probably very simple differential equation :D

**sedam7** · September 12th 2010, 02:30 PM

hello,

i have problem solving this one

$y'' - 2y' = e^{-x} (2\cos {x} + 3 \sin {x} )$

okay i don't have any problem with homogeneous part

$r^2-2r=0$

$r(r-2)=0$

$r_1=0 \; ; \; r_2=2$

$y_h = C_1 e^{0x} + C_2 e^{2x}$

$y_h = C_1 + C_2 e^{2x}$

i have problem with particular part now... (never encountered with similar till now) and really don't know where to go from here ...

i know that if instead of what i have here is :

$y'' -2y' = \cos {x} + e^x$

particular part would be

$y_p = A\cos{x} + B \sin {x} + Ce^x$

but with that up there i'm just stuck ...

any help is very much appreciated

**yeKciM** · September 12th 2010, 03:39 PM

Originally Posted by sedam7

hello,

i have problem solving this one

$y'' - 2y' = e^{-x} (2\cos {x} + 3 \sin {x} )$

okay i don't have any problem with homogeneous part

$r^2-2r=0$

$r(r-2)=0$

$r_1=0 \; ; \; r_2=2$

$y_h = C_1 e^{0x} + C_2 e^{2x}$

$y_h = C_1 + C_2 e^{2x}$

i have problem with particular part now... (never encountered with similar till now) and really don't know where to go from here ...

i know that if instead of what i have here is :

$y'' -2y' = \cos {x} + e^x$

particular part would be

$y_p = A\cos{x} + B \sin {x} + Ce^x$

but with that up there i'm just stuck ...

any help is very much appreciated

for solving differential equation formed like this

$a_0(x)y^{(n)} + a_1(x) y^{(n-1)}+ ... + a_n(x)y = e^{\alpha x } (a \cos{\beta}x + b \sin {\beta} x )$

than you should do next ...

as you know $y = y_h +y_p$

so you should solve for $y_h$ first (as you did)

now if solutions of homogeneous part are not $r = \alpha \pm i \beta$ (means that you don't have them in your $f(x)$ solution of particular part is searched as :

$y_p(x) = e^{\alpha x } (A \cos{\beta}x + B \sin {\beta} x )$

but if you have $r = \alpha \pm i \beta$ and are the same as $f(x)$ than it's (because of multiplicity)

$y_p(x) =x^m e^{\alpha x } (A \cos{\beta}x + B \sin {\beta} x )$

**sedam7** · September 12th 2010, 10:06 PM

meaning that i just look if "r" appear in homogeneous solutions.I don't look at separate $\alpha$ and $\beta$ ?

lol that would mean it's the same i'm just to dumb to see it

thank you very much !

**yeKciM** · September 12th 2010, 10:09 PM

Originally Posted by sedam7

meaning that i just look if "r" appear in homogeneous solutions.I don't look at separate $\alpha$ and $\beta$ ?

lol that would mean it's the same i'm just to dumb to see it

thank you very much !

yes it's the same. As you look at the real solutions, you look at the complex solutions

good luck

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