Express the general solution of x'=(3, 4, -2, -1)x in terms of real-valued functions. (this is 2x2 matrix, 3 and 4 on the left, -2 and -1 on the right.)

Sun,

As you will have noticed, the characteristic polynomial,

λ²-2λ+5=0,

will give you two complex conjugate eigenvalues, λ=1±2i, so the corresponding eigenvectors,

x

_{1,2}=[1±i,2],are also complex.

You could write the general solution as the complex-valued vector function

x=c

_{1}[1+i,2]e^{(1+2i)t}+c_{2}[1-i,2]e^{(1-2i)t}and leave it at that, but I suspect you are supposed to give the final answer in real form. For this, you would break the exponentials into a real and complex part,

e

^{(1±2i)t}=e^{t}e^{±2it},and use Euler's identity,

e

^{±iθ}=cosθ ± i sinθon the complex exponential to get

e

^{(1±2i)t}=e^{t}(cos (2t) ± i sin (2t) ),so that your solution becomes

x=c

_{1}[1+i,2]e^{t}(cos (2t) +i sin (2t) ) + c_{2}[1-i,2] e^{t}(cos (2t) - i sin (2t) ).Finally, you can collect the real and imaginary parts of this expression and redefine the c's so that everything is real (no more i's). I'm not sure if you are supposed to go this far, so I leave this part to you. :)

## Comments

_{1}e^{t}(cos(2t), cos(2t)+sin(2t))+c_{2}e^{t}(sin(2t), -cos(2t)+sin(2t)) is the answer. Where cos(2t) and sin(2t) are at the top, and cos(2t)+sin(2t) and -cos(2t)+sin(2t) are at the bottom._{1}(e^{αt}cos(βt) a - e^{αt}sin(βt) b) + c_{2}(e^{αt}sin(βt) a + e^{αt}cos(βt) b)._{1}(e^{t}cos(2t) [1,2] - e^{t}sin(2t) [1,0]) + c_{2}(e^{t }sin(2t) [1,2] + e^{t}cos(2t) [1,0])._{1}and c_{2}switched, and top and bottom switched.