Question #4e226

If `M(x,y)` is homogeneous then

`M(lambda x, lambda y) = lambda^alpha M(x,y)` with `lambda in RR` constant, so considering

`{(x = lambda xi),(y= lambda eta):}` we have

`{(dx = lambda d xi),(dy= lambda d eta):}`

and substituting

`M(x,y) dx+N(x,y)dy=0 rArr M(lambda xi, lambda eta)lambda d xi+N(lambda xi, lambda eta) lambda eta = 0 rArr lamda^alpha M(xi, eta)d xi+lambda^alpha N(xi,eta) d eta = 0` then

a) `M(xi, eta)d xi+ N(xi,eta) d eta = 0`

Considering now `y=lambda(x) x` we have

`dy/dx = lambda + x (d lambda)/(dx)` and

`dy/dx = -(M(x,y))/(N(x,y)) rArr lambda+x (d lambda)/(dx) = -x^alpha/x^alpha(M(1,lambda))/(N(1,lambda))= -(M(1,lambda))/(N(1,lambda)) = psi(lambda)` or grouping variables

`(dx)/x=(d lambda)/(psi(lambda)-lambda)` and integrating

b) `x = c e^(theta(lambda)) = c e^(theta(y/x)) = c phi(y/x)`

The remaining items are left to the proficient reader.