Question

How to solve second order ODE with RHS side that is a product of an exponential and another function, e.g. Part I problems 2 and 3 in Linear Operators section?

(something like p(D)y = xe^(ax))

The lecture video goes through how the exponential shift rule can be used to simplify linear operators acting on a RHS of this form, but even after repeated viewing I can't follow how to use this to solve the ODE itself.

Answer 1

You can use the method of undetermined coefficients. Your guess would be (A'x+B')e^(ax), and you would solve for A' and B' by plugging this in to the DE. This comes from guessing Ax+B for the x part and Ce^(ax) for the exponential part. When you multiple the two together, you get (Ax+B)(Ce^(ax)). Note that when you multiple C through, you get (ACx+BC)e^(ax), where AC and BC are also just constants, so we just replace "AC" with A' and "BC" with B' and solve for these two constants.

Answer 2

It is simple.
Let the function we want find is y.then,
y=(e^ax)(1/f(D))(x),
Here f(D) is the differential function. For example, if ,
f(D)=D^2+2D+2,
y=(e^ax)(1/(D^2+2D+2)) (x)
Put D=D+a in f(D)
Then f(D) changes to f(D+a).
Let's take f(D+a)=D^2+4D+4
**The equation y can be solved using binomial theorams of the form,
1/(1+D)=1-D+D^2-........ And,
1/(1-D)=1+D+D^2+........
All of the equations in this format can be solved using FOLLOWING METHOD
*make the constant term in f(D) as 1 (here take 4 outside the bracket)
y=(e^ax)(1/4((D^2/2)+D+1))
*Now the denominator of f(D) is either in the form of 1/(1+D) or 1/(1-D)
Here,by above binomial expansion,
1/(1+[D+(D^2)/2])=1-(D+[D^2]/2)
*#* Note that the number of terms want to expand are the power of the x in the differential equation(why?).
Here,
y=(e^ax)*1/4*(1-[D+(D^2)/2])x
y=(e^ax)*1/4*[x-Dx-((D^2)x)/2]
y=(e^ax)*1/4*[x-1]
y=((e^ax)(x-1))/4