Question

What have I achieved?

\[y''-2y'+y=e^{2x}\]
\[y_c=c_1e^x+c_2xe^x\]

\[y=Ae^{2x}\]
\[y'=2Ae^{2x}\]
\[y''=4Ae^{2x}\]

\[4Ae^{2x}-4Ae^{2x}+Ae^{2x}=e^{2x}\]
A=1

Answer 1

why is the solution y=e^{2x}
?
@hartnn

Answer 2

This is a nonhomogeneous equation.

Answer 3

yes
what do I with the A that I solved for?

Answer 4

Weill I think they chose A becuase you're using undetermined coefficients to solve, correct?

Answer 5

yes

Answer 6

what's the next step after solving for "A"?

Answer 7

They used A as a"guess" Since g(t) is an exponential and we know that exponentials never just appear or disappear in the differentiation process it seems that a likely form of the particular solution would be Ae^2t

Answer 8

The next step is to take the 1st and 2nd derivative of your Ae function. Plug it back in for y' and y''. and solve for "A".

Answer 9

Which is your particular solution.

Answer 10

Unless you were given initial conditions, your particular solution is as far as you need to go.

Answer 11

ok this is how far I've gotten
\[4Ae^{2x}-4Ae^{2x}+Ae^{2x}=e^{2x}\]
A=1
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- what
- happened
- here?
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the solution manual tells me that y=e^2x

Answer 12

Well, since you got A = 1 that's why your answer is e^2x
\[y = 1 (e^{2x})\]

Answer 13

All that u need to do is look at g(t) and make a guess as to the form of YP(t) leaving the coefficient(s) undetermined (and hence the name of the method).
Plug the "guess" into the differential equation and see if we can determine values of the coefficients. If we can determine values for the coefficients then we guessed correctly, if we can’t find values for the coefficients then we guessed incorrectly.

Answer 14

so the whole point of this problem was to guess what y=A(e^2x)

so when I'm given an equation similar to \[y''-2y'+y=e^{2x}\]
my goal is to get a solution in a form similar to\[y= A(e^{2x})\]

ok makes sense

Answer 15

Yes. You make a "guess" which is what your:
\[Ae^{2x} \]
that was your guess for g(t). Remember that a nonhomogeneous equation is in the form:
\[y'' + p(t)y' + q(t)y = g(t)\]
Let me summarize how your guesses should be for other functions:
\[\alpha \cos(\beta t) = A \cos(\beta t) + B \sin(\beta t)\]
that guess works for not only cos, but sin AND also for
a cos(Bt) + b sin(Bt)! Also,
BUT, for an Nth degree polynomial, you have to guess:
\[A_n t^n+ A_{n-1} t^{n-1}+...+A_1 t+ A_0\]