Chapter 9: Primitives & Differential Equations

• Chapter 9 focuses on primitives and differential equations.
• The chapter aims to calculate primitives of functions using reference primitives and functions of the form (v'ou) × u', and determine the set of solutions for a differential equation of the type y' = ay + f, given a particular solution when f isn't constant.

Rugby Ball Trajectory Example

• Considers the trajectory of a rugby ball during a penalty kick.
• At the moment of the kick (t=0), the ball is 40 m from the goalposts and is kicked with an initial velocity (vo) of 21 m/s at an angle of 45° to the horizontal.
• Only the weight of the ball is considered, ignoring friction.
• The coordinate system is orthonormal (O; i, j), with ||i|| = ||j|| = 1 m.
• The penalty is successful if the ball passes over the crossbar, which is 3 m high.
• According to the fundamental principle of dynamics, since the ball is only subject to its weight (P), then P = mg = ma, therefore a = g.
• Acceleration vector coordinates in the (O; i, j) frame are represented.
• The ball's velocity vector coordinates are solutions to differential equations.

Parachute Trajectory Example

• Analyzes the velocity of a parachute with mass m, subjected to its weight and air friction forces.
• The fundamental principle of dynamics yields the equation F + P = ma = m(dv(t)/dt) = mv'(t).
• Air friction (F) is proportional to the object's velocity and opposes displacement, F = -kv(t).
• Projecting forces on a vertical axis gives mg - kv(t) = mv'(t).
• Object's velocity is a solution to the differential equation (E2): v'(t) = -(k/m)v(t) + g.
• Both equations are differential equations where the unknown is a function (v(t)) with its successive derivatives (v'(t)).

Differential Equation Types and Solutions

• Mathematical function is generally denoted by 'y', with derivatives y', y", and so on.
• The first differential equation is of the form y' = g, where g is a function.
• The second is of the form y' = ay + b, where 'a' and 'b' are real numbers.
• Function 'f' is a solution to the differential equation if replacing 'y' with 'f' in the equation makes the equation true over its domain.

Differential Equation y' = f

• Definition: Given a function 'f' defined on an interval I of R, function F is a solution to the differential equation y' = f on I if F is differentiable, F'(x) = f(x) for all x in I, meaning F is a primitive of f on I.
• Solving y' = f on I means finding all differentiable functions F on I such that F' = f.
• Example: x → 3x² - x, x → 3x² - x + 2 are primitives of x → 6x - 1 on R and are solutions to the differential equation y' = 6x - 1.

Properties of Continuous Functions on an Interval

• Every function continuous on an interval I admits primitives on I.
• If F is a primitive of a function f on an interval I, then the set of all primitives of f on I is defined by G(x) = F(x) + C, where C is in R.
• There exists a primitive of G on I that satisfies G(xo) = yo, where xo is a real number in I, and yo is a given real number.

Demonstrations

• If F is a primitive of f, G(x) = F(x) + C (C ∈ R), so G'(x) = F'(x) + 0 = f(x).
• Conversely, if G is another primitive of f on I, F - G is differentiable with derivative F'(x) - G'(x) = f(x) - f(x) = 0, implying F - G is constant.
• Given F - G = C, where C is a real number, F(x) = G(x) + C for all x ∈ I.
• G(xo) = yo if and only if F(xo) + C = yo, implying C = yo - F(xo); thus G(x) = F(x) + yo - F(xo), ensuring a unique function.

Additional Notes and Primitive Table

• Condition G(xo) = yo is called initial condition in physical sciences.
• The graphs of the primitives of f are derived from each other by translations of vector kj, with k as a real number.

Common Primitives

• f(x) = a, then F(x) = ax on R.
• f(x) = xⁿ (n ∈ Z*), then F(x) = (1/(n+1))x^(n+1) on R if n > 0, or on ]-∞; 0[ or ]0; +∞[ if n < -1.
• f(x) = 1/x, then F(x) = ln(x) on ]0; +∞[.
• f(x) = e^x, then F(x) = e^x on R.
• f(x) = 1/√x, then F(x) = 2√x on ]0; +∞[.
• f(x) = cos(x), then F(x) = sin(x) on R.
• f(x) = sin(x), then F(x) = -cos(x) on R.

Primitives and Operations Table

• If f and g have primitives F and G, respectively, on an interval I, and k is a real number:
• For f + g, a primitive is F + G on I.
• For k × f, a primitive is k × F on I.

Composite function primitives

• For a differentiable function u on an interval I:
• If f has the form u'eu, then primitive is eu on I.
• If f has the form (u' x uⁿ) where n ∈ Z, n ≠ 1, then primitive is (uⁿ⁺¹)/(n+1) on I.
• If f has the form u'/2√u with u(x) > 0, then primitive is √u on I.
• If f has the form u’/u with u(x) > 0, primitive is ln(u) on I.
• If f has the form u' x (v'ou), then primitive is vou.

