# Complex repeated roots for ODEs

• fred2028
In summary, the solution to a complex repeated root in a second order differential equation with complex coefficients would be written in terms of complex numbers, using the form y(t)= Ce^{-it}+ Dte^{-it}. This is different from the solutions for repeated real roots and a single complex root, which involve cosines and sines. Quadratics in real numbers do not have repeated complex roots, but in a fourth order linear differential equation with real constant coefficients, repeated roots can occur if they are complex conjugates. In this case, the solution would involve both sine and cosine terms, multiplied by the independent variable.
fred2028
I know that a 2nd order homo ordinary differential equation's solution is in the form of

$$$f(x) = {C_1}{e^{{a}t}} + {C_2}t{e^{{a}t}}$$$

for repeated real roots of the characteristic equation, and that the solution for a single complex root (and its conjugate) involves a cosine. I'm curious as to what the solution to a complex repeated root would look like?

If you have, say, a fourth order linear differential equation with (real) constant coefficients, then you can have repeated roots. If, say, $a+ bi$ and $a- bi$ are double roots, then the general solution to the differential equation is
$$y(x)= e^{ax}(A cos(bx)+ B sin(bx))+ xe^{ax}(C cos(bx)+ Dsin(bx))$$
that is, you multiply the first solutin by the independent variable just as for real roots.

(By the way, you wrote your solution as "y(x)" but then wrote the independent variable as "t". Surely, your teacher wouldn't let you get away with that on a test!)

You can have "repeated complex roots" to a second order equation if it has complex coefficients. For example, the second order, linear, differential equation with constant coefficients, y"+ 2iy'- y= 0 has characteristic equation $r^2+ 2ir- 1= (r+ i)^2= 0$ and so has r= -i as a double characteristic root. In that case, it would be more common to write the solution in terms of complex numbers. That is, rather than trying to use sine and cosine (which does not work if your roots are not complex conjugate), the general solution would be written $y(t)= Ce^{-it}+ Dte^{-it}$.

## 1. What are complex repeated roots for ODEs?

Complex repeated roots for ODEs refer to a situation where the characteristic equation of a linear differential equation has complex roots that are repeated. This means that the roots of the characteristic equation are complex numbers and they appear more than once.

## 2. How are complex repeated roots different from real repeated roots?

Complex repeated roots differ from real repeated roots in that they involve complex numbers instead of real numbers. Real repeated roots occur when the characteristic equation has real roots that are repeated, while complex repeated roots occur when the characteristic equation has complex roots that are repeated.

## 3. What does it mean for an ODE to have complex repeated roots?

If an ODE has complex repeated roots, it means that the solution to the differential equation will involve complex numbers. This is because the repeated complex roots will result in a general solution containing terms with complex exponentials.

## 4. How can I solve an ODE with complex repeated roots?

To solve an ODE with complex repeated roots, you can use the general solution for repeated roots, which involves terms with complex exponentials. You can also use the method of reduction of order or variation of parameters to find the particular solution for the given initial conditions.

## 5. Why are complex repeated roots important in ODEs?

Complex repeated roots are important in ODEs because they can lead to behaviors in the system that cannot be achieved with real roots. They also provide a way to model systems with oscillatory behavior, which is commonly seen in physical and biological systems.

• Calculus and Beyond Homework Help
Replies
1
Views
496
• Calculus and Beyond Homework Help
Replies
4
Views
426
• Calculus and Beyond Homework Help
Replies
17
Views
2K
• Calculus and Beyond Homework Help
Replies
9
Views
1K
• Calculus and Beyond Homework Help
Replies
8
Views
452
• Calculus and Beyond Homework Help
Replies
3
Views
346
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
16
Views
539
• Calculus and Beyond Homework Help
Replies
6
Views
4K