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

8 years agoSo 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.

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

8 years agookay so do you need to use the roots to find 1 equation first?

8 years agoHence 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)

8 years agoThere's the homogeneous solution as well; we'll come back to that.

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

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

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

8 years agoNow 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.

8 years agostill there?

8 years agosorry 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)?

8 years agoi meant equation not answer

8 years agoyes. With y = Ae^(2x), substitute this into the equation y'+y=8e^(2x) and solve for A

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

8 years agoIn 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.

8 years agoCan you give me an example of using variation of parameters with a first order equation?

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

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

8 years agoI guess but it seems pointless since they are the same thing

8 years agoWe can do that if you want.

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

8 years agoI just don't see what the difference is between the two approaches for this case

8 years agoIf 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.

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

8 years agoOhhhh okay i think I see the difference now

8 years agoThanks for your help!

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

8 years agoPedagogically, this is not a bad exercise your professor is giving you.

8 years agoyeah it was the simplest problem on the hw and it gave me the most trouble.

8 years agook. Well, good luck with the rest.

8 years ago