Thread: *Urgent* Differential Equations

Need to find the general solution of:
y" + (1+2i)y' + (i-1)y = 0

Originally Posted by gidget

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

yeah he uses linear D operators I just have no clue how to attempt this or where to go with this problem.
thank you!!

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

D + (1+2i)D + i -1 =0

This is what was also given to us.

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