Solve dz/dx-z(2yx+x^2)=x where y =1
pls also illustrate the procedure
dz/dx - (2x + x^2)z = x
If you can't think of any other way, you can always find an integrating factor. The equation is in the form
z' + P(x)z = Q(x)
so the integrating factor will be of the form:
exp[ (integral) P(x) dx ]
Edit: This leaves one heck of an integral to try and solve, though. You might do better to do a series solution...