x''+x' = 100sin5t
Find a particular solution xp(t) of the given differential equation using the complex function method

Question

x''+x' = 100sin5t
Find a particular solution xp(t) of the given differential equation using the complex function method

anonymous · Answer

@Neemo do u find anything ?

anonymous · Answer

yeaaaah ! I have connection problem...equation editor isn't working well...W8 pleaase ! okk ?

anonymous · Answer

ok of course

anonymous · Answer

I can guide you if you want to ! my equation editor is not working anymore !!!! :(

anonymous · Answer

seriously :/

anonymous · Answer

i want of course but my english is not perfect. i may have a problem

anonymous · Answer

we will use complex functions....Asin(wt)------------>Aexp(iwt) 
now let's find xp(t)=Aexp(iwt+phi)

anonymous · Answer

' prime is equivalent to multiply this function by iw

anonymous · Answer

still with me @SkykhanFalcon

anonymous · Answer

sorry i had a phone call :/ now i am with u

anonymous · Answer

what other language can you talk better ? 
so xp'(t)=iwxp(t) and xp"(t)=-w^2xp(t)

anonymous · Answer

Turkish but i guess you cannot talk

anonymous · Answer

No sorry ! then , A(-w^2+iw)exp(iwt+phi)=100exp(i5t)

anonymous · Answer

it is really hard to understand like that :/

anonymous · Answer

yeaaah ! I know :( ! It's confusing !

anonymous · Answer

I'll try to use my equation editor !

anonymous · Answer

thank u if it works its gonna be perfect for me

anonymous · Answer

$$A \exp(i \phi)=(1/(- \omega ^2+i \omega))\exp(i(5- \omega )t)$$
then $$\omega = 5$$ 
so $$Aexp(i \phi)=1/5(-5+i)$$

anonymous · Answer

exp(i) is meaning $$e^{i}$$ right ?

anonymous · Answer

so we can choose $$A=1/(5\sqrt{26})$$
and phi the argument of that complex  number so xp(t)=Asin(omega t+phi)
yeaaah right!
an BE right back kk !?

anonymous · Answer

okey

anonymous · Answer

still heere ?!

anonymous · Answer

@Neemo yeah i am here

anonymous · Answer

did you understand or not ?!

anonymous · Answer

i am lost neemo

anonymous · Answer

maybe i have to study much more

anonymous · Answer

Sure ; this is useful for physics right ?!

anonymous · Answer

right :)

anonymous · Answer

$$\sin(\omega t+\phi)-----> \exp (i \omega t+\phi)$$

anonymous · Answer

now the operation will be done to exp ...
If for physics ; the we got to find a xp 
$$xp(t)=A \exp (i \omega t+\phi)$$you will calculcate the first deviative ·
did you get it now ?!

anonymous · Answer

now the operation will be done to exp ... If for physics ; then we got to find a xp
xp(t)=Aexp(iωt+ϕ)
you will calculcate the first deviative · did you get it now ?!

anonymous · Answer

it is very different than i check into the book's answer key

anonymous · Answer

it is -10/13(5sin5t+cos5t)

anonymous · Answer

I think it's the same ! we will find xp(t)=Asin(wt+phi) and we will use trig formula's to put sin(wt+phit) that way ! ! But certainly there is something wrong with my answer ! probably a miscalculation !