Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Factorization wikipedia , lookup
History of algebra wikipedia , lookup
System of polynomial equations wikipedia , lookup
Fundamental theorem of algebra wikipedia , lookup
System of linear equations wikipedia , lookup
Elementary algebra wikipedia , lookup
Quadratic equation wikipedia , lookup
Quartic function wikipedia , lookup
SECOND ORDER LINEAR Des WITH CONSTANT COEFFICIENTS Second order homogenous linear differential equation with constant coefficients The general formula for such equation is To solve this equation we assume the solution in the form of exponential function: If y e x then y' e and the equation will change into x ay' 'by 'cy 0 ye x y' ' 2 e x and a2 e x b e x ce x 0 x e (a b c) 0 after dividing by the eλx we obtain We obtained a quadratic characteristic equation. The roots are y ( x) ? 2 a 2 b c 0 b b 2 4ac 12 2a There exist three types of solutions according to the discriminant D D b 2 4ac 1) If D>0, the roots λ1, λ2 are real and distinct y C1e1x C2e2 x 2) If D=0, the roots are real and identical λ12 =λ y C1ex C2 xex 3) If D<0, the roots are complex conjugate λ1, λ2 where α and ω are real and imaginary parts of the root 1 i 2 i y K1e1x K 2e2 x K1e x i x K 2e x i x i x x i x i x e cos x i sin x y e ( K1e K 2e ) Eulers formula x y e [( K1 K 2 ) cos x i( K1 K 2 ) sin x] If we substitute we obtain C1 ( K1 K 2 ); C2 i ( K1 K 2 ) y( x) e x [C1 cos x C2 sin x] This is general solution in some cases, but … Further substitution is sometimes used C1 A sin ; C2 A cos and then y( x) e x [ A sin cos x A cos sin x] considering formula sin( ) sin cos cos sin we finally obtain y( x) e x A sin( x ) where amplitude A and phase φ are constants which can be obtained from initial conditions and ω is angular frequency. This example leads to an oscillatory motion. Example of the second order LDE – a simple harmonic oscillator Evaluate the displacement x(t) of a body of mass m on a horizontal spring with spring constant k. There are no passive resistances. If the body is displaced from its equilibrium position (x=0), it experiences a restoring force F, proportional to the displacement x: From the second Newtons law of motion we know m x kx x F k x d 2x F ma m 2 m x dt k x0 m We have two complex conjugate roots with no real part Characteristic equation is k 12 i m k 0 m 2 x(t ) e t A sin( t ) The general solution for our symbols is No real part of λ means α=0, and omega in our case The final general solution of this example is k m x(t ) A sin( t ) Answer: the body performs simple harmonic motion with amplitude A and phase φ. We need two initial conditions for determination of these constants. x (0) 0 x(0) 2 These conditions can be for example From the first condition From the second condition The particular solution is A cos(0 ) 0 cos 0 2 A sin( 0 ) 2 2 x(t ) 2 sin( t 2 ) A2 x(t ) 2 cos( t ) Example 2 of the second order LDE – a damped harmonic oscillator The basic theory is the same like in case of the simple harmonic oscillator, but this time we take into account also damping. The damping is represented by the frictional force Ff, which is proportional to the velocity v. The total force acting on the body is F ma m x x c k x x 0 m m x 2 x 2 x 0 F f c v c dx cx dt F kx F f kx c x m x kx cx m x cx kx 0 The following substitutions are commonly used Characteristic equation is k c ; 2 m m 2 2 2 0 2 2 2 4 4 Solution of the characteristic 12 2 2 equation 2 where δ is damping constant and ω is angular frequency There are three basic solutions according to the δ and ω. 1) δ>ω. Overdamped oscillator. The roots are real and distinct x(t ) C1e1t C2e2t 2) δ=ω. Critical damping. The roots are real and identical. 1 2 2 2 2 2 12 x(t ) C1e t C2te t 3) δ<ω. Underdamped oscillator. The roots are complex conjugate. x(t ) Ae t sin( ' t ) 1 i 2 2 i ' 2 i 2 2 i ' Damped harmonic oscillator in the Mathematica All three basic solutions together for ω=10 s-1 Overdamped oscillator, δ=20 s-1 Critically damped oscillator, δ=10 s-1 Underdamped oscillator, δ=1 s-1