Question #4e226

1 Answer
Jun 29, 2017

See below.

Explanation:

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.