Differential Equations Plotting scalar functions of complex

Differential Equations

Plotting scalar functions of complex variables Given the function f(z) = 1/z^2 + 1 with complex variable input z = x + jy, create a contour plot in Matlab/Octave that has Re(z) as the x-axis, Im(z) as the y-axis, and |f(z)| as the z-axis. Plot it for the range - 1

Solution

1.(D+1)X=sint given intial values x(0)=-1

f(d)x=0, then the auxilary equation is f(m)=0

i.e m+1=0 , then m=-1

the complementary function is x=c₁e^-t

the complementary function is given by

x_p=1/f(d) *Q(t)

= 1/(D+1)sint.

rationalise the denometor , we get

=(D-1)/D^2-1.sint

put D^2=-1^2, then

x_p=(-1/2)sint

the general solution is given by

x=x_c+x_p

x=c₁e^-t+(-1/2)sint where c₁ is a arbitary constant

at t=0 and x=-1 , then c₁=-1/2

x=-(1/2)e^-t+(-1/2)sint

(2) 1/3Dx+x=0, then (D+3)X=0,

the auxilary equation is f(m)=0,

m+3=0, then m=-3,then x_c=c₁e^-3t

given initial values t=0 and x=1/3, c₁=1/3

the general solution is x=1/3e^1/3t

(6).given differential equation

(D²⁺2D+1)x=e^-2t

the auxilary equation is given by f(m)=0

m²+2m+1=0, then m=-1,-1

x_c=(c₁+c₂t)e^-t, then x_p=1/(D²⁺2D+1).e^-2t

^{put D=-2, then} x_p=1/(D²⁺2D+1),

x_p=e^-2t,

the general solution is given by

x=(c₁+c₂t)e^-t+e^{-2t -------------(1)}

given initial values x(0)=1 and x¹(0)=1

put t=0 and x=1 in equation , we get c₁=0 and differencating equation (1) w.r.t t and substituting t=0 and x=1 , we get c₂=3,

the general solution is given by

x=3te^-t+e^-2t

(7) Given differential equation is

(D²+4D)=17cost

the auxilary equation is f(m)=0

m²+4m=0, then m=0 ,and m=-4

x_c=c₁+c₂e^-4t, the x_p=1/(D²+4D).17cost,

put D²=-1²

^{X_P=-1/17[-68sint+17cost]}

^{the general solution is given by}

x=c₁+c₂e^-4t+1/17[-68sint+17cost]