Anonymous function in MATLAB/Octave

f1=@(x) x^2-x +1 ;
f2=@(x,y) x^2-3*y + x*y;

f1(5)
f2(4,5)

Why would you still want to use C after learning MATLAB/Octave ?

Differential equations

1. Launch angle of a projectile (Newton's second law)
   
   
   

   m a = F,

   (3)

   or
   

   m ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ d2 x d t2 d2 y d t2 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎝ Fx Fy ⎞ ⎟ ⎟ ⎠ ,

   (4)

   or
   

   m ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ d2 x d t2 d2 y d t2 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎝ 0 − m g ⎞ ⎟ ⎟ ⎠ .

   (5)

   The differential equations to be solved are
   
   

   d2 x(t) d t2 = 0,     d2 y(t) d t2 = − g.

   (6)

   The initial conditions (t = 0) are
   
   

   x(0) = 0,     y(0) = 0,     d x dt t = 0  = v0 cos θ,     d y dt t = 0  = v0 sin θ.

   (7)

   By integrating eq.(6)
   
   

   d x(t) d t = c1,     d y(t) d t = − g t + c2.

   (8)

   By integrating eq.(8) again, one obtains
   
   

   x(t) = c1 t + c3,     y(t) = − 1 2 g t2 + c2 t + c4.

   (9)

   The unknown constants, c₁  ∼ c₄, can be determined from eq.(7) as
   
   

   c1 = v0 cos θ,     c2 = v0 sin θ,     c3 = 0,     c4 = 0.

   (10)

   So the solutions for x(t) and y(t) are
   
   

   x(t) = v0 cos θ t     y(t) = − 1 2 g t2 + v0 sin θ t.

   (11)

   When the projectile is on the ground,
   
   

   0 = − 1 2 g t2 + v0 sin θ t,

   (12)

   from which
   

   t = 2 v0 sin θ g .

   (13)

   Therefore,
   

   x ⎛ ⎝ 2 v0 sin θ g ⎞ ⎠ = v0 cos θ × 2 v0 sin θ g = vo2 g sin 2 θ.

   (14)

   This quantity is maximized when θ = π/4 (link).
2. Population growth (logistic equation)
   Population growth can be modeled by
   

   dU(t) dt = αU (t),

   (15)

   where U(t) is the population of a country at time t and α is the growth rate.
   Equation (15) can be re-written as
   

   1 U d U = α d t

   (16)

   By integrating equation (16) with respect to each variable, one obtains (separation of variables)
   
   

   ln U = αt + C,

   (17)

   where C is an integral constant.
   Equation (17) can be written as
   
   

   U(t) = eαt + C = eC  eαt.

   (18)

   If the initial population is U₀ at t = 0, equation (18) can be written as
   
   

   U0 = eC.

   (19)

   So finally, one obtains
   
   

   U(t) = U0 eαt.

   (20)

   
   Correction to the above equation can be made as
   
   

   dU dt = αU (1 − U β )

   (21)

   where β is known as the carrying capacity.

1. Linear differential equations with constant coefficients
   Examples:
   1. y"(t) − 3 y′(t) + 2 y(t) = 0,     y(0) = a, y′(0) = b.
      (Solution): Solve λ² − 3 λ+ 2 = 0 as λ₁ = 1, λ₂ = 2. Then, express y(t) = c₁ e^t + c₂ e^{2 t}. Determine c₁ and c₂ to satisfy the initial conditions.
   2. y"(t) + 4 y(t) = 0,     y(0) = a, y′(0) = b.
      (Solution): Solve λ² + 4 = 0 as λ₁ = 2 i , λ₂ = − 2 i. Then, express y(t) = c₁ e^{2 i t} + c₂ e^{− 2 i t} = c′₁ cos 2 t + c′₂ sin 2 t. Determine c′₁ and c′₂ to satisfy the initial conditions.
2. Separation of variables
   
   

   dy dt = f(y) g(t) ⇒ 1 f(y) d y = g(t) d t.

   Integrating both sides of the equation with respect to each variable, one gets
   

   ⌠ ⌡ 1 f(y) d y = ⌠ ⌡ g(t) dt.

   Example:
   

   d y d t = y (1 − y).

   
   

   1 y (1 − y) d y = 1  d t, ⇒ ⎛ ⎝ 1 y − 1 y − 1 ⎞ ⎠ d y = d t ⇒ (ln y − ln (y − 1) ) = t + C

   The general solution is
   

   ln  y y − 1 = t + C.