Question

How to find the particular solution from the differential equation?

Answer 1

(1-t)y''+ty'-y=2(t-1)^2e^-t
y_1 and y_2 are given as e^t and t respectively

Answer 2

By using determining coefficient.
\[y_{p} = e ^{-t}(At^2+Bt+D)\]
\[y'_{p} = e ^{-t}(2At+B-At^2-Bt-D)\]
\[y''_{p} = e ^{-t}(2A-4At-2B+At^2+Bt+D)\]

I substituted these back into the original equation, but I am not sure of my answer.

Answer 3

http://www.sosmath.com/diffeq/second/variation/variation.html
isn't this the method you want to use?

Answer 4

i didn't attempt the variation of parameters.

Answer 5

well if you wanted to go with the method suggested...
\[y''+\frac{t}{1-t}y'-\frac{1}{1-t} y=\frac{2(t-1)^2e^{-t}}{1-t} \\ y''+\frac{t}{1-t}y'-\frac{1}{1-t}y=-\frac{2(t-1)^2e^{-t}}{t-1} \\ y''+\frac{t}{1-t}y'-\frac{1}{1-t}y=-2(t-1)e^{-t} \\ y_p=u_1 y_1+u_2 y_2 \\ y_p=u_1 e^t+u_2 t \\ \text{ we want to solve the following system } \\ u_1' e^{t}+u_2't=0 \\ u_1'e^{t}+u_2'=-2(t-1)e^{-t}\]
just trying to follow their method there...

\[\text{ first equation solved for } u_2' \text{ gives } \\ u_2'=-\frac{u_1' e^{t}}{t} \\ \text{ plug this into second equation } \\ u_1'e^{t}-\frac{u_1'e^{t}}{t}=-2(t-1)e^{-t} \\ \text{ multiply both sides by } t \\ t e^{t} u_1'-e^{t} u_1'=-2t (t-1)e^{-t} \\ \text{ solve for } u_1' \\ u_1'=\frac{-2t(t-1)e^{-t}}{te^{t}-e^{t}} \\ u_1' =\frac{-2t (t-1)e^{-t}}{e^t(t-1)} \\ u_1'=\frac{-2te^{-t}}{e^{t}} \\ u_1'=-2te^{-2t} \\ u_1=\frac{1}{2}e^{-2t}(2t+1)+C \\ \text{ then you can plug into } \\ u_2'=-\frac{u_1'e^{t}}{t} \text{ and find } u_2\]

Answer 6

u2 is a bit easier to find
(no integration by parts required for that one)

Answer 7

Thanks. The first method I was using was time consuming.

Answer 8

I'm checking my work

Answer 9

it looks good by that method...

Answer 10

neat method

Answer 11

did you find u_2 yet?

Answer 12

\[u_2'=-\frac{u_1'e^{t}}{t} \\ u'_1=-2te^{-2t} \text{ plug this in} \\ u_2'=-\frac{-2te^{-2t} e^{t}}{t}\]

Answer 13

yes, u_2 is 2e^-t+another constant.

Answer 14

-2e^(-t)+c right?

Answer 15

i meant +ve

Answer 16

The wolf gave me something else. I will look at this later. I'm exhausted.

Answer 17

so the particular solution is:
\[y_p=u_1 e^t+u_2 t \\ y_p=[\frac{1}{2}e^{-2t}(2t+1)+k_1]e^{t}+[-2e^{-t}+k_2]t\]

Answer 18

and we should be able to make our solution look like what they have

Answer 19

we just have to do some distributing
and collecting like terms

Answer 20

\[y_p=\frac{1}{2}e^{-t}(2t+1)+k_1e^{t}-2te^{-t}+k_2 t \\ y_p=te^{-t}+\frac{1}{2}e^{-t}+k_1e^{t}-2te^{-t}+k_2t \\ \\ y_p=-te^{-t}+\frac{1}{2}e^{-t}+k_1e^{t}+k_2t\]

Answer 21

the only like terms was t*exp(-t) and -2t*exp(-t)

Answer 22

this is exactly what wolfram has

Answer 23

@freckles Thanks so much. I want to look this over.

Answer 24

k ask any questions

Answer 25

but just so you I followed that method exactly on that one page I refer to you up there