Answer to Question #58422 in Differential Equations for Joseph

Question

6 Solve
(x3+y3)dx−3xy2dy=0
x5−2y3=Cx
x3−2y3=Cx
x3−3y3=Cx
x3−2y2=Cx

7 Solve
(1+2exy)dx+2exy(1−xy)dy=0
5x+2yexy=C
x+2ye2xy=C
x+2yexy=C
5x+3yexy=C

8 Solve
y(xy+1)dx+x(1+xy+x2y2)dy=0
2x2y2lny−2xy−1=Cx2y2
3x2y2lny−2xy−3=Cx2y2
2x2y2lny−xy=Cx2y2
2x3y2lny−2xy−1=Cx2y5

9 Solve
xdy−ydx−x2−y2−−−−−−√dx=0
Cx=2earcsinyx
Cx=earcsinyx
Cx=earcsin2y3x
Cx=earccosyx

10 The population of student P at NOUN increases at a rate proportional to the population and to the addition of 150,250 and the population divided by 3, the differential equation of this statement is
dPdT=3kP(150,250+P)4
dPdT=2kP(150,250+P)3
5dPdT=kP(150,250+P)3
dPdT=kP(150,250+P)3