Need to find the general solution of:

y" + (1+2i)y' + (i-1)y = 0

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- Nov 21st 2008, 04:14 AMgidget*Urgent* Differential Equations
Need to find the general solution of:

y" + (1+2*i*)y' + (*i*-1)y = 0 - Nov 21st 2008, 06:52 AMDavid24
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 - Nov 21st 2008, 06:53 AMgidget
yeah he uses linear D operators I just have no clue how to attempt this or where to go with this problem.

thank you!! - Nov 21st 2008, 07:04 AMDavid24
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 - Nov 21st 2008, 07:10 AMgidget
D$\displaystyle ^2$ + (1+2

*i*)D +*i*-1 =0

This is what was also given to us. - Nov 21st 2008, 07:46 AMDavid24
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