Question

what is the yp in \[(D^4 + 2D^3 + 5D^2)y = x^2 - e^{-2x}\]

i got \[y_p = \frac{e^{-2x}}{20}\]
or something like that

Answer 1

solve for \(y_c\)

\[ D^4+2D^3+5D^2=D^2(D^2+2D+5)=0 \] it gives \(D=0,0, -1\pm2i \)

so \(y_c=c_1+c_2x+e^{-x}(c_3\cos 2x+c_4 \sin 2x ) \)

now for \(y_p\)

see \(R(x)=x^2-e^{-2x}\) for \(x^2\) u should apply \(ax^2+bx+c\) but \(bx+c\) is similiar to \(c_1x+c_2\) in \(y_c\) so u must multiply \(ax^2+bx+c\) by \(x^2\) for Removing similarity...for \(e^{-2x}\) u must apply \(de^{-2x}\) so \(y_p=x^2(ax^2+bx+c)+de^{-2x}\) plug it in the original equation and find a,b,c and d

final answer: \[y_p=x^2(\frac{25}{2}x^2-20x-6)-\frac{75}{2} e^{-2x} \]

Answer 2

im not familiar witht he method you used o.O whatever it is...

Answer 3

i just know undetermined coefficients and variation of paramaters...and that doesnt look like either

Answer 4

variation of paramaters will work here...

Answer 5

i think undetermined coefficients will too...can you show me the process using that?

Answer 6

sorry igba almost forgot them...

Answer 7

and wait actually what i did is undetermined coefficients i think...

Answer 8

really? they dont look familiar to me o.O

Answer 9

@lgbasallote what mukushla did is method of undetermined coefficients.
just take the respective derivative
\[y_{p}=x^2(a^2+bx+c)+de^{-2x}\]
and put in the original equation
then compare the coefficients .

Answer 10

what the heck is that ax^2 + bx + c? o.O they didnt teach that to us in undetermined coefficients

Answer 11

what they taught you how to take guess for polynomial?

Answer 12

kind of....well they just taught us the two basic forms \[c_1 e^{mx}\] and then \[e^{ax} (c_1 \cos bx + c_2 \sin bx)\]
and then the yp is whichever looks like one of those

Answer 13

some guess for particular functions are given below.Taken from Paul online DE notes

Answer 14

so i took a guess and i got \[Ax^2 - Be^{-2x}\]

Answer 15

i think the following example will be good to understand .taken from Paul online notes .

Answer 16

\[Dy_p = 2Ax + 2Be^{-2x}\]
\[D^2y_p = 2A - 4Be^{-2x}\]
\[D^3 y_p = 2 + 8Be^{-2x}\]
\[D^4 y_p = -16e^{-2x}\]

Answer 17

\[-16e^{-2x} +2(2 + 8De^{-2x}) + 5(2A - 4Be^{-2x}) = x^2 - e^{-2x}\]

Answer 18

\[4 + 10A - 20 Be^{-2x} = x^2 - e^{-2x}\]

Answer 19

\[A = 0\]
\[B = \frac{1}{20}\]

Answer 20

that's my solution

Answer 21

whenever you have something like polynomial we generally write the full form of that degree polynomial for example if you have x^2 term in the right side then in guess for yp you must
write
\[y_{P}=ax^2+bx+c\] (general form of second degree polynomial)
if you have x^3 in the write side then
\[y_{p}=ax^3+bx^2+cx+d\] (general form of third degree polynomial)
if you have x^4 in the write side then
\[y_{p}=ax^4+bx^3+cx^2+dx+e\] (general form of 4th degree polynomial)

i hope now you can guess it correctly for polynomials !

Answer 22

...still not making sense....

Answer 23

it should make sense now because
you have x^2 terms in the right hand side and you must use yp guess following
\[y_{p}=ax^2+bx+c+de^{-2x}\]

but to remove duplication (terms are also in the complementary solution) multiply with x^2
as mukushla did

you are using guess
\[Ax^2-Be^{-2x}\] which is incorrect

Answer 24

so bottomline here is im incorrect and i might have failed the "easiest" quiz in differential equations??!!! o.O

Answer 25

i am sorry to say yes:(

Answer 26

<--going into avatar state now

Answer 27

do not worry you wouldn't repeat this in your Examinations after this :)
what if the polynomial appears in the final exam :! m sure you will be able to do it better than any one :)

Answer 28

too late to stop the avatar state >:(