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Linear Constant Coefficients Equations (§ 1.1) I Overview of Differential Equations I Linear Ordinary Differential Equations I Integrating a Differential Equation Linear Constant Coefficients Equations I I I The Integrating Factor Method The Initial Value Problem Linear Constant Coefficients Equations (§ 1.1) I Overview of Differential Equations I Linear Ordinary Differential Equations I Integrating a Differential Equation Linear Constant Coefficients Equations I I I The Integrating Factor Method The Initial Value Problem Overview of Differential Equations Definition A differential equation is an equation, the unknown is a function, and both the function and its derivatives appear in the equation. Example (Newton’s Second Law of Motion) Mass × Acceleration = Force: m a = f . The unknown is x(t), the particle position as function of time t and the equation is d2 1 x(t) = f (t, x(t)), dt 2 m with m the particle mass and f the force acting on the particle. Remark: This is an Ordinary Differential Equations (ODE): Derivatives with respect to only one variable appear in the equation. Overview of Differential Equations Definition A differential equation is an equation, the unknown is a function, and both the function and its derivatives appear in the equation. Example (Wave Equation) Sound propagation in the air is described by the wave equation 2 ∂2 2 ∂ u(t, x) = v u(t, x), ∂t 2 ∂x 2 where function u represents air pressure and v the wave speed. The unknown is u, which depends on two variables (t, x). Remark: This is a Partial Differential Equations (PDE): Partial derivatives of two or more variables appear in the equation. Overview of Differential Equations Remark: Differential equations are a central part in a physical description of nature: I Classical Mechanics: I I I Fluid Dynamics: I I Einstein equation. (PDE) Quantum Mechanics: I I Maxwell’s equations. (PDE) General Relativity: I I Navier-Stokes equation. (PDE) Electromagnetism: I I Newton second law of motion. (ODE) Euler-Lagrange equations. (ODE) Schrödinger’s equation. (PDE) Standard Model of Particle Physics. (PDE) Linear Constant Coefficients Equations (§ 1.1) I Overview of Differential Equations I Linear Ordinary Differential Equations I Integrating a Differential Equation Linear Constant Coefficients Equations I I I The Integrating Factor Method The Initial Value Problem Linear Ordinary Differential Equations Remark: We use the notation y0 = dy , dt y 0 (t) = dy (t). dt Definition An ODE determined by a functions f of two variables, with unknown function y , is the equation y 0 (t) = f (t, y (t)). This equation is called linear iff there exist functions a, b such that f (t, y ) = a(t) y + b(t). That is, f is linear on its argument y , hence a first order linear ODE is given by y 0 (t) = a(t) y (t) + b(t). Linear Ordinary Differential Equations Example A first order linear ODE is given by y 0 (t) = −2 y (t) + 3. Here a(t) = −2 and b(t) = 3. Since these functions do not depend on t, the equation is called of constant coefficients or autonomous. Example A first order linear ODE is given by 2 y 0 (t) = − y (t) + 4t. t Here a(t) = −2/t and b(t) = 4t. Since these functions depend on t, the equation is called of variable coefficients or nonautonomous. Linear Constant Coefficients Equations (§ 1.1) I Overview of Differential equations I Linear Ordinary Differential Equations I Integrating a Differential Equation Linear Constant Coefficients Equations I I I The Integrating Factor Method The Initial Value Problem Integrating a Differential Equation Remarks: I To integrate a differential equation y 0 = ay + b is to find all functions y such that y 0 is equal to ay + b. I Integrating on both sides of the equations fails. For example, integrate on t on a constant coefficients equation: Z Z Z 0 y (t) dt = ay (t) + b dt ⇒ y (t) = a y (t) dt + bt + c where c ∈ R is any integration constant. Remarks: I We have R not solved the equation, because we do not know what y dt actually is. I We need a new idea to integrate the equation. Linear Constant Coefficients Equations (§ 1.1) I Overview of Differential equations I Linear Ordinary Differential Equations I Integrating a Differential Equation Linear Constant Coefficients Equations I I I The integrating Factor Method The Initial Value Problem The Integrating Factor Method New idea: Transform the differential equation y 0 = ay + b into the total derivative of a potential function ψ depending on (t, y ), y 0 = ay + b ⇔ ψ 0 = 0. The last equation is easy to integrate: ψ(t, y ) = c, for any c ∈ R. How can the potential function ψ be obtained? Multiplying the differential equation y 0 = ay + b by a nonzero function, called the integrating factor. Example Find all functions y solution of the ODE y 0 = 2y + 3. Solution: Write down the differential equation as y 0 − 2 y = 3. Key idea: The left hand side above is a total derivative if we multiply it by an exponential, e −2t (y 0 − 2 y ) = 3e −2t . The Integrating Factor Method Example Find all functions y solution of the ODE y 0 = 2y + 3. Solution: Recall: e −2t (y 0 − 2 y ) = 3e −2t . 