Second order differential equation

Question

How do you solve the second order differential equation below?

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

Answer 1

You first need to solve for the homogeneous solution (that is, the solution y_h=c_1y_1+c_2y_2, c_1 and c_2 arbitrary constants, to the equation y"+3y'+2y=0) with the characteristic polynomial. Then, you will add to that a particular solution to the original equation. y"+3y'+2y=1/(1+e^x), which we need to find. To find that particular solution, I would rewrite the fraction 1/(1+e^x) as (1+e^x-e^x)/(1+e^x)=[(1+e^x)/(1+e^x)]-e^x/(1+e^x)=1-[e^x/(1+e^x)] because the final bracketed quantity is conducive to u-substitution u=(1+e^x) if we have to integrate it. From there, though, I'm stuck too. Laplace Transforms don't seem to do anything (unless you have a more informative table than I do), and picking a trial solution y_p=A/(1+e^x) with undetermined coefficient A (that's my first impulse) doesn't work. Sorry. :(

However, the homogeneous solution is worth something.