Browse Questions

Show that the differential equation is homogeneous and solve$(x-y)\;dy-(x+y)\;dx=0$

 Solution: $tan^{-1}(y/x) = (1/2)log(x^2+y^2) + C$

Toolbox:

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

Step1

The given equation can be rearranged and written as$dy/dx = (x+y)/(x-y)$
$F(x,y) = (x+y)/(x-y)$
$F(kx,ky) = (kx+ky)/(kx-ky) = k^0.f(x,y)$
Hence it is a homogenous equation with degree zero.

Step2

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

Bringing $v$ from the LHS to RHS we get,
$x.dv/dx = (1+v^2)/(1-v)$

Step3:

Separating the variables we get
$(1-v)dv/(1+v^2) = dx/x$
$dv/1+v^2 - vdv/1+v^2 = dx/x$

Integrating on both sides we get,
$tan^{-1}v - (1/2)log(1+v^2) = logx + C$

Step4:

Substituting back for v as y/x we get
$tan^{-1}(y/x) - (1/2)log(x^2+y^2)/x^2 = logx + C$

This can be rewritten as ,
$tan^{-1}(y/x) - (1/2)x^2+y^2 + (1/2)log x^2 = logx + C$
$tan^{-1}(y/x) = (1/2)log(x^2+y^2) + C$
This is the required solution.

edited Dec 24, 2012