y''+y' -2y=2-2e^-2x ...... find general soln

ay''+by'+cy= G(x)

y''+y'-2y=0

r^2+r-2=(r-1)(r+2)=0

roots = 1 & -2

Yc = c1 e ^x + c2 e^-2x

Yp???? stuck!

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- Dec 17th 2009, 03:59 PMharveyononhomogeneous 2nd order linear eqn
y''+y' -2y=2-2e^-2x ...... find general soln

ay''+by'+cy= G(x)

y''+y'-2y=0

r^2+r-2=(r-1)(r+2)=0

roots = 1 & -2

Yc = c1 e ^x + c2 e^-2x

Yp???? stuck! - Dec 17th 2009, 05:19 PMkjchauhan
Use method of variation of parameters,

You get,

- Dec 18th 2009, 05:09 AMHallsofIvy
"Variation of parameters" always works but is typically much harder than "undetermined coefficients" when that works- which it will as long as the right hand side consists of the kind of functions we get for homogeneous linear equations with constant coefficients, polynomials, exponentials, sine and cosine, or products of those.

Here, your right hand side is . Because of the constant "2", try some constant A. Because of the " ", your first thought should be " ". However, is already a solution to the associated homogeneous equation so try " " instead.

In other words, try . - Dec 18th 2009, 09:19 AMkjchauhan
- Jan 4th 2010, 06:52 PMharveyo
could somebody show me the next line combining yp to yc

- Jan 4th 2010, 08:14 PMharveyo
i am now at 10 posts.

- Jan 4th 2010, 09:33 PMharveyo
yp'= -2(A+Bxe^-2x)

yp''= 4(A+Bxe^-2x)

therefore from orginal equation

y''+y' -2y

4(A+Bxe^-2x) + -2(A+Bxe^-2x) - 2*(A+Bxe^-2x) = 2-2e-2x

???? - Jan 5th 2010, 04:22 AMHallsofIvy
What are you taking as yp? I suggested but that certainly is not its derivative.

If , then

Quote:

yp''= 4(A+Bxe^-2x)

You should have

Quote:

therefore from orginal equation

y''+y' -2y

4(A+Bxe^-2x) + -2(A+Bxe^-2x) - 2*(A+Bxe^-2x) = 2-2e-2x

????

- Jan 6th 2010, 01:26 PMharveyo
makes complete sense now. (Doh)