Q1

DSolve[{
n'[t]== k n[t] (50000- n[t]),
n[0]==500
},
n[t],
t]

Solve[
((50000 (E^(50000 k )))/(99+(E^(50000 k ))))==1000.,
k
]

Q2

xSol[t_]=x[t]/.DSolve[{
x'[t]==k(40- 2/3 x[t])(50- 1/3 x[t]),
x[0]==0},
x[t],
t
][[1]]

Solve[xSol[6]==10.0, k]

k=0.0009444057108916939`
xSol[12]
Plot[xSol[t],{t, 0, 123}]

Q3

g=32;
DSolve[{
h'[t]== - δ(Sqrt[2 g h[t]]),
h[0]==H
},
h[t],
t]

{H, δ}={10, 0.01}
Plot[
{H-8 (Sqrt[H]) t δ+16 (t^2) (δ^2)},
{t, 0, 80},
PlotLegends->Automatic]

Q4

Clear[g,k]
DSolve[{
m v'[t]== m g - k v[t]^2,
v[0]==v0},
v[t],
t]

Integrate[(((Sqrt[g]) (Sqrt[m]) Tanh[(((Sqrt[g]) (Sqrt[k]) t+(Sqrt[m]) ArcTanh[(((Sqrt[k]) v0)/((Sqrt[g]) (Sqrt[m])))])/(Sqrt[m]))])/(Sqrt[k])),{t, 0, T}]

Tanh[123456.7]

Q5

DSolve[ W'[t] == W[t] (Sqrt[4 - 2 W[t]]), W[t],t]//Simplify

Exam 2 Results

27 out of 135 folks (20%) got 100

42 out of 135 folks (31%) got 95+

49 out of 135 folks (36%) got 90+

19 out of 135 folks (15%) got 40-

Web Assign

xSol[t_]=x[t]/.DSolve[ {
x'[t]== k (60-2x[t])(150-x[t]),
x[0]==40
},
x[t],
t][[1]]
Solve[xSol[20]==32.0]

4.1 Higher Order Theory

Linear nth order IVP

(Subscript[a, n])[x](((d^n)y)/(dx^n))+…+(Subscript[a, 1])[x]y'[x]+(Subscript[a, 0])[x]y[x]==g[x]
 y[(Subscript[x, 0])]=(Subscript[y, 0]),y'[(Subscript[x, 0])]=(Subscript[y, 1]),… , (y^((n-1)))[(Subscript[x, 0])]=(Subscript[y, n-1])

Thm: 4.1.1
If all the (Subscript[a, i]) are continuous and (Subscript[a, n])≠0 on an interval I then there is exactly one solution of the IVP.

If the IC are not all at the same x value it is a Boundary Value Problem. No such thm is possible for BVPs.

Terminology

The IVP is homogeneous if g[x]=0.

If (Subscript[y, 1]),(Subscript[y, 2]),…,(Subscript[y, n]) are solns of a homogeneous ODE then the superposition 
y=(Subscript[c, 1])(Subscript[y, 1])+ (Subscript[c, 2])(Subscript[y, 2])+…+(Subscript[c, n])(Subscript[y, n])
is also a soln of the ODE.

Functions (Subscript[y, 1]),(Subscript[y, 2]),…,(Subscript[y, n]) are Linearly Dependent (LD) if there are constants (not all zero) satisfying.
0=(Subscript[c, 1])(Subscript[y, 1])+ (Subscript[c, 2])(Subscript[y, 2])+…+(Subscript[c, n])(Subscript[y, n]) 
Linearly Independent means not LD.

The Wronskian of functions (Subscript[y, 1]),(Subscript[y, 2]),…,(Subscript[y, n]) is 
W=Det(({{(Subscript[y, 1]), (Subscript[y, 2]), …, (Subscript[y, n])}, {((Subscript[dy, 1])/dx), ((Subscript[dy, 2])/dx), …, ((Subscript[dy, n])/dx)}, {⋮, ⋮,  , ⋮}, {(((d^n)(Subscript[y, 1]))/((d^n)x)), (((d^n)(Subscript[y, 2]))/((d^n)x)), …, (((d^n)(Subscript[y, n]))/((d^n)x))}})) 
If W≠0 on an interval ℐ for solutions of a homogenous ODE then the solns are LI.

A Fundamental Set (FS) of solutions of an nth order linear ODE is a LI set of n solutions {(Subscript[y, 1]),(Subscript[y, 2]),…,(Subscript[y, n])}

The general solution of an nth order linear homogeneous ODE is the superposition of an FS
y=(Subscript[c, 1])(Subscript[y, 1])+ (Subscript[c, 2])(Subscript[y, 2])+…+(Subscript[c, n])(Subscript[y, n])

Thm 4.1.6. If (Subscript[y, p]) is any solution of the non-nth order inhomogeneous ODE 
(Subscript[a, n])[x](((d^n)y)/(dx^n))+…+(Subscript[a, 1])[x]y'[x]+(Subscript[a, 0])[x]y[x]=g[x] 
the general solution is
y=(Subscript[y, p])+(Subscript[c, 1])(Subscript[y, 1])+ (Subscript[c, 2])(Subscript[y, 2])+…+(Subscript[c, n])(Subscript[y, n]) 
where {(Subscript[y, 1]),(Subscript[y, 2]),…,(Subscript[y, n])} is a FS for
(Subscript[a, n])[x](((d^n)y)/(dx^n))+…+(Subscript[a, 1])[x]y'[x]+(Subscript[a, 0])[x]y[x]=0

4.2 Reduction of Order

We can use a known solution of a linear second order ODE to simplify the equation.

We are going to substitute 
	y=(Subscript[y, 1])[x]u[x] 
into the standard form ODE 
  y''[x]+P[x]y'[x]+Q[x]y[x]=0.

So 
	({{y, ->, (Subscript[y, 1])u}, {y', ->, (Subscript[y, 1])'u+(Subscript[y, 1])u'}, {y'', ->, (Subscript[y, 1])''u+2(Subscript[y, 1])'u'+(Subscript[y, 1])u''}})

Plug in and organize
({{(Subscript[y, 1])''u    , +, P (Subscript[y, 1])'u, +, Q (Subscript[y, 1])u, =, 0}, {2 (Subscript[y, 1])'u', +, P (Subscript[y, 1])u',  ,  ,  ,  }, {(Subscript[y, 1])u''        ,  ,  ,  ,  ,  ,  }})
If (Subscript[y, 1]) is a solution the top row is zero.
({{(Subscript[y, 1])u''    , +, (2(Subscript[y, 1])'+P (Subscript[y, 1]))u', +, 0u, =, 0}})
Which is a separable 1st order ODE for w=u'.
({{(Subscript[y, 1])w'    , +, (2(Subscript[y, 1])'+P (Subscript[y, 1]))w, =, 0}})