If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Differential equations

Course: Differential equations>Unit 2

Lesson 2: Complex and repeated roots of characteristic equation

Complex roots of the characteristic equations 2

What happens when the characteristic equation has complex roots? Created by Sal Khan.

Want to join the conversation?

• What video, or in what part of the website is the deduction of Euler's formula?
Thank you
• The very famous video which concludes with the solution and sal saying " IF THIS DOESN'T BLOW YOUR MIND, THEN YOU HAVE NO EMOTION". Infact watching that particular video on youtube brought me here.
• I thought C1*i - C2*i = (C1-C2)*i = C*i where C is some arbitrary constant. How did Sal take the i away?
• I assume you are referring to where C1*i - C2*i is turned into C4.
C4 is just a constant that could have any value, even imaginary or complex. Therefore there could be an i hidden inside C4, so the i is not lost, it's just not shown.
• How does this work when you have something greater than a quadratic (order>2)? Because you won't be able to use the quadratic formula and find the Lambda and Mu?
• All else constant, you will end up with a characteristic polynomial of order > 2 . Beyond order 4 is when things become interesting because, in general, exact solutions cannot be found according to any convenient equation like the quadratic formula. :D
Thumbs up!
• why can you just drop the relationship of c3 being the sum of 2 arbitrary #s and c4, their difference? doesnt c4 have to fall within the range -c3<=c4<=c3? I mean the absolute value of c4 is less than or equal to the absolute value of c3.actually the abs value of c4 could be anything, if the two constants are conjugates. but still, why can you drop the realationship?nevermind,I just realized that i answered my own question, if they are conjugates, then c3 is an abitrary real number, depending on choice of rel part, and c4 is also a arbitray real #, because a pure imaginary # times i is a real #.
• From a linear algebra standpoint, if you set it up as a matrix equation:

Ax = b
[ 1 1 ][ c1 ] = [ c3 ]
[ i -i ][ c2 ] = [ c4 ]

A is nonsingular.
(Multiply bottom row2 by i; replace row2 with row2+row1; multiply row2 by 1/2; replace row1 with row1-row2).

Therefore A is also invertible. That means that c1 and c2 can be solved for in terms of c3 and c4, and vice versa. Since c1 and c2 are independent of eac hother, this would imply that there are no dependencies between the variables c3 and c4... I think.
• what if I have an equation with a complex root but not its conjugate root?
(1 vote)
• To explain, any quadratic equation with complex roots is going to have the form -b/2a (the real part) plus or minus (b^2 - 4ac)^(1/2) / 2a (The part that can be imaginary). This means that the roots will always be complex conjugates, because if the roots are complex, the form of the solution is always the real part plus or minus the imaginary part.
• Is any chance, finally after 5 years from first comment about lost imaginary part from particular solution, that you correct this video tutorial and next too.
Someone can say its only particular solution and not full solution of the problem,
or the simply way to avoid complex numbers.
I think its important to show people who interested in second order differential equations that particular solution goes to complex numbers and then one can better
imagine complexity of possible implications.
• at around he forget a paranthesis to multiply the e-term with the rest
• "The standard pronunciation in English has always been something like oy-ler. We don't have a vowel that corresponds to eu in German, but the next best thing to it is oi. I've never known a mathematician to pronounce Euler as yoo-ler."

I have now! :)