Thread: probably very simple differential equation :D

1. probably very simple differential equation :D

hello,

i have problem solving this one

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

okay i don't have any problem with homogeneous part $\displaystyle r^2-2r=0$

$\displaystyle r(r-2)=0$

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

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

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

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 :

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

particular part would be

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

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

any help is very much appreciated 2. Originally Posted by sedam7 hello,

i have problem solving this one

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

okay i don't have any problem with homogeneous part $\displaystyle r^2-2r=0$

$\displaystyle r(r-2)=0$

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

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

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

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 :

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

particular part would be

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

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

any help is very much appreciated for solving differential equation formed like this

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

than you should do next ...

as you know $\displaystyle y = y_h +y_p$

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

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

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

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

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

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

lol that would mean it's the same i'm just to dumb to see it thank you very much !

4. Originally Posted by sedam7 meaning that i just look if "r" appear in homogeneous solutions.I don't look at separate $\displaystyle \alpha$ and $\displaystyle \beta$ ?

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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