0 e −2t y 0 − 2 e −2t y = 3 e −2t ⇔ e −2t y 0 + e −2t y = 3 e −2t . 0 The derivative of a product implies e −2t y = 3 e −2t , hence h −2t 0 h 3 −2t i0 3 −2t i0 e y = − e ⇔ y+ e = 0. 2 2 So e −2t is an Integrating Factor. A Potential Function is 3 −2t and the equation y 0 = 2y + 3 is: ψ 0 = 0. ψ= y+ e 2 3 −2t 3 Integrating, ψ = c ⇒ y + e = c ⇒ y (t) = c e 2t − . C 2 2 The Integrating Factor Method Example Find all functions y solution of the ODE y 0 = 2y + 3. Solution: We concluded that the ODE has infinitely many solutions, given by y c>0 3 y (t) = c e 2t − , 2 c ∈ R. Since we did one integration, it is reasonable that the solution contains a constant of integration, c ∈ R. − 32 Verification: y 0 = 2c e 2t , but we know that 2y + 3 = 2c e 2t , therefore we conclude that y satisfies the ODE y 0 = 2y + 3. t c=0 c<0 C The Integrating Factor Method Remark: Solutions to first order linear ODE can be obtained with the Integrating Factor Method. Theorem (Constant Coefficients) If the constants a, b ∈ R satisfy a 6= 0, then the linear equation y 0 (t) = a y (t) + b has infinitely many solutions, one for each value of c ∈ R, given by b y (t) = c e at − . a Remarks: I This expression for y is called the General Solution. I Here we present the main idea of the proof using an exponential integrating factor. I In the Lecture Notes we show that this is essentially the only integrating factor. The Integrating Factor Method Main ideas of the Proof: Write down the differential equation as y 0 − a y = b. Key idea: The left-hand side above is a total derivative if we multiply it by an exponential e −at (y 0 − a y ) = b e −at . 0 e −at y 0 − a e −at y = b e −at ⇔ e −at y 0 + e −at y = b e −at . 0 The derivative of a product implies e −at y = b e −at , hence h −at 0 h b −at i0 b −at i0 e y = − e ⇔ y+ e = 0. a a b −at Introducing a potential function ψ = y + e , the equation a y 0 = ay + b is now ψ 0 = 0. We can now integrate the equation, b −at b ψ=c ⇔ y+ e = c ⇔ y (t) = c e at − . a a Linear Constant Coefficients Equations (§ 1.1) I Overview of Differential Equations I Linear Ordinary Differential Equations I Integrating a Differential Equation Linear Constant Coefficients Equations I I I The Integrating Factor Method The Initial Value Problem The Initial Value Problem Definition The Initial Value Problem (IVP) for a linear ODE is the following: Given functions a, b and a constant y0 ∈ R, find a solution y of y 0 = a(t) y + b(t), y (0) = y0 . Remark: The initial condition selects one solution of the ODE. Theorem (IVP Constant Coefficients) Given constants a, b, y0 ∈ R, with a 6= 0, the initial value problem y 0 = a y + b, y (0) = y0 has the unique solution b at b y (t) = y0 + e − . a a The Initial Value Problem Example Find the solution to the initial value problem y 0 = 2y + 3, y (0) = 1. Solution: Every solution of the ODE above is given by 3 y (t) = c e 2t − , 2 c ∈ R. The initial condition y (0) = 1 selects only one solution: 1 = y (0) = c − We conclude that y (t) = 3 2 ⇒ 5 c= . 2 5 2t 3 e − . 2 2 The Initial Value Problem Example Find the solution y to the IVP y 0 = −3y + 1, y (0) = 1. Solution: Write down the differential equation as y 0 + 3 y = 1. Key idea: The left-hand side above is a total derivative if we multiply it by the exponential e 3t . Indeed, 0 e 3t y 0 + 3 e 3t y = e 3t ⇔ e 3t y 0 + e 3t y = e 3t . This is the key idea, because the derivative of a product implies 3t 0 e y = e 3t . The exponential e 3t is called an integrating factor. Integrating, 1 1 e 3t y = e 3t + c ⇔ y (t) = c e −3t + . 3 3 C The Initial Value Problem Example Find the solution y to the IVP y 0 = −3y + 1, y (0) = 1. Solution: Every solution of the ODE above is given by 1 y (t) = c e −3t + , 3 c ∈ R. The initial condition y (0) = 1 selects only one solution: 1 = y (0) = c + We conclude that y (t) = 1 3 2 −3t 1 e + . 3 3 ⇒ 2 c= . 3 C