#### Trenters4325

##### Junior Member

- Joined
- Apr 8, 2006

- Messages
- 122

- Thread starter Trenters4325
- Start date

- Joined
- Apr 8, 2006

- Messages
- 122

This seems to work, although i'm not sure if it is a valid derivation. Maybe someone can check this over.Trenters4325 said:

Since m1 = m2 = m

a^2 - 4b = 0

a^2 = 4b

a = 2*SQRT(b)

also, m = -a/2 = -SQRT(b), so b = m^2

y'' - 2my' + m^2y = 0

Using Laplace:

L(y'') - 2m*L(y') + m^2*L(y) = 0

Using basic rules for Laplace: (note now y is a function of s.)

s^2*y - sy(0) - y'(0) -2m(s*y - y(0)) + m^2*y = 0

(s^2-2ms)y - (s + 2m)y(0) - y'(0) = -ym^2

Let C1 = y(0) and C2 = y'(0)...

-(s+2m)C1 - C2 = -ym^2-(s^2-2ms)y

(s+2m)C1 + C2 = (m^2 - 2ms + s^2)y

(s+2m)C1 + C2 = y(m-s)^2

y = [(s+2m)C1 + C2]/(m-s)^2

to get y(x):

y(x) = L<sup>-1</sup>{ ([s+2m]C1 + C2)/(m-s)^2 }

y(x) = C1* L<sup>-1</sup>{ (s+2m)/(s-m)^2 } + C2L<sup>-1</sup>{C2/(s-m)^2}

Note s+2m = s - m + 3m:

L<sup>-1</sup>{ (s+2m)/(s-m)^2 } = L<sup>-1</sup>{ (s-m)/(s-m)^2 + 3m/(s-m)^2}

= e^(mx)*L<sup>-1</sup>{ (s)/(s)^2 } + 3m*e^(mx)*L<sup>-1</sup>{1/(s)^2}

= e^(mx)(L<sup>-1</sup>{1/(s)} + 3m*L<sup>-1</sup>{1/s^2})

= e^(mx)(1 + 3mx)

Also,

L-1{1/(s-m)^2} = xe^(mx)

So, y(x) = e^(mx)[C1(1+3mx) + C2x]

= e^(mx)[C1 + C1*3mx + C2x]

= e^(mx)[C1 + x(C1*3m + C2)]

Let C2' = C1*3m + C2

Then y(x) = e^(mx)[C1 + C2'x]

G

Daon: For a junion member you are quite bright.

I could not have worked that out myself!

I could not have worked that out myself!