linear D.E of nth order with constant coefficient...help guys .

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linear D.E of nth order with constant coefficient...help guys .

Differential Equations
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This WILL be messy... are you up for it? :D
yes

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For these kinds of differential equations, the general form of the solution is given by \[\Large f^{(h)}(x) + f^{(p)}(x)\] where the f^h is the general solution to the homogeneous equation while f^p is any particular solution of this equation. Can you find the general solution to the homogeneous equation?
\[m ^{^{2}}=11\]
What does that mean? D:
\[y=e ^{a}\]
Oh wait... there's no y' term?
get the higher order first D.E
So basically, it's \[\Large y'' - 11y = xe^x\]?
\[m ^{2} +1 -12 =0\]
then m= square root of 11
then get the vlue of y
\[y=C{1}^{+\sqrt{11}x}+C{2}^{-\sqrt{11}x}\]
Im right? what is the next get the other y which from Xe^x
whic is y= Ae^u
take derivative until second derivative
ang get the value of A I stocked here
help guys if Im wrong :(
Well, if it doesn't work with only one undetermined coefficient, did you try 2? Specifically this: \[\Large y = \color{blue}Axe^x + \color{red}Be^x\]
subst the value of Y" ,Y' and Y to the y′′+ y -12y
where did you get this y=Axex+Bex
are you sure the second term in the question isn't a derivative?
Odd, isn't it :D
Oh by the way, I didn't actually "get" that Axe^x + Be^x, for all intents and purposes, it IS just a guess... see if it gets you the coefficients you want, since Axe^x alone doesn't seem to do it.
if not (yes strange), then i think you (@melmel ) got the complementary homogenous solution \[\large f^h(x)=y_c(x)=C_{1}e^{+\sqrt{11}x}+C_{2}e^{-\sqrt{11}x}\] {forgot to type the e's in the bases }
Which method to calculate the particular solution \[f^h(x)=y_p(x)\] are you going to use; undetermined coefficients, operator D , or variation of parameters, @melmel ?
apparently offline :( Oh well, onwards and upwards XD
^_^ thanks

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