Question

y''+3y'-10y=t+e^(-t) . . . I can find the complimentary function but am struggling to find the particular integral! Any help appreciated!:-D

Answer 1

whats the comp?

Answer 2

it also might be helpful to know what methods you have been taught

there is a table method (trial and error really)
there is the variation of parameters
and a wronskin

Answer 3

Ae^(-4x) + Be^(2x)

Answer 4

the long version, which i learned on ... is to assume, and since your setup is wrt "t" and not "x"

\[y_h=Ae^{-4t} + Be^{2t}\]and take the derivatives, assuming A and B are functions of t

Answer 5

what is y'h?

Answer 6

make that yp :) im used to homogenous (yh) for "compliment"

Answer 7

Thats the part I'm struggling with. Yp=Ct+D I think!

Answer 8

yp, not yh, to fix any ambiguity in this
\[y_p=Ae^{-4t} + Be^{2t}\]
\[y'_p=-4Ae^{-4t}+A'e^{-4t}+ 2Be^{2t} + B'e^{2t}\]
let A'+B' = 0 to account for the homogenous results
\[y'_p=-4Ae^{-4t}+2Be^{2t}\]
\[y''_p=16Ae^{-4t}-4'Ae^{-4t}+4Be^{2t}+2B'e^{2t}\]
now insert these original setup into the

Answer 9

...insert those into the original setup :/

Answer 10

y''+3y'-10y
\[~~~~~~~y''_p:~~16Ae^{-4t}-4'Ae^{-4t}+4Be^{2t}+2B'e^{2t}\\
~~~~~~3y'_p:-12Ae^{-4t}~~~~~~~~~~~~~~~+6Be^{2t}\\
-10y_p:-10Ae^{-4t} ~~~~~~~~~~~~~~-10 Be^{2t}\]
well the Bs are nice enough to zero out for us :/ whats up with those As?

Answer 11

am i going to have to dbl chk your characteristic?

Answer 12

y''+3y'-10y

r^2 + 3r - 10 = 0

r = -2 and 5

Answer 13

sigh .... why would youwant to lie to me?

Answer 14

apologies, got mixed up between to different examples!

Answer 15

\[~~~~~~~y''_p:~~~25Ae^{5t}+5A'e^{5t}+4Be^{2t}-2B'e^{-2t}\\
~~~~~~3y'_p:~~~15Ae^{5t}~~~~~~~~~~~~~~~-6Be^{-2t}\\
-10y_p:-10Ae^{5t} ~~~~~~~~~~~~~~-10 Be^{-2t}\]
is that fixed better?

Answer 16

got me signs mixed, roots are -5 and 2 :)

Answer 17

\[~~~~~~~y''_p:~~~25Ae^{-5t}-5A'e^{5t}+4Be^{2t}+2B'e^{2t}\\
~~~~~~3y'_p:-15Ae^{-5t}~~~~~~~~~~~~~~~+6Be^{2t}\\
-10y_p:-10Ae^{-5t} ~~~~~~~~~~~~~~-10 Be^{2t}\\----------------------\\
t+e^{-t}=~~~-5A'e^{5t}+2B'e^{2t}\\\]\[0=A'e^{-5t}+B'e^{2t}\]
now we have a system of equations

Answer 18

this system presents a Cramer rule that is equivalent to the Wronskian method