Recent content by metalscot

Thanksfor your help Mute I think I understand now

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 -...

Thanks Hootenanny, I have a more complex forcing term to move onto now but with this basis I think I will get there.

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?

the real number is a root, so my solution must be in the form y= e^{at}[/itex]. ?

the real number is a root, so my solution must be in the form y= e^{at}[/itex]. ?

Im sorry ehild I am confused by this, does this mean that my A and B constants are correct making x(t) = cos2t a solution?

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...

Thanks ehild, do I then use x=1 and and t=0 to find constant A and x'=1 and t=0 to find constant B for both questions?