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# Second Order Differential Eqn. watch

1. http://www.mediafire.com/view/?grhasyd8ilzemsw

I'm not completely sure what I'm being asked to do in the first part.

This is what I've done :

I let y = e^(mx) and the substitute into the given equation to get a quadratic:

m2-bm+4 Since b<0 it must be negative

I then complete the square to obtain two solutions :

The problem that I now face is how do I express the general solution. Depending on the value of b^2 in the square root I may have something of the form:

y=Acoswx + Bsinwx

or

y=Aex+Bex
2. The critical case is when the quadratic has a repeated root. For which value negative of b does this occur? You should also know that when the root is repeated (call it r), the solution to the second-order linear homogenous ODE is (A+Bx)exp(rx).
3. (Original post by Ari Ben Canaan)
http://www.mediafire.com/view/?grhasyd8ilzemsw

I'm not completely sure what I'm being asked to do in the first part.

This is what I've done :

I let y = e^(mx) and the substitute into the given equation to get a quadratic:

m2-bm+4 Since b<0 it must be negative

I then complete the square to obtain two solutions :

The problem that I now face is how do I express the general solution. Depending on the value of b^2 in the square root I may have something of the form:

y=Acoswx + Bsinwx

or

y=Aex+Bex
"critical case" means double root.

to have real, equal roots.

If

is the value for which the quadratic holds, then it is the repeated root so

For the second part, try

You could (and I would) use complex variables to make the algebra less messy.
4. (Original post by Lord of the Flies)
The critical case is when the quadratic has a repeated root. For which value negative of b does this occur? You should also know that when the root is repeated (call it r), the solution to the second-order linear homogenous ODE is (A+Bx)exp(rx).
Ah, I see. I never knew that referred to a repeated root case. So when b^2 = 4ac b^2 = 16 Hence, my answer to the quadratic simplifies to :

(16)^0.5 / 2 = ±2 However, we are told b<0 hence the general solution is :

y = (A+Bx)e^2x

Is that correct ?

Thank you everyone ! +rep to all.
5. (Original post by Ari Ben Canaan)
Ah, I see. I never knew that referred to a repeated root case. So when b^2 = 4ac b^2 = 16 Hence, my answer to the quadratic simplifies to :

(16)^0.5 / 2 = ±2 However, we are told b<0 hence the general solution is :

y = (A+Bx)e^2x

Is that correct ?

Thank you everyone ! +rep to all.
Yes

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