

substituting for v
$y/x = logx + C$
Therefore, $y = x.logx + Cx$

This is the required solution for the differential equation.

edited Dec 24, 2012