# Finding a diff. eq.

1. Sep 1, 2007

### kasse

1. The problem statement, all variables and given/known data

Find a differential eq. y''+ay'+by=0 with this basis: exp(-1-i)x , exp(-1+i)x

2. The attempt at a solution

I see from this that the solutions of the characteristic eq. is two complex conjugates, and that a^2-4b=-4.

Substitution gives me the eq. -2i-a+ai+b=0, and combining these two gives me

a=(1/2) v (1/2)-i

Then I calcutale b when a=(1/2)

b=(17/16)

So that the eq. is

y''+0.5y'+(17/16)y=0

Looks quite nice, but when I transform the bases (exp(-x)*sinx and exp(-x)*cosx) and substitute, it proves not to be correct.

What's my mistake?

2. Sep 1, 2007

### learningphysics

Can you explain how you got the above equation?

3. Sep 1, 2007

### kasse

The roots of the characteristic eq. are 0.5(-a (+/-) sqrt(a^2-4c))

Since the two solutions are complex numbers, a^2-4c must be negative, more precisely -4, since the imaginary part of the complex solutions is (+/-) 1.

4. Sep 1, 2007

### learningphysics

I see... you must have made a mistake when you solved the two equations... I'd just take the sum and product of the roots... then you can write out the equation:

s^2 - qs + p... where q is the sum of the roots and p is the product...

then you get a = -q and b = p for the differential equation

or you can just multiply out:

(s - (-1 - i))(s - (-1 + i))

5. Sep 1, 2007

### kasse

Ah, so I use the superposition principle and substitute the sum of Y1 and Y2 into the original eq?

6. Sep 1, 2007

### learningphysics

No, I'm not using the superposition principle here. Since you know that the basis is exp(-1-i)x , exp(-1+i)x... you immediately know that the characteristic equation has roots (-1-i) and (-1+i).

or another way to think about it... you know that the characterstic equation is:
s^2 + as + b

This means that any function of the form e^sx that satisfies the equation must have the property that s^2 + as + b =0... ie s=(-1-i) must satisfy this equation, and s = (-1+i) must satisfy this equation... ie these are the two roots of the equation... Once you know the roots of a degree-2 equation r1 and r2... then you immediately know the form of the equation... s^2 - (r1+r2)s + r1r2... which you can also get by (s-r1)(s-r2)

You can also plug in -1-i into the equation and -1+i into the equation, then get 2 equations in 2 unknowns (a and b) and solve... but this is a little more roundabout way to do the problem I think...