Higher-order differential equations

[jackalope1234]

Homework Statement

y'' + 2y' +y = 5e^6x

use undetermined coefficients to solve this equation.

Homework Equations

The Attempt at a Solution

what I would normally do is

m^2 + 2m + 1 = 0

but then I get stuck I can't find the general equation as this won't factor. I tried several different methods so I'm just wondering what everyone else would do. After I get the general it's easy but I'm stuck at this point. I have several other problems similar to this making me think there is some method I can't remember for fixing this but I can't seem to find it.

 

[Mute]

Do you remember the quadratic equation?

if ax^2 + bx + c = 0,

x = \frac{1}{2a}\left(-b \pm \sqrt{b^2 - 4ac}\right)

Use that to get the values of m you need for your homogeneous solution.

 

[jackalope1234]

can I leave a negative within the squareroot sign and that would give me my m1 and m2? I'm also just worried about this for later on I can see it getting very messing when i Have to differentiate and integrate these for the other questions in my homework.

 

[Mark44]

Homework Statement

y'' + 2y' +y = 5e^6x

use undetermined coefficients to solve this equation.

Homework Equations

The Attempt at a Solution

what I would normally do is

m^2 + 2m + 1 = 0

but then I get stuck I can't find the general equation as this won't factor.

Sure it factors! This is a perfect square trinomial, and you shouldn't need to use the Quadratic Formula. Give it another try.

I tried several different methods so I'm just wondering what everyone else would do. After I get the general it's easy but I'm stuck at this point. I have several other problems similar to this making me think there is some method I can't remember for fixing this but I can't seem to find it.

 

[jackalope1234]

sorry that I misplaced the 2 the equation should read
2m^2 + 2m + 1

 

[Mark44]

Well, that makes a difference. Your characteristic values are going to be complex.

 

[jackalope1234]

so something like
-2/4 + isqrt(4)/4 and -2/4 - isqrt(4)/4

I have a similar problem when I am doing variation of parameter using this same one as I have to use the Wronskian method which I don't know if it would work.

 

[Mark44]

No, you're way off. I would advise you to review using the Quadratic Formula, especially when the roots are complex. Here's a link to a site that might be helpful: http://www.purplemath.com/modules/quadform.htm. Look at all three modules.

 

[jackalope1234]

-2/4 + or - sqrt(4)i/4 is the answer I got after reviewing those pages. Also wondering if you have a good page on complex differentiation and integrating?

 

[Mark44]

That's correct, but should be simplified. The solutions to 2m² + 2m + 1 = 0 are m = (-1/2) +/- i/2.

This means that two solutions to the homogeneous problem are e^{(-1/2 + i/2)x} and e^{(-1/2 - i/2)x}.

These are a bit inconvenient to work with, so the usual trick is to break them up into products like this:
u₁ = e^-x/2e^{i/2 * x} and u₂ = e^-x/2e^{-i/2 * x}

By choosing suitable linear combinations of these you can eventually arrive at
y₁ = e^x/2sin(x/2) and
y₂ = e^x/2sin(x/2)

Next, you need to find a solution of the nonhomogeneous problem
2y'' + 2y' + y = 5e^6x