1. Show that the family of curves x^2 + y^2 = ax is orthogonal to the family of curves
x^2 + y^2 = by at every point of intersection. (These are called orthogonal trajectories.)
Sketch several curves from each family on the same axes.


5. Find the orthogonal trajectories of the family of curves x^2+y^2 = r^2. Sketch several
curves from each family on the same axes.



13. Using mv' = -pv - mg:
(a) Solve the differential equation using the fact that it is a separable equation.
(b) Solve the differential equation using the fact that it is first-order linear equation.
(c) Solve the differential equation using the fact that the substitution P = v + [(mg)/p] yields a new differential equation that follows the Law of Natural Growth.

(Note the equation: dP/dt = kP - m. This equation accounts for "harvesting" or
"emigration". For Newton's Law of Heating/Cooling it could also account for the
ambient temperature.)


Thanks to everyone that is helping me, i REALLY appreciate it. You are ALL awesome!

Just need help with this one, figured the other ones out.
I need this before 8:00am tommorow (so if its past that, dont bother doing lolol)

Well I am going back to my internet-less aprtment and will check back in in the morning when I have internet. Have a goodnite guys!

13. Using mv' = -pv - mg:
(a) Solve the differential equation using the fact that it is a separable equation.
(b) Solve the differential equation using the fact that it is first-order linear equation.
(c) Solve the differential equation using the fact that the substitution P = v + [(mg)/p] yields a new differential equation that follows the Law of Natural Growth.

(Note the equation: dP/dt = kP - m. This equation accounts for "harvesting" or
"emigration". For Newton's Law of Heating/Cooling it could also account for the
ambient temperature.)


Thanks to everyone that is helping me, i REALLY appreciate it. You are ALL awesome!

Your differential equation falls under the general form :

dy/dt = a.y + b

where a, b are some constants (a non zero).

The general solution is :

y(t) = K.exp(at) - b/a

with K any constant.

You are asked to solve this equation using 3 different methods ? I can only see 1 method, but maybe it can be written under 3 different forms :

(a) Let's first solve :
mv' = - pv
v' = - pv/m
v(t) = K.exp(- pt/m) for any constant K

Now let's solve :
0 = -pv - mg
pv = - mg
v = - mg/p

And finally let's add together our 2 solutions :

v(t) = K.exp( - pt/m) - mg/p

(b) It is a known fact that a first-order linear equation dy/dt = a.y + b has solutions of the form : y(t) = K.exp(at) - b/a (see above)
mv' = - pv - mg
v' = - pv/m - g

Let's denote a = - p/m and b = - g :

v(t) = K.exp( pt/m ) - (-g) / ( - p/m)
v(t) = K.exp( - pt/m) - gm/p

(c) Let's P = v + mg / p
v = P - mg / p
P' = v'

Let's substitute v with P in our differential equation :
m P' = - p( P - mg/p ) - mg
m P' = - p P + mg - mg
m P' = - p P
P' = - p P / m

Hence P = K. exp( - pt/m )

Now let's get back to v :

v = P - mg / p = K.exp ( -pt/m) - mg / p


I'm not sure I did the work as it was expected, since, for me, those "3" methods are exactly similar...
Is it a maths course or a physics course ?