Need to find the general solution of:
y" + (1+2i)y' + (i-1)y = 0
There are numerous methods to solve 2nd Order Linear DE's, has your teacher specified a method to use? i.e. Method of Un-Determined Coefficients, Linear D operators (Forming an Auxillary eqn etc) ?
If none is specified, for simplicity I would recommend the Linear D operator technique,
If you let me know what method your meant to use, I will post a solution using that method,
Regards,
David
Okay, just for complete clarrification, in your notes does he convert the above D into the form
(D+ a)(D + b)y = 0 to solve ??
and then solve from here using the properties of the linear D operator or does your teacher then further impose that y must take the form Ae^mx ? and solve that way?
Sorry, just want to get it 100% clear before I go ahead with the solution,
Regards,
David
ahh okay cool
Basically what your teacher is saying to do is to form what is known as the auxillary eqn,
here you have a linear DE of the form
a* D^2y + b*Dy + cy = 0
or
(a*D^2 + b*D + c)y = 0
let y = Ae^(mx)
Subsitutuion yeilds,
Ae^(mx)*(a*m^2 + b*m + c) = 0
As Ae^(mx)~= 0 unless A = 0 (which of course would be the trivial solution) we are arrive at the Auxillary equation,
a*m^2 + b*m + c = 0
here a = 1; b = (1 + 2i); c = i - 1
obviously this is a mere quadratic and thus form the quadratic formula is used to solve the Auxillary eqn,
m = (- (1+2i) +- sqrt( (1+2i)^2 - 4(1)(i-1) ))/(2(1))
note (1 + 2i)^2 - 4(i-1) = 4i^2 + 4i + 1 - 4i + 4 = 1
thus,
m = (-(1 + 2i) +- 1)/2
Hence
m1 = i , m2 = -(1+i)
as y = e^(mx) and by employing the superposition principle, the solution takes the form,
y = Ae^(m1*x) + Be^(m2*x); where A, B are any complex values
= Ae^(ix) + Be^(-(1+i)x)
And viola your done!
Hope this helps,
Regards,
David
note - you could put the expression of e^aix into terms of sin(ax), cos(ax) using Eulers Identity - but once again it all depends on what your teacher wants, the solution provided is perfectly correct