# 2nd order differential equation - variation of parameters

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• March 30th 2009, 11:50 PM
zxcv
2nd order differential equation - variation of parameters
Hi,

I would like to know how to solve the following:

Solve the differential equation using the method of varaition of parameters.

y" - 3y' + 2y = 1/(1+e^-x)

What I have done till now (not really able to continue):

- solution to y" - 3y' + 2y = 0 is c1e^x+c2e^2x
-y1 = e^x & y2 = e^2x
-yp = u1e^x + u2e^2x

However, I am not sure if I have u2 correctly:

u1 = ln (1+e^-x)
u2 = -(1+e^-x) + ln (1+e^-x) where I have integral of (e^-2x)/(1+e^-x)

can someone please check with me and help me solve the following please.

Thanks in advance!
• March 31st 2009, 06:25 AM
Jester
Quote:

Originally Posted by zxcv
Hi,

I would like to know how to solve the following:

Solve the differential equation using the method of varaition of parameters.

y" - 3y' + 2y = 1/(1+e^-x)

What I have done till now (not really able to continue):

- solution to y" - 3y' + 2y = 0 is c1e^x+c2e^2x
-y1 = e^x & y2 = e^2x
-yp = u1e^x + u2e^2x

However, I am not sure if I have u2 correctly:

u1 = ln (1+e^-x)
u2 = -(1+e^-x) + ln (1+e^-x) where I have integral of (e^-2x)/(1+e^-x)

can someone please check with me and help me solve the following please.

Thanks in advance!

When I did it, I got your u1 but for u2

$u_2 = -e^{-x} + \ln | 1 + e^{-x}|$

but I don't think it matters in the end b/c when you expand

$u_1 e^x + u_2 e^{2x}$

the first term in u2 (either yours or mine) can be absorbed into the complementary solution.