Applying Primitives

• Determine the primitives on R for f(x) = 2x(x² + 1)⁴.
• Setting u(x) = (x² + 1), u'(x) = 2x and f(x) = u'(x) × (u(x))⁴.
• The primitives of f are F(x) = (1/5)(x² + 1)⁵ + C, with C ∈ R.
• Determine the primitives on R for f(x) = (e^(3x+2))/(e^(3x+2) + 1)
• Setting u(x) = e^(3x+2) + 1, u'(x) = 3e^(3x+2).
• Then f(x) = (1/3) × (u'(x)/u(x)), and F(x) = (⅓)ln(e^(3x+2) + 1) + C, with C ∈ R.

Further Application of Primitives

• Determine primitive of g(x) = 2e^(-5x/4) given G(0) = 3.
• Setting u(x) = (-5/4)x, u'(x) = -5/4, so g(x) = 2 × (-8/5) × u'(x)eu(x)
• Primitives are G(x) = (-8/5)e^(-5x/4)+ C, with C ∈ R.
• If G(0) = 3, G(0) = (-8/5) + C = 3, so C = 3 + (8/5) = 23/5, thus, G(x) = (-8/5)e^(-5x/4) + 23/5.

Solving the Rugby Ball Trajectory Problem

• Initial Conditions
• Vx(t) = 0, Vy(t) = -g
• Horizontal velocity component: Vx(t) = 0 has solutions of the form Vx(t) = k.
  • At t = 0, Vx(0) = Vo * cos(π/4) = 21√2 / 2 = k. Therefore Vx(t) = 21√2 / 2
• If v(t) = x'(t), the horizontal position is x(t) = (21√2 / 2)t + C with C from R.
  • Since x(0) = 0 for initial position, C = 0, so x(t) = (21√2 / 2)t for t ∈ [0 ; +∞[.
• Vertical velocity component: Vy(t) = -g leads to solutions of the form Vy(t) = -gt + C₁ with C₁ ∈ R.
  • At t = 0, Vy(0) = Vo * sin(π/4) = 21√2 / 2. Therefore 0 + C₁ = 21√2 / 2, Thus Vy(t) = -gt + 21√2 / 2
• Ball's vertical position satisfies y'(t) = Vy(t) leading to solutions of the form y(t) = -g(t²/2) + (21√2 / 2)t + C₂ with C₂ ∈ R.
  • With y(0) = 0 meaning 0 + 0 + C₂ = 0 so C₂ = 0, so, y(t) = -g(t²/2) + (21√2 / 2)t
• Overall Trajectory Equations
  • x(t) = (21√2 / 2)t. y(t) = -(g*t²)/2 + (21√2 / 2)t (for t ∈ [0; +∞[)

Determining if the Penalty Kick is Successful

• Pass level of the crossbar, x(t) = (21√2 / 2)t = 40, then t = (40*2)/21√2
• Subbing derived t value y(t) = -(g*(40√2 / 21)²) / 2 + (21√2 / 2)*(40√2 / 21)
• With g = 9.81 m/s² , then y(t) ≈ 4.41 m.
• The ball is approximately 4.41 m high when it passes the plane of the crossbar, so height with this is around 3m, therefore, kick is successful.

Differential Equations of 1st order

• Homogeneous equation y' = ay
• Solutions to the diff equation y' = ay, are the functions defined R by y:x = Ceax where C is a real number
• y' = -y -> has -1 in the equation therefore solution y(x) = Ce-x with C in R
• y’ = 2y -> has 2 in equation therefore solution y(x) = Ce2x with C in R
• 3y’ - y = o, 1/3, then y = 1/3, therefore has solution y(x) = Ce1/3x with C in R

Initial Condition Considerations and RC Circuits

• The particular condition, y(x₀) = k, is used to find the specific solution.
• For a capacitor initially charged, the equation u(t) = -RCu'(t) holds, relating voltage (u) to current (i).
• Given R = 10kΩ and C = 10³µF, u(t) satisfies y' = -(1/RC)y, with solutions u(t) = Ke^(Rt/C).
• Given the 10^3 µF initial voltage of 4V: u(0) = 4,
• Find u(x) = 4e^(-0.1t)

Solutions

• If f and g are solutions of diff equation (E): y’ = ay
• Then f + g and k x f ( k in R) are solutions to the diff equation (E)
• Equation of the form y’ = ay + b
• Set of solutions to the diff equation are, y(x) = Ceax -b/a where C in R.
• Note, a constant function -b/a is a particular solution of equation y’ ay + b.,
• To solve solutions add a particular homogenous solutions together
• If you want a real number add the number
• Example
• Parachutist with the mass m velocity
• Parachutist with the mass m velocity with the g = 9.81 m/s^2
• Parachutist with the mass m velocity with the K = 180 SI and an initial velocity of zero
• Find V(t):
• V’(t) = -(k/m)v(t+g) = v’ = ay+b
• V(t) = Ce^(-kt/m) + (gm/k)
• So, V(t) = ((gm)/k){ 1- e (kt/M)}

Equation of the Form y' = ay + f and Associated Properties

• Consider solutions to the form: y(x) = Ce^ax + g(x); with contansts
• Consider the differential equations where the curve intersects
• y = (x+1) *ex - Therefore = y – 2y = ex( _x2) - 2x (-x1) ex
• Since y’ = 2y, therefore function g has a particular solution
• Function g is a particular solution If y' - 2y =o and then = to 2 where a = 2: equation equation should be: F(x) = Ce2x+ (x+1) ex
• For one to look for that fits equation 3 =4