Help with particular integrals

SoSolid

SoSolid

SoSolid

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I am stuck solving this second order differential equation:

2y" -y' -3y = 10xe^-x


I can easily find the complimentary function. The problem is that the x on the RHS mucks everything up, I do not know which standard format to use to solve for the particular integral. Can someone please advise me on just this part.
 
First thing that I'd try is the obvious y = Axe^-x.

If that didn't work (probably the case) I'd try a combination of xe^-x and x^2e^-x. The coefficient of x^2e^-x magically disappears, you use one constant to fix the coefficient of e^-x to zero, and then fiddle the other one to get the right coefficient (10) for the coefficient of xe^-x.
 
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Hey yeh I had originally tried the first method suggested. The thing is, how do you know which is the correct method to substitute in. I mean if I am in an exam I do not want to be messing around trying to get the correct format to use. Any hints / tips on how to do this?
 
Say you have something that looks like

ay'' + by' + cy = f(x)

and you've already solved to find the complementary solution, which looks like Ae^mx + Be^nx for some values of n and m.

The best guess for the particular integral is y = Af(x) where A is a constant. If that doesn't work then the next best guess is y = Af(x) + Bxf(x) where A and B are constants.

The most common situation when your first guess won't work is when f(x) looks like one of the solutions to the differential equation. For example, in the equation you posted, y = e^-x was a complementary solution. Therefore whenever you stick y = xe^-x in as a guess for the particular integral the coefficient of xe^-x will always work out to zero - this is because e^-x solves the differential equation. This hints that you need to look at particular integrals that involve more factors of x, i.e. x^2e^-x.

So in general, always guess y = f(x) for a particular integral, UNLESS f(x) looks like one of your complementary solutions. In that case guess a combination of f(x) and xf(x).
 
Put it back into the differential equation and see if you get the correct RHS, and in this case you do not. You get your 10xe^-x but you also get an extra -4e^-x.

I guess you tried bx e^-x as your particular integral. Do as Arcade Fire suggests and try (ax + bx^2)e^-x.
 
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Your solution is composed of a complimentary function and a particular integral. The point of the complimentary function is that the LHS of your differential equation, in this case 2y" -y' -3y, annihilates it, ie. takes it to zero. Then, your particular integral is the part that provides the RHS of the differential equation. So when you put your particular integral y, into 2y" -y' -3y, you must get 10xe^-x. So, you are putting y = your particular integral into 2y" -y' -3y, and you want to get 10xe^-x.

So to find the PI, we can just put y = (ax+bx^2)e^-x (my previous post contains a mistake, which I will edit) into 2y" -y' -3y. Then compare this result to 10xe^-x and you should be able to match up coefficients of xe^-x and x^2 e^-x and possibly e^-x to find out what a and b are.

I hope that helps.
 
Psiko said:
Your solution is composed of a complimentary function and a particular integral. The point of the complimentary function is that the LHS of your differential equation, in this case 2y" -y' -3y, annihilates it, ie. takes it to zero. Then, your particular integral is the part that provides the RHS of the differential equation. So when you put your particular integral y, into 2y" -y' -3y, you must get 10xe^-x. So, you are putting y = your particular integral into 2y" -y' -3y, and you want to get 10xe^-x.

So to find the PI, we can just put y = (ax+bx^2)e^-x (my previous post contains a mistake, which I will edit) into 2y" -y' -3y. Then compare this result to 10xe^-x and you should be able to match up coefficients of xe^-x and x^2 e^-x and possibly e^-x to find out what a and b are.

I hope that helps.
Click to expand...

Hey yeh that helps thanks, I am just tired and being slow. It makes sense about substituting the answer back in and seeing if it is correct. I will give it a whirl.
 
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