Using Eulers formula for e^±iθ , obtain the trigometric identities for cos(θ1, θ2) and sin(θ1, θ2)
The above is the only information that I am given
Using eulers formula for both exponentials I get
eiθ1= cosθ1 + isinθ2
e-iθ2=cosθ2-isinθ2
eiθ2=cosθ2+isinθ2
combing i get...
Thank you Mute
e^(θ1±θ2)=cos(θ1±θ2)+isin(θ1±θ2)
Gives the below for +
e^(θ1+θ2)=cos(θ1+θ2)+isin(θ1+θ2)
Using the rule cos(A+B) = cosAcosB - sinAsinB
sin(A+B) = sinAcosB + sinBcosA
I get real part cos(θ1 + θ2) = cosθ1 cosθ2 -...
The question is: Using Eulers formula for e^±iθ , obtain the trigometric identities for cos(θ1, θ2) and sin(θ1, θ2)
I think I have completed the real and imaginary solutions for the base e^+iθ using the real part cos(θ1+θ2)and imaginary isin(θ1+θ2)
Gaining
cos(θ1 + θ2) = cosθ1 cosθ2 -...
Thank you LCKurtz, in this example because I have x''+2x'+x=3
If I plug a value of x=3 into my original equation I end up with x=3 as a final answer so
x(t)particular=3
giving me x(t)general= x(t)complimentary + x(t)particular:
x(t)=Ae^-1t + Bte^-1t + 3 ?
I am having trouble after finding the complimentary solution to this problem, if possible I would like some guidance on how to proceed with the particular solution
Find a general solution to the following equation: x'' + 2x' + x = 3
x'' + 2x' + x = 3
complimentary solution when = 0...
Thank you ehild. I believe I have now managed the other problem, can you confirm the solution to be x(t) = e^2t - 2te^2t
I really appreciate all the help you have given me.
I think ehild I have finally arrived at the correct solution
when x(t) = Ae^2t + Be^-2t
x(0)=1
Ae^2t + Be^-2t=1
and x'(0)=0
2Ae^2t - 2Be^2t =0
so with simultaneous equations A = 1/2 AND B = 1/2
I arrive at
x(t)=1/2 e^2t + 1/2 e^-2t
I hope you can tell me this is correct?
The solutions of the characteristic equation m^2-4=0 are real m=±2.
does the above not mean that -2 and 2 are roots?
So that then m1 and m2 are -2and 2?
Working through my first problem x''-4x=0 ; x(0)=1 x'(0)=0
characteristic equation is m^2-4=0
solutions m=+/-2i
so complex solution (e^2it, e^-2it)
since (cos2t, sin2t)
x(t)=Acos2t-Bsin2t
For x(0)=1
1=A-0
so
A=1
x'(t)= -2Asin2t-2Bcos2t
for x'(0)=0
0=0-2B...