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Mathematics 11 Online
OpenStudy (lgbasallote):

how to find the \(y_p\) of this using undetermined coefficients \[(D^2 - 1)y = e^x +1\]

OpenStudy (unklerhaukus):

have you found \(y_c\) already?

OpenStudy (anonymous):

Have you found \(y_c\) ???

OpenStudy (unklerhaukus):

\[(D^2 - 1)y_c = 0\]\[m^2-1=0\]\[m=\pm1\]\[y_c=Ae^x+Be^{-x}\]

OpenStudy (lgbasallote):

^that's what i got for yc too

OpenStudy (unklerhaukus):

notice how the non-homogenous term \(e^x\) on the right hand side of the original DE is a scalar multiple of one of the complementary terms \(Ae^x\), this corresponds to RESONANCE , so you multiply what you might first think for \(y_p\) by x \[y_p=(A+Be^x)x\]

OpenStudy (lgbasallote):

huh?

OpenStudy (anonymous):

The particular solution will be the sum of the particular solutions. \[y''-y=e^x\] and \[y''-y=1\]

OpenStudy (lgbasallote):

and that means?

OpenStudy (lgbasallote):

go slow =_=

OpenStudy (lgbasallote):

im not following

OpenStudy (unklerhaukus):

\[y_p=y_1+y_2\]\[y_{p_1}=A\]\[y_{p_2}=Bxe^x\]

OpenStudy (lgbasallote):

...i dont think that was any slower o.O

OpenStudy (unklerhaukus):

what bit are you stuck on?

OpenStudy (lgbasallote):

just imagine i dont know how to use undetermined coefficients

OpenStudy (lgbasallote):

i dont have much practice yet so im not yet good at it

OpenStudy (unklerhaukus):

ok so you have a table right/?

OpenStudy (lgbasallote):

table?

OpenStudy (lgbasallote):

maybe showing my solution will help...im not sure if this is right \[y_p = Ae^x + B\] \[Dy_p = Ae^x\] \[D^2 y_p = ae^x\] \[(D^2 - 1)y = e^x + 1\] \[Ae^x - (Ae^x + B) = e^x + 1\] \[-B = e^x + 1\] \[B = 1 - ^x\] is this right? if yes what's next?

OpenStudy (lgbasallote):

that should be \[B = 1 - e^x\]

OpenStudy (unklerhaukus):

yeah , that y_p wont work because you already have a e^x term in the complementary solution

OpenStudy (lgbasallote):

so what's the right way?

OpenStudy (lgbasallote):

\[Ax e^x + B?\]

OpenStudy (unklerhaukus):

yeah i guess that could work, (i havent tried it)

OpenStudy (unklerhaukus):

That should work

OpenStudy (lgbasallote):

so what's next?

OpenStudy (unklerhaukus):

\[y_p=Axe^x+B\]\[y^\prime_p=\]\[y^{\prime\prime}_p=\]

OpenStudy (lgbasallote):

\[Dy_p = A(xe^x + e^x)\] \[D^2y_p = A(xe^x + 2e^x)\] \[(D^2 - 1)y = e^x +1\] \[Axe^x + 2Ae^x - Axe^x - B = e^x+1\] is that right?

OpenStudy (unklerhaukus):

looks good

OpenStudy (lgbasallote):

so what's next?

OpenStudy (unklerhaukus):

now find A,B

OpenStudy (lgbasallote):

how?

OpenStudy (lgbasallote):

gaussian method?

OpenStudy (unklerhaukus):

ok \[Axe^x + 2Ae^x - Axe^x - B = e^x+1\] \[2Ae^x - B = e^x+1\]

OpenStudy (lgbasallote):

A = 1/2 B = -1?

OpenStudy (unklerhaukus):

simply equate co efficients ., yeap

OpenStudy (lgbasallote):

so then....sub into yp?

OpenStudy (unklerhaukus):

so you now have A and B, plug them into y_p, yep

OpenStudy (lgbasallote):

\[\frac{xe^x}{2} +1\]

OpenStudy (unklerhaukus):

and finally \[y(x)=y_c+y_p\]

OpenStudy (lgbasallote):

oh wait...it needs yc

OpenStudy (unklerhaukus):

-1 not +1

OpenStudy (lgbasallote):

\[C-1 e^x + c_2e^{-x} + \frac{xe^x}{2} - 1\]

OpenStudy (lgbasallote):

ugh darn latex

OpenStudy (lgbasallote):

\[C_1 e^x + C_2 e^{-x} + \frac{xe^x}{2} - 1\]

OpenStudy (unklerhaukus):

you know how to copy paste right?

OpenStudy (unklerhaukus):

with the right click-show math as-tex commands

OpenStudy (lgbasallote):

yeah..just too lazy

OpenStudy (unklerhaukus):

anyhoo, \[\Huge\color{red} \checkmark\]

OpenStudy (lgbasallote):

nice thanks

OpenStudy (unklerhaukus):

more/?

OpenStudy (lgbasallote):

haha lol yes please

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