Question

need help using variation of parameters to solve simple D.E. y'+y=8e^(2x) need help using variation of parameters to solve simple D.E. y'+y=8e^(2x) @Mathematics

Answer 1

So by the variation of parameters method, we want to try and find a particular solution that is a linear combination
c1.y1 + c2.y2 + .... + cn.yn
where the cj are constants and the yn are functions that are the derivatives of terms on the right hand side.

Answer 2

In this case, there only one term on the RHS, e^(2x), and what's more its derivative is the same up to a constant.

Answer 3

okay so do you need to use the roots to find 1 equation first?

Answer 4

Hence the only term will need to try in the variation of parameters method is e^(2x); namely, try and find a solution of the form
y = Ae^(2x)

Answer 5

There's the homogeneous solution as well; we'll come back to that.

Answer 6

That is, the solution of y' + y = 0; that is the homogeneous solution, call it yh (or sometimes it's called the complimentary solution yc).

Answer 7

Once we've found the particular solution yp, the general solution to the equation is the sum
y = c.yh + yp

Answer 8

yh is easy with y' + y = 0. You've probably found it already; it's yh = e^(-x)

Answer 9

Now coming back to the particular solution ... plug in our general form of the answer
y = Ae^(2x) into the inhomogeneous equation and derive an equation in A. You do that step; I'll wait a minute for you.

Answer 10

still there?

Answer 11

sorry I had to switch computers.... so you use the general answers Ae^(2x) take the derivative and plug it back into the equation and set it equal to 8e^(2x)?

Answer 12

i meant equation not answer

Answer 13

yes. With y = Ae^(2x), substitute this into the equation
y'+y=8e^(2x)

and solve for A

Answer 14

isn't that the same steps as the method of undermined coefficients? or do I just have them backwards

Answer 15

In a way undetermined coefficients is a special case of variation of parameters; I guess I've been advising you here to use that special case, especially as the equation is linear and first order.

Answer 16

Can you give me an example of using variation of parameters with a first order equation?

Answer 17

I need to solve the same equation using both methods but when I tried I basically use the same steps.

Answer 18

Are you suggesting we write y = u(x).e^(-x) and then solve for u?

Answer 19

I guess but it seems pointless since they are the same thing

Answer 20

We can do that if you want.

Answer 21

More or less; but let's just do it. It's actually not a bad exercise.

Answer 22

I just don't see what the difference is between the two approaches for this case

Answer 23

If y = u(x)e^-x, then
y' = u' e^-x - u e^-x

Hence
y' + y = 8e^(2x)
implies
u' e^-x - u.e^-x + u.e^-x = 8e^2x
=>
u' e^-x = 8e^2x
=>
u' = 8e^3x
=>
u = (8/3)e^3x + c
=>
y = (8/3)e^2x + c.e^-x

And this indeed is the general solution.

Answer 24

A third way to do it is with an integrating factor. That method would most closely imitate what I just wrote above.

Answer 25

Ohhhh okay i think I see the difference now

Answer 26

Thanks for your help!

Answer 27

Sure. I actually had never thought of using VOP with first order equations before, so I was interested to see how this worked out.

Answer 28

Pedagogically, this is not a bad exercise your professor is giving you.

Answer 29

yeah it was the simplest problem on the hw and it gave me the most trouble.

Answer 30

ok. Well, good luck with the rest.