Linear Differential Equation (1st Order)

Transcription

1 Linear Differential Equation (1st Order) First Order Linear Differential Equation is an equation of the form y' Cp y = q For q = 0, this is a homogeneous equation and the solution is in the form of: y = Ce Kp d = Ce Kh ; where h = p d For q s 0, this is a nonhomogeneous equation and the solution is in the form of: y = Ce Kh Ce Kh qe h d ; where h = p d In both cases, we find "C" by specifying 1 initial condition. Summary of steps: Write an ODE in a standard form. Identify p and q Find h = p d and qe h d Find y from y = Ce Kh Ce Kh qe h d Substitute the intial condition = 0 and y= y 0 (this is given) and solve for C Plug C back into y Eample: y'cy = This is a linear nonhomogeneous ODE. p = 1, q = p d = 1 d = y = C$e K Ce K $ $e d = Ce K Ce K e = Ce K C This is a class of solutions. Different C produces different results. plot 6$e K C, 3$e K C, 0$e K C,K6$e K C,K3$e K C, =K1..5

2 K K5 K10 C=0 C=6 C=3 C=-6 C=-3 If we specify intial condition, say y 0 = 5, then y 0 =5 CC$e 0 =5 C =3 Thus, the solution for y'cy = given y 0 =5 is y =3e K C

3 Linear Differential Equation (nd Order) Second Order Linear Differential Equation is an equation of the form y" Cp y'cq y = f with initial conditions y 0 = A, y' 0 Solutions for Homogeneous equation (f =0 The ODE y" Cp y'cq y =0 has two solutions: and y. The general solution is a linear combination of these two solutions. y = C 1 CC The constants C 1 and C can be found from the initial conditions y 0 = A, y' 0 To check whether the two solutions are linearly independent on an interval, we define Wronskian of solutions and y to be: W = y' 1 y y' = y' Ky' 1 y If we can find an point (any one at all!) on the interval that makes W s 0 then we can conclude that and y are linearly independent. In fact, W s 0 4 and y are linearly independent Eample: y" Cy =0 We claim that =cos and y =sin are the solutions of this ODE. W = y' 1 y y' = cos sin Ksin cos =cos Csin =1s 0 r These are linearly independent solutions. The general solution is y = C 1 cos CC sin.

4 Solutions for Nonhomogeneous equation (f s 0 The ODE y" Cp y'cq y = f has a general solution in the form of: y = y h Cy p = C 1 CC Cy p Where y h is the solution of the homogeneous equation (y" Cp y'cq y =0 y p is the particular solution due to f The constants C 1 and C can be found from the initial conditions y 0 = A, y' 0 Solutions for Constant Coefficient Homogeneous Linear Differential Equation The differential equation is in the form of y" CAy'CBy =0 ;A, B = constant The solution of this ODE would be in the from of y =e λ Thus, y'= λe λ and y" = λ e λ Substitute this into the equation to get λ e λ CAλe λ Ce λ =0 λ CAλCB =0 This is called "Characteristic Equation". The roots of this polynomial are λ= KA G A K4 B. There are 3 possible cases: Case 1: A K 4 B O 0 we have λ 1 = KA C A K4 B and λ = KAK A K4 B which both are Real numbers. The solutions are =e λ 1 and y =e λ The general solution is y = C 1 CC = C 1 $e λ 1 CC $e λ

5 Case : A K 4 B =0 we have λ= KA and we have =e λ =e KA We still need another solution that is also linearly independent to the first solution. This is done by assuming another solution in the form of y = $e KA The general solution is y = C 1 CC = C 1 $e KA CC $$e KA =e KA C 1 CC Case 3: A K 4 B! 0 we have comple roots λ 1 = KA C 4 B KA i and λ = KAK 4 B KA i where i = K1 Let p =K A, q = 1 4 B KA λ 1 =p Ciq and λ =p Kiq =e p Ciq and y =e pkiq The general solution is y = C 1 CC = C 1 $e p Ciq CC $e pkiq Alternatively, we can rewrite y in a different form using Euler's formula: e i =cos Ci$sin and e Ki =cos Ki$sin to get y =e p C 3 cos q CC 4 sin q We can also rewrite it as y = C 5 e p cos q Kδ where C 5 is often called "Amplitude" and can be found from C 5 = C 3 CC 4 δ is called "Phase" and δ = C 4 C3