If I use z=(y+3)/x I get to dy/dx=z2+z
When I take dz/dx I get -(y+3)/x2, so dz=-(y+3)/x2 dx,
How do I go about substituting this into the equation so I can integrate wrt z?
Hello, having a lot of trouble with a dodgy question one of my lecturers has set us before teaching us how to do it, none of my course can seem to work out what to do. The question is:
dy/dx=(x(y+3)+(y+3)2)/x2
where y(1)=4, and x>0
I tried a substitution of z=y/x to eventually give...