linear second order differential equation

Trenters4325

Junior Member
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Apr 8, 2006
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122
The solution of y'' + a* y' + b*y =0 takes the form of (C1 + C2*x)*exp(mx) when there is only one solution to the characteristic equation. Can someone show me how this works i.e. how you derive the solution form.
 

daon

Senior Member
Joined
Jan 27, 2006
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Trenters4325 said:
The solution of y'' + a* y' + b*y =0 takes the form of (C1 + C2*x)*exp(mx) when there is only one solution to the characteristic equation. Can someone show me how this works i.e. how you derive the solution form.
This seems to work, although i'm not sure if it is a valid derivation. Maybe someone can check this over.


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

Guest

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Daon: For a junion member you are quite bright.

I could not have worked that out myself!
 
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