Thread: 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!

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



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



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