Question

How do I find the particular solution for [e^(2x)]/(1+e^x)

/i have a differential equation : y''' - 3y''+2y' = [e^(2x)]/(1+e^x)

I understand how to find the complimentary solution but I am unsure of how to solve for the particular solution ( [e^(2x)]/(1+e^x) )

Answer 1

I heard of particular solutions for differential equations
what is the particular solution meaning here?

Answer 2

The particular solution used to determine a solution. Solution containing both complimentary and particular answers

Answer 3

homogenous solution: de = 0
particular solution: de = f(x)

when we know a homogenous solution, we can use a wronskian or variation of parameters to determine the particular solution

Answer 4

not too sure how 'simple' that is to do here :)

Answer 5

have you determined the homogenous solution?

Answer 6

.

Answer 7

I found the complimentary (homogeneous) solution to be:
x(t) = c1 e^x +c2 e^2x

Answer 8

it y(t) that you are trying to determine

y''' - 3y''+2y' = 0, let y' = u

u'' - 3u'+2u = 0

r^2 -3r +2 = 0; when r=2 or 1

u = c0 e^(x) + c1 e^(2x)

y' = c0 e^(x) + c1 e^(2x)

y = c0 e^(x) + c1 e^(2x) + c3

Answer 9

Thank you for the help!!!

Would you be able to help me solve this for the non-homogeneous equation? I have no idea what to do :(

Answer 10

sure, its a variation in the parameters .... let the undetermined constants be defined as undetermined functions of t

then find y' y'' and y''' and insert them into the setup

if you want an overview of what we are to do then i have a sort of tutorial:

http://openstudy.com/users/amistre64#/updates/552fd761e4b0af026591bbbe

Answer 11

Thank you so much for the tutorial!

Answer 12

functions of x ... that is. your x(t) threw me :) we want to find y(x) which we did and can now use it to determine a particular solution.

\[\hat y(x) = A e^x + B e^{2x} + C\]
\[\hat y'(x) = A e^x + B \frac12e^{2x} +C+ (A' e^x + B' e^{2x} + C':=0)\]
\[\hat y''(x) = A e^x + B \frac14e^{2x}+C+(A' e^x + B' \frac12e^{2x}+C':=0)\]
\[\hat y'''(x) = A e^x + B \frac18e^{2x}+C+A' e^x + B' \frac14e^{2x}+C'\]
now when we organize the data collected the tutrial i posted shows why we end up with a system of equations defined as:
\[A' e^x + B' e^{2x} + C':=0\]
\[A' e^x + B' \frac12e^{2x}+C':=0\]
\[A' e^x + B' \frac14e^{2x}+C':= \frac{e^{2x}}{1+e^x}\]

Answer 13

forgot to include that C e^(0x) derives to 0C and i posted all C instead ... but thats immaterial since it bears nothing on the solution.

Answer 14

we could have done undetermined coefficients, but that tends to invoke an 'educated guess' that i simply do not have the tables for

Answer 15

sigh .. seems i was thinking integration instead of derivative while trying to make some pretty coding.....

1/2 to 2
1/4 to 4

Answer 16

We could also attempt to solve via reduction of order given that we know two linearly independent solutions.

Answer 17

u = co e^x + c1 e^(2x)

A' e^x + B' e^(2x) := 0
A' e^x + B' 2e^(2x) := e^(2x)/ (1+e^x)

A' = -B' e^x

-B' e^(2x) + B' 2e^(2x) := e^(2x)/ (1+e^x)

B' [2e^(2x)-e^(2x)] := e^(2x)/ (1+e^x)

B' e^(2x) := e^(2x)/ (1+e^x)
-----------------------------

A' = -e^x/(1+e^x)
B' = 1/(1+e^x)
----------------------------

A = -ln(1+e^x)

ive go no good idea on B lol, wolf says: x - log(1+e^x)
------------------------------

u_p = -e^x ln(1+e^x) + (x - log(1+e^x)) e^(2x)

y'_p = -e^x ln(1+e^x) + x e^(2x) - e^(2x) log(1+e^x)

maybe .....

Answer 18

working those integrals may or maynot be simple :)