# *Urgent* Differential Equations

• Nov 21st 2008, 04:14 AM
gidget
*Urgent* Differential Equations
Need to find the general solution of:
y" + (1+2i)y' + (i-1)y = 0
• Nov 21st 2008, 06:52 AM
David24
Quote:

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
• Nov 21st 2008, 06:53 AM
gidget
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 AM
David24
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 AM
gidget
D\$\displaystyle ^2\$ + (1+2i)D + i -1 =0

This is what was also given to us.
• Nov 21st 2008, 07:46 AM
David24
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)