Second-Order Differential Equations.ppt

Differential Equations Chapter 3 Second-Order Differential Equations §1 Particle Mechanics §2 Linear Equations with Constant Coefficients §3 The Nonhomogeneous Equation §4 Variable Coefficients §5 Boundary Value Problems and Heat Flow §6 Higher-Order Equations §1 Particle Mechanics Some second-order differential equations can be reduced essentially to a single first-order equation that can be handled by methods from Chapter 2. The general form of Newton’s law is where x = x(t) is the displacement from equilibrium. Make the velocity substitution y = x’ to obtain which is a first-order DE (1) The force does not depend on the position x Once the velocity y = y(t) is found, It yields Example 1 DE: Solution: Letting y’=p, we obtain namely, Therefore the general solution is Introduce y = x’ , Then (2) The force does not depend explicitly on time t, which is a first-order differential equation for the velocity y in terms of the position x. If we solve this equation to obtain y = y(x), then we can recover x(t) by solving the equation x = y(x) by separation of variables. It yields Example 2 DE: Solution: Letting y’=p, we have namely, Therefore the general solution is we obtain Introduce y = x’ , Then (3) F is a conservative force, i.e. , F depends only on the position x which is a separable equation integrate both sides with respect to x to get or Note that the left side is the kinetic energy Use the symbol E for the constant of integration because it must have dimensions of energy Potential energy function V (x) is defined by Then which is the energy conservation theorem: the kinetic plus potential energy for a conservative system is constant. It may be recast into This equation is separable, and its solution would give x = x(t). EXAMPLE 3 Consider a spring-mass system without damping Solution The governing equation is where k is the spring constant The force is ?kx We have picked the constant of integration to be zero, i.e., V (0) = 0 Th