- #1

shamieh

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Use Variation of Parameters to find a particular solution to $y'' - y = e^t$

Solution:

$y_p = \frac{1}{2}te^t - \frac{1}{4} e^t$

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- MHB
- Thread starter shamieh
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- #1

shamieh

- 538

- 0

Use Variation of Parameters to find a particular solution to $y'' - y = e^t$

Solution:

$y_p = \frac{1}{2}te^t - \frac{1}{4} e^t$

- #2

MarkFL

Gold Member

MHB

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(a) A fundamental solution set \(\displaystyle \{y_1(t),y_2(t)\}\) for the corresponding homogeneous equation are:

\(\displaystyle y_1(t)=e^{t}\)

\(\displaystyle y_2(t)=e^{-t}\)

and so we take:

\(\displaystyle y_p(t)=v_1(t)y_1(t)+v_2(t)y_2(t)=v_1(t)e^{t}+v_2(t)e^{-t}\)

(b) Solve the system:

\(\displaystyle e^{t}v_1'+e^{-t}v_2'=0\tag{1}\)

\(\displaystyle e^{t}v_1'-e^{-t}v_2'=e^{t}\tag{2}\)

From (1), we find:

\(\displaystyle v_2'=-e^{2t}v_1'\)

And, then substituting into (2), we get:

\(\displaystyle v_1'=\frac{1}{2}\implies v_1=\frac{1}{2}t\)

\(\displaystyle v_2'=-\frac{1}{2}e^{2t}\implies v_2=-\frac{1}{4}e^{2t}\)

Thus:

\(\displaystyle y_p(t)=\frac{1}{2}te^{t}-\frac{1}{4}e^t\)

We may discard the term that simply adds a solution to the homogeneous equation, and state:

\(\displaystyle y_p(t)=\frac{1}{2}te^{t}\)

Because no work was shown, I didn't know until I actually worked the problem that this was the case. :D

- #3

shamieh

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We may discard the term that simply adds a solution to the homogeneous equation, and state:

\(\displaystyle y_p(t)=\frac{1}{2}te^{t}\)

Because no work was shown, I didn't know until I actually worked the problem that this was the case. :D

Mark, if my particular solution $y_p = ue^t + ve^{-t}$ then why are you "discarding" the solution for $v$ ? I mean isn't that part of my particular solution? Why would you throw $v$ away? If that was the case then why wouldn't I just set my $y_p$ (particular solution) to $y_p = ue^t$ in the first place?

Thanks for all your help.

- #4

MarkFL

Gold Member

MHB

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Adding:

\(\displaystyle -\frac{1}{4}e^t\)

to the general solution when we already have:

\(\displaystyle c_1e^t\)

as one of the terms results in us still just having an arbitrary constant times $e^t$ in the general solution, so we may just discard it.

- #5

shamieh